To solve the problem, we require an understanding of physics concepts like angular velocity, moment of inertia, and torque. The catcher catching the ball changes its angular momentum, resulting in an angular velocity. The torque experienced when the arm stops the rotation can be computed using known equations.
Explanation:This question involves concepts of physics like angular velocity, moment of inertia, and torque. Initially, with the catcher's arm at the ready position, the system (arm and ball) has zero angular velocity. Then when the catcher catches the ball, he applies an impulse to it and changes not just the linear momentum but the angular momentum about the shoulder as well.
The change in angular momentum (angular impulse) will be equal to the product of the mass of the baseball, its velocity, and the arm's length, i.e., 0.145kg × 40m/s × 0.5m= 2.9 kg m²/s. This change in angular momentum over time will induce an angular velocity, which can be calculated by dividing the change in angular momentum by the moment of inertia of the system (arm and ball).
For part (b), the torque experienced by the arm when it stops the rotation can be computed from the known equation Torque = (Moment of Inertia × Angular Acceleration). The angular acceleration is determined by the change in angular velocity divided by the time taken which in this case is 0.3 seconds. Taking all these physics concepts into account will yield the correct numerical solutions for parts (a) and (b).
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To find the angular velocity of the arm after catching the ball, conservation of angular momentum was used, resulting in an angular velocity of 7.85 rad/s. The torque needed to stop the arm's rotation in 0.3 seconds is -9.66 N - m, calculated using the angular deceleration. The answer involves concepts of angular momentum and torque.
A baseball catcher extends his arm straight up to catch a fast ball with a speed of 40 m/s. The baseball is 0.145 kg and the catcher’s arm length is 0.5 m and mass 4.0 kg.
(a) What is the angular velocity of the arm immediately after catching the ball?
To find the angular velocity of the arm immediately after catching the ball, we need to use the principle of conservation of angular momentum. The initial angular momentum of the ball can be calculated using:
L_initial = m_ball * v_ball * r_arm
where m_ball = 0.145 kg, v_ball = 40 m/s, and r_arm = 0.5 m.
L_initial = 0.145 kg * 40 m/s * 0.5 m = 2.9 kg·m²/s
The moment of inertia of the arm plus the ball (approximated as point mass at the end) is:
I_total = I_arm + m_ball * r_arm²
Using the formula for the moment of inertia of a rod about one end: I_arm = (1/3) * m_arm * (r_arm)², where m_arm = 4.0 kg and r_arm = 0.5 m:
I_arm = (1/3) * 4.0 kg * (0.5 m)² = 0.333 kg·m²
Adding the moment of inertia of the ball:
I_total = 0.333 kg·m² + 0.145 kg * (0.5 m)² = 0.333 kg·m² + 0.03625 kg·m² = 0.36925 kg·m²
Since angular momentum is conserved, L_initial = I_total * ω, where ω is the angular velocity:
ω = L_initial / I_total = 2.9 kg·m²/s / 0.36925 kg·m² = 7.85 rad/s
(b) What is the torque applied if the catcher stops the rotation of his arm 0.3 s after catching the ball?
Torque (τ) can be calculated using the relationship between torque, angular deceleration (α), and moment of inertia (I):
τ = I_total * α
First, we find the angular deceleration. The arm stops, meaning final angular velocity is 0.
Using the angular kinematic equation: ω_final = ω_initial + α * t, where ω_final = 0 and t = 0.3 s:
0 = 7.85 rad/s + α * 0.3 s
α = -7.85 rad/s / 0.3 s = -26.17 rad/s²
Now, calculate torque:
τ = 0.36925 kg·m² * (-26.17 rad/s²) = -9.66 N·m
The negative sign indicates that the torque is in the direction opposite to the rotation.
Interactive Solution 9.63 illustrates one way of solving a problem similar to this one. A thin rod has a length of 0.620 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.185 rad/s and a moment of inertia of 1.43 x 10-3 kg·m2. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (whose mass is 5 x 10-3 kg) gets where it's going, what is the change in the angular velocity of the rod?
Answer:
w = 7.89 10⁻² rad/s
Explanation:
We will solve this exercise with the conservation of the annular moment, let's write it in two moments
Initial. With the insect in the center
L₀ = I w₀
End with the bug on the edge
[tex]L_{f}[/tex]= I w + [tex]I_{bug}[/tex] w
The moments of inertia are
For a rod
I = 1/3 M L²
For the insect, taken as a particle
I = m L²
The system is formed by the rod and the insect, this is isolated, therefore the external torque is zero and the angular momentum is conserved
L₀ = [tex]L_{f}[/tex]
I w₀ = I w + [tex]I_{bug}[/tex] w
w = I / (I + [tex]I_{bug}[/tex]) w₀
w = I / (I + m L²) w₀
Let's calculate
w = 1.43 10⁻³ / (1.43 10⁻³ + 5 10⁻³ 0.620²)² 0.185
w = 1.43 10⁻³ / 3.352 10³ 0.185
w = 7.89 10⁻² rad/s
You throw a ball of mass 1 kg straight up. You observe that it takes 2.2 s to go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up to the top is half the total time, 1.1 s. Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward.
(a) Use the momentum principle to determine the speed that the ball had just AFTER it left your hand.
vinitial = ?? m/s
(b) Use the Energy Principle to determine the maximum height above your hand reached by the ball.
h = ?? m
Answer:
10.791 m/s
5.93505 m
Explanation:
m = Mass of ball
[tex]v_f[/tex] = Final velocity
[tex]v_i[/tex] = Initial velocity
[tex]t_f[/tex] = Final time
[tex]t_i[/tex] = Initial time
g = Acceleration due to gravity = 9.81 m/s²
From the momentum principle we have
[tex]\Delta P=F\Delta t[/tex]
Force
[tex]F=mg[/tex]
So,
[tex]m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s[/tex]
The speed that the ball had just after it left the hand is 10.791 m/s
As the energy of the system is conserved
[tex]K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m[/tex]
The maximum height above your hand reached by the ball is 5.93505 m
The speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.
What is momentum principal?When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the principal of momentum.
The net force using the momentum principle can be given as,
[tex]F_{net}=\dfrac{\Delta P}{\Delta t}[/tex]
Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. Therefore, the above equation for initial and final velocity and time can be written as,
[tex]F_{net}=\dfrac{m(v_f-v_i)}{(t_f-t_i)}[/tex]
(a) The speed that the ball had just AFTER it left your hand-The mass of the ball is 1 kg, and it takes 2.2 s to go up and down. The net force acting on the body is mass times gravitational force.
Therefore, the above equation can be written as,
[tex]mg=\dfrac{m(v_f-v_i)}{(t_f-t_i)}\\g=\dfrac{(v_f-v_i)}{(t_f-t_i)}\\[/tex]
As the final velocity is zero and initial time is also zero. Therefore,
[tex](-9.81)=\dfrac{(0-v_i)}{(1.1-0)}\\v_i=10.791 \rm m/s[/tex]
(b) The maximum height above your hand reached by the ball-Using the energy principle, we can equate the kinetic energy of the system to the potential energy of the system as,
[tex]\dfrac{1}{2}mv_i^2=mgh\\\dfrac{1}{2}(10.791)^2=(9.81)h\\h=5.94\rm m[/tex]
Thus, the speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.
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An electromagnetic wave is transporting energy in the negative y direction. At one point and one instant the magnetic field is in the positive x direction. The electric field at that point and instant is:Why is (D) the answer? I thought it's (C)A) positive y directionB) negative y directionC) positive z directionD) negative z direction
Answer:negative Z direction
Explanation:
It is given that EM wave is travelling in negative y direction i.e. -[tex]\hat{j}[/tex]
Magnetic Field is in positive x direction i.e. [tex]\hat{i}[/tex]
We know that Electric field and magnetic field are perpendicular to each other and they are mutually perpendicular to direction of wave propagation.
Mathematically if Electric field is in negative z direction it will yield the direction of wave propagation
[tex]=(-\hat{k})\times \hat{i}[/tex]
[tex]=-\hat{j}[/tex]
For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equation
1/?= 2p2 mk2 e4/ h3c (Z2) (1/n2f - 1/n2i)
Using this equation, show that the ranges of wavelengths in these two series overlap.
Shortest Wavelength (m) Longest Wavelength (m)
Paschen Series
Final answer:
The Paschen and Brackett series of lines in atomic hydrogen can overlap in terms of their wavelength ranges.
Explanation:
The Paschen series of lines in atomic hydrogen occurs when nf = 3, and the Brackett series occurs when nf = 4.
The range of wavelengths in these two series can overlap because the wavelength equation 1/λ = 2π²mk²e⁴/(h³c)(Z²)(1/nf² - 1/ni²) depends on the values of nf and ni, which can be different for each series
For example, if nf = 3 in the Paschen series and nf = 4 in the Brackett series, the wavelengths in these two series can overlap depending on the values of ni.
Billiard ball A of mass mA = 0.117 kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.135 kg . As a result of the collision, ball A is deflected off at an angle of θ′A = 30.0∘ with a speed v′A = 2.10 m/s, and ball B moves with a speed v′B at an angle of θ′B to original direction of motion of ball A.Part A Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the x direction.A. 0=mAv′Asinθ′A−mBv′Bsinθ′BB. mAvA=mAv′Acosθ′A+mBv′Bcosθ′BC. mAvA=mAv′Acosθ′A−mBv′Bsinθ′BD. 0=(mAvA+mBv′B)sinθ′BPart B Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the y direction.A. mAvA=mAv′Acosθ′A−mBv′Bsinθ′BB. 0=(mAvA+mBv′B)sinθ′BC. 0=mAv′Asinθ′A−mBv′Bsinθ′BD. mAvA=mAv′Acosθ′A+mBv′Bcosθ′BPart C Solve these equations for the angle, θ′B , of ball B after the collision. Do not assume the collision is elastic.Part D Solve these equations for the speed, v′B , of ball B after the collision. Do not assume the collision is elastic.
Answer:
Part A:
B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B
Part B:
C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B
since vAy = 0 m/s
Part C:
θ′B = tan⁻¹(1.0699) = 46.94°
Part D:
v′B = 1.246 m/s
Explanation:
Given:
mA = 0.117 kg
vA = vAx = 2.80 m/s
mB = 0.135 kg
vB = 0 m/s
θ′A = 30.0°
v′A = 2.10 m/s
Part A: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the x direction is
B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B
Part B: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the y direction is
C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B
since vAy = 0 m/s
Part C: Solving these equations for the angle, θ′B , of ball B after the collision and assuming that the collision is not elastic:
mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B
⇒ (0.117)(2.80) = (0.117)(2.10)Cos 30° + (0.135)*v′B*cosθ′B
⇒ v′B*cosθ′B = 0.8505 ⇒ v′B = 0.8505/cosθ′B
then
0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B
⇒ 0 = (0.117)(2.10)Sin 30° - (0.135)*v′B*sinθ′B
⇒ v′B*sinθ′B = 0.91 ⇒ v′B = 0.91/sinθ′B
if we apply
0.8505/cosθ′B = 0.91/sinθ′B
⇒ tanθ′B = 0.91/0.8505 = 1.0699
⇒ θ′B = tan⁻¹(1.0699) = 46.94°
Part D: Solving these equations for the speed, v′B , of ball B after the collision and assuming that the collision is not elastic:
if v′B = 0.91/sinθ′B
⇒ v′B = 0.91/sin 46.94°
⇒ v′B = 1.246 m/s
Conservation of momentum can be used to solve the equations for the angle and the speed of ball B after the collision, using the initial conditions, final conditions, and trigonometric identities. This involves the application of physics concepts, combined with the mathematics of trigonometry.
Explanation:Part A: The correct equation expressing the conservation of momentum for the components in the x direction would be option B, mAvA=mAv′Acosθ′A+mBv′Bcosθ′B. This equation shows that the initial momentum of ball A (mAvA) equals the sum of the momentum of ball A and ball B after the collision in the x direction.
Part B: For the components in the y direction, the right answer is C, 0=mAv′Asinθ′A−mBv′Bsinθ′B. Since ball B was initially at rest and ball A was moving along the x-axis, there was no momentum in the y direction before the collision. Therefore, the total momentum in the y direction after the collision should also be 0.
Part C and D: To solve these equations for the angle and the speed of ball B after the collision, you need to use these equations in combination with the conservation of kinetic energy formula (1/2*m*v^2) and the trigonometric identities. Detailed solution steps require knowledge of the involved mathematics.
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In an atom, an electron is confined to a space of roughly 10?10 meters. If we take this to be the uncertainty in the electron's position, what is the minimum uncertainty ?p in its momentum?
Answer:
[tex]5.2728\times 10^{-25}\ kgm/s[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
[tex]\Delta x[/tex] = Uncertainity in position = [tex]10^{-10}\ m[/tex]
[tex]\Delta p[/tex] = Uncertainty in momentum
According to the Heisenberg uncertainity principle we have
[tex]\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta p=\dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{4\pi\times 10^{-10}}\\\Rightarrow \Delta p=5.2728\times 10^{-25}\ kgm/s[/tex]
The minimum uncertainty in its momentum is [tex]5.2728\times 10^{-25}\ kgm/s[/tex]
The two cars collide at right angles in the intersection of two icy roads. Car A has a mass of 1965 kg and car B has a mass of 1245 kg. The cars become entangled and move off together with a common velocity in the direction indicated. If car A was traveling 52 km/h at the instant of impact, compute the corresponding velocity of car B just before impact.
Answer:
U2 = 47.38m/s = initial velocity of B before impact
Explanation:
An example of the diagram is shown in the attached file because of missing angle of direction in the question
Mass A, B are mass of cars
A = 1965
B =1245
U1 = initial velocity of A = 52km/hr
U2 = initial velocity of B
V = common final velocity of two cars
BU2 = (A + B)*V sin ¤ ...eq1 y plane
AU1 = (A + B) *V cos ¤ ....equ 2plane
From equ 2
V = AU1/(A + B)*cos ¤
Substitute V into equation 1
We have
U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°
Substitute all parameters to get
U2 = (1965/1245)*52 * tan 30°
U2 = 47.38m/s
To find the corresponding velocity of car B just before impact, we can use the principle of conservation of momentum. By equating the total momentum before the collision to the total momentum after the collision, we can solve for the velocity of car B. Plugging in the given values and calculating, we can find the corresponding velocity of car B just before impact.
Explanation:To solve this problem, we can use the principle of conservation of momentum. Since the two cars stick together after the collision, the total momentum before the collision is equal to the total momentum after the collision.
Let's define the velocity of car A just before impact as vA and the velocity of car B just before impact as vB.
Using the principle of conservation of momentum, we can write:
(mass of car A * velocity of car A) + (mass of car B * velocity of car B) = (total mass * common velocity)
Plugging in the given values, we have:
(1965 kg * 52 km/h) + (1245 kg * vB) = (3210 kg * common velocity) (Note: Convert km/h to m/s)
Simplifying the equation, we get:
101820 kg * km/h + 1245 kg * vB = 3210 kg * common velocity
Now, we can solve for the velocity of car B just before impact:
vB = (3210 kg * common velocity - 101820 kg * km/h) / 1245 kg
By substituting the values and calculating, we can find the corresponding velocity of car B just before impact.
A 1.6-kg block is attached to the end of a 2.0-m string to form a pendulum. The pendulum is released from rest when the string is horizontal. At the lowest point of its swing when it is moving horizontally, the block is hit by a 10-g bullet moving horizontally in the opposite direction. The bullet remains in the block and causes the block to come to rest at the low point of its swing. What was the magnitude of the bullet's velocity just before hitting the block?
Answer:
1002.2688 m/s
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
h = The length of a string = 2 m
m = Mass of block = 1.6 kg
[tex]m_2[/tex] = Mass of bullet = 0.01 kg
Here, the potential energy of the fall will balance the kinetic energy of the bullet
[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 2}\\\Rightarrow v=6.26418\ m/s[/tex]
Velocity of block is 6.26418 m/s
As the momentum of system is conserved we have
[tex]mv=m_2u\\\Rightarrow u=\dfrac{mv}{m_2}\\\Rightarrow u=\dfrac{1.6\times 6.26418}{0.01}\\\Rightarrow u=1002.2688\ m/s[/tex]
The magnitude of velocity just before hitting the block is 1002.2688 m/s
A small object is attached to the end of a relaxed, horizontal spring whose opposite end is fixed. The spring rests on a frictionless surface. Let the initial position of the object be defined as x-0. The object is pulled to position x = A and then released, after which it undergoes simple harmonic motion. In one full cycle of its motion, the total distance traveled by the object is 2 4 O 4A
Answer:
The total distance traveled by an object attached to a spring that is pulled to position x=A and then released is 4 A.
Explanation:
We have an small object attached to a relaxed spring whose opposite end is fixed. The spring rests on a frictionless surface. This means the only force acting on the object is the elastic force of the spring, a conservative force, since the weight and the normal force compensate between them.
The initial position of the object is x=0 then is pulled to position x=A and released. After which it undergoes a simple harmonic motion with an amplitude A. From the position x=A to the equilibrium position in x=0 the object travels a distance A. From the equilibrium position x=0 to maximum negative position in x= -A the object travels again a distance A. Then to return to the original position the object should travel a distance 2 A in reverse direction.
In one full cycle of its motion the object travels a distance 4 A.
You are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 m long, and its top end is 1.4 m above the ground. As the refrigerator moves down the ramp, you are on the down side of the ramp trying to slow the motion by pushing horizontally against the refrigerator with a force of 270N .
Part A
How much work do you do on the refrigerator during its trip down the ramp?
W = ???
Final answer:
The work done on the refrigerator during its trip down the ramp is 1350 J.
Explanation:
To calculate the work done while unloading the refrigerator, we need to determine the displacement of the refrigerator and the force applied.
The displacement is given by the length of the ramp, which is 5.0 m.
The force applied is given as 270 N.
To calculate the work done, we use the formula:
Work = Force x Displacement x Cosine(angle)
In this case, the angle is 0 degrees since the force is applied horizontally.
Therefore, the work done on the refrigerator is:
Work = 270 N x 5.0 m x Cos(0°) = 1350 J.
A steam of hot dry nitrogen flows through a process unit that contains liquid acetone. A substantial part of the acetone vaporizes and is carried off by the nitrogen. The combined gases leave the unit at 205 ◦C and 1.1 bar, are compressed and then enter a condenser in which a portion of the acetone is liquefied. The remaining gas leaves the condenser at 10 ◦C and 40 bar. The partial pressure of acetone in the feed to the condenser is 0.100 bar, and that in the effluent gas from the condenser is 0.379 bar. Assume ideal gas behavior.
(a) Calculate, for a basis of m3 of gas fed to the condenser, the mass of ace- tone condensed (kg) and the volume of gas leaving the condenser (m3).
(b) Suppose the volumetric flow rate of the gas leaving the condenser is 20.0 m3/h. Calculate the rate (kg/h) at which acetone is vaporized in the solvent recovery unit.
The problem is solved by applying the Ideal Gas Law and Dalton's Law of Partial Pressures. First, we calculate the mass of acetone condensed using these laws and then determine the rate at which acetone is vaporized using the given volumetric flow rate.
Explanation:The given problem involves a number of gas law principles, but its main focus is on the application of the Ideal Gas Law and Dalton's Law of partial pressures. Initially, we calculate the moles of acetone in the feed using the Ideal Gas Law, and then we find out the moles of acetone in the effluent using Dalton's law. Subtracting gives us the moles of acetone condensed.
(a) Using Ideal Gas Law, we have PV=nRT. Hence, n (acetone, feed) = P (acetone, feed) * V(feed) /RT(feed). To find the moles of acetone in the effluent, we use Dalton's law and the Ideal Gas Law to get n (acetone, effluent) = P(acetone, effluent) * V (effluent) / RT (effluent). Subtracting moles in effluent from moles in feed gives moles condensed. Multiplying by the molar mass of acetone, we get mass of acetone condensed.
(b) The question tells us the volumetric flow rate of the gas leaving the condenser. Therefore, number of moles of acetone vaporized per hour can be calculated using the Ideal Gas Law and then can be converted into mass by using the molar mass of acetone. Hence, rate at which acetone is vaporized = moles (acetone, vaporized per hour) * molar mass (acetone).
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A spring that is stretched 23 cm from its equilibrium point experiences a force of 103 N.a.) How much energy is stored in the spring?
Answer:
U = 11.85 J
Explanation:
given,
spring is stretched = 23 cm
x = 0.23 m
Force experiences = 103 N
we know,
F = k x
where k is the spring constant
[tex]k =\dfrac{F}{x}[/tex]
[tex]k =\dfrac{103}{0.23}[/tex]
k = 447.83 N/m
energy stored in the spring
[tex]U =\dfrac{1}{2}kx^2[/tex]
[tex]U =\dfrac{1}{2}\times 447.83 \times 0.23^2[/tex]
U = 11.85 J
hence, energy stored in the spring is equal to U = 11.85 J
The elastic potential energy stored in the spring when stretched to a length of 23 cm, given a spring constant of 4 N/cm, is 0.18 J.
Explanation:The question pertains to how much energy is stored in a spring when it is stretched 23 cm from its equilibrium point, where it experiences a force of 103 N. The potential energy stored in the spring can be calculated using the formula U = 1/2kx². Given that the spring constant is 4 N/cm and the displacement of the spring from its unstretched length is 3 cm (since the unstretched length is 20 cm), the potential energy can be calculated as follows: U = 1/2 * 4 N/cm * (3 cm)² = 0.18 J. Thus, the elastic potential energy contributed by the spring when it is stretched to a length of 23 cm is 0.18 J.
A backyard swimming pool with a circular base of diameter 6.00 m is filled to depth 1.50 m. (a) Find the absolute pres- sure at the bottom of the pool. (b) Two persons with com- bined mass 150 kg enter the pool and float quietly there. No water overflows. Find the pressure increase at the bottom of the pool after they enter the pool and float.
Answer
given,
diameter = 6 m
depth = h = 1.5 m
Atmospheric pressure = P₀ = 10⁵ Pa
a) absolute pressure
P = P₀ + ρ g h
P = 10⁵ + 1000 x 10 x 1.5
P = 1.15 x 10⁵ Pa
b) When two person enters into the pool
mass of the two person = 150 Kg
weight of water level displaced is equal to the weight of person
ρ V g = 2 m g
[tex]V = \dfrac{2 m}{\rho}[/tex]
[tex]V = \dfrac{2\times 150}{1000}[/tex]
V = 0.3 m³
Area of pool = [tex]\dfrac{\pi}{4}d^2[/tex]
=[tex]\dfrac{\pi}{4}\times 6^2[/tex]
= 28.27 m²
height of the water rise
[tex]h = \dfrac{V}{A}[/tex]
[tex]h = \dfrac{0.3}{28.27}[/tex]
h = 0.0106 m
pressure increased
P = ρ g h
P = 1000 x 10 x 0.0106
P = 106.103 Pa
a. the absolute pressure at the bottom of the pool is approximately 116.02 kPa.
b. the pressure increase at the bottom of the pool after the persons enter and float is approximately 0.147 kPa.
To find the absolute pressure at the bottom of the pool, we can use the hydrostatic pressure formula:
(a) Absolute pressure at a depth in a fluid:
P=P_0 +ρ⋅g⋅h
Where:
P is the absolute pressure at the given depth.
P_0 is the atmospheric pressure (which we'll assume to be 1 atm, or approximately 101.3 kPa).
ρ is the density of the fluid (water), which is about 1000 kg/m³.
g is the acceleration due to gravity, approximately 9.81 m/s².
h is the depth below the surface.
Given that the diameter of the pool is 6.00 m, the radius (r) is 3.00 m, and the depth (h) is 1.50 m, we can find the pressure at the bottom of the pool:
P=101.3kPa+(1000kg/m³)⋅(9.81m/s²)⋅(1.50m)
Solve for P:
P≈101.3kPa+14,715Pa≈116,015Pa≈116.02kPa
So, the absolute pressure at the bottom of the pool is approximately 116.02 kPa.
(b) To find the pressure increase at the bottom of the pool after two persons with a combined mass of 150 kg enter and float, we need to consider the additional weight of the water displaced by the persons. This additional weight will create an increase in pressure at the bottom of the pool.
The buoyant force (Fb) on an object submerged in a fluid is given by Archimedes' principle:
Fb =ρ⋅V⋅g
Where:
Fb is the buoyant force.
ρ is the density of the fluid (water), which is 1000 kg/m³.
V is the volume of water displaced, which is equal to the volume of the submerged part of the persons.
g is the acceleration due to gravity (9.81 m/s²).
The volume of water displaced is equal to the volume of the persons submerged, which is given by the cross-sectional area (A) of the pool base times the depth (h) to which they are submerged:
V=A⋅h
A=π⋅r^2
Given that the radius (r) is 3.00 m and the depth (h) is 1.50 m, we can calculate the area and then the volume:
A=π⋅(3.00m)^2 ≈28.27m²
V=28.27m²⋅1.50m≈42.41m³
Now, we can calculate the buoyant force:
Fb=(1000kg/m³)⋅(42.41m³)⋅(9.81m/s²)≈416,068N
Since pressure is force per unit area, we need to calculate the additional pressure (ΔP) at the bottom of the pool:
ΔP= Fb/A
A=π⋅(3.00m)^2 ≈28.27m²
ΔP= 416,068N / 28.27m² ≈14,699Pa≈0.147kPa
So, the pressure increase at the bottom of the pool after the persons enter and float is approximately 0.147 kPa.
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A 10-kg block on a rough horizontal surface is attached to a light spring (force constant = 1400 N/m). The block is pulled 10.0 cm to the right from its equilibrium position and released from rest. The frictional force between the block and surface has a magnitude of 30 N. What is the kinetic energy of the block as it passes through its equilibrium position?
To develop this problem it is necessary to apply the concepts related to the conservation of Energy. In this case the definition concerning kinetic energy from the simple harmonic movement.
From the conservation of energy we know that the kinetic energy would be conserved through the work done by the frictional force and the simple harmonic potential energy, in other words:
[tex]KE = PE_s +W_f[/tex]
[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]
Where,
[tex]KE =[/tex] Kinetic Energy
[tex]PE_s =[/tex]Potential Harmonic Simple Energy
[tex]W_f =[/tex] Work made by friction.
Our values are given as,
[tex]m = 10Kg \rightarrow[/tex] mass
[tex]k = 1400N/m \rightarrow[/tex] Spring constant
[tex]A = 0.1m \rightarrow[/tex]Amplitude
[tex]f_f = 30N \rightarrow[/tex] Frictional Force
Replacing we have,
[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]
[tex]KE =\frac{1}{2} 1400 * 0.1^2 + ( - 30 * 0.1)[/tex]
[tex]KE = 4 J[/tex]
Therefore the Kinetic Energy of the block as it passes through its equlibrium position is 4J.
The kinetic energy of the 10-kg block as it passes through its equilibrium position is 4 Joules. This is calculated by converting the potential energy stored in the spring to kinetic energy and then subtracted the energy lost due to friction.
Explanation:The first step in solving this problem is to understand the two forces acting on the block in this question: the spring force and the frictional force. The spring potential energy when the block is pulled 10 cm to the right is given by the formula U = 1/2kx^2, where k is the force constant and x is the displacement. Substituting the given values, we have U = 1/2(1400 N/m)(0.1 m)^2 = 7 Joules. This is the initial potential energy stored in the spring. As the block passes through its equilibrium position, this potential energy is fully converted to kinetic energy.
We also need to take into account the work done against frictional force which is equal to the frictional force times the displacement, i.e., W_friction = Friction * displacement = 30N * 0.1m = 3 Joules. This is the energy lost due to friction.
Finally, we subtract the work done by the frictional force from the potential energy to achieve the kinetic energy. Therefore, the kinetic energy of the block as it passes through its equilibrium position is K = U - W_friction = 7J - 3J = 4 Joules.
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A "Barrel of Fun" consists of a large vertical cylinder that spins about its axis fast enough so that any person inside will be held against the wall. At an angular speed ω1, an upward frictional force F1 holds a person against the wall without slippingIf you double the angular speed (i.e., ω2 = 2 ω1) what is the new friction force F2?
Answer:
Friction force F₂ after doubling the angular speed is same as the friction force at angular speed ω₁
Explanation:
Consider the fig attached below. Forces acted on person are Centripetal force (-mv²/r) exerted in x direction and reversal normal force N wall exerted on person.
[tex]\sum F_{x} =0\\N+ ma_{x}\\-N=m(-\frac{v^{2}}{r})\\-N=m(-r\omega^{2})N=m(r\omega^{2})[/tex] ---(1)
In y direction there is frictional force Fs exerted in upward direction that keeps the person standing without falling which is balanced by weight of person in downward direction.
[tex]\sum F_{y} = 0\\F_{s}-mg=0\\F_{s}=mg[/tex]----(2)
from eq 2 it can e seen that frictional force is equal to weight of person exerted in upward direction, it does not depends on angular speed ω₁. So when the angular speed is doubled i.e ω₂ = 2ω₁, frictional force Fs remains same.
An automatic coffee maker uses a resistive heating element to boil the 2.4 kg of water that was poured into it at 21 °C. The current delivered to the coffee pot is 8.5 A when it is plugged into a 120 V electrical outlet. If the specific heat capacity of water is 4186 J/kgC° , approximately how long does it take to boil all of the water?
Answer:
Explanation:
The expression for the calculation of the enthalpy change of a process is shown below as:-
[tex]\Delta H=m\times C\times \Delta T[/tex]
Where,
[tex]\Delta H[/tex]
is the enthalpy change
m is the mass
C is the specific heat capacity
[tex]\Delta T[/tex]
is the temperature change
Thus, given that:-
Mass of water = 2.4 kg
Specific heat = 4.18 J/g°C
[tex]\Delta T=100-21\ ^0C=79\ ^0C[/tex]
So,
[tex]\Delta H=2.4\times 4.18\times 79\ J=792.52\ kJ[/tex]
Heat Supplied [tex]Q=VIt[/tex]
where [tex]i=current[/tex]
[tex]V=Voltage[/tex]
[tex]8.5\times 120\times t=2.4\times 4186\times 79[/tex]
[tex]t=778.10 s[/tex]
Final answer:
To determine the time required to boil the water in a coffee maker, calculate the power using P = IV, then the heat needed to raise the water temperature to its boiling point with Q = mcΔT, and finally divide Q by P to find the time.
Explanation:
The question deals with calculating how long it takes an automatic coffee maker to boil 2.4 kg of water using a resistive heating element. The initial temperature of the water is 21 °C, and we are given that the coffee maker is plugged into a 120 V electrical outlet with a current of 8.5 A. The specific heat capacity of water is 4186 J/kg°C. To find the time to boil the water, we'll first calculate the power delivered by the heating element using the relationship P = IV, where P is power, I is current, and V is voltage. Once power is known, we can use the equation Q = mcΔT to find the heat Q required to raise the temperature of water to its boiling point (100 °C), where m is water's mass, c is the specific heat capacity, and ΔT is the change in temperature. Finally, we'll divide quantified heat Q by the power P to find the time t: t = Q/P.
For a new TV series "Stupidity Factor contestants are dropped into the ocean (p 1030 kg/ms) along with a Styrofoam soo kg/ms block that is mm x 3 mm X 20 cm thick. If too many contestants climb aboard the block and it sinks below the water surface, they are declared "stupid" and abandoned at sea. Find the maximum number of 7o kg contestants that the block can hold. (No fractional contestants, please. They hate it when that happens.
Answer:
Explanation:
Heres the correct and full question:
For a new TV series "Stupidity Factor contestants are dropped into the ocean (p=1030 kg/m³) along with a Styrofoam p=3oo kg/m³ block that is 2m x 3 m X 20 cm thick. If too many contestants climb aboard the block and it sinks below the water surface, they are declared "stupid" and abandoned at sea. Find the maximum number of 7o kg contestants that the block can hold. (No fractional contestants, please. They hate it when that happens.
answer:
volume of styrofoam block=V=2m x 3m x 0.20m =1.2m³
density of styrofoam=ρs=300kg/m³
mass of styrofoam=ms=v(ρs)=1.2 x 300=360kg
weight of styrofoam = ws=(ms)g=360 x 9.8=3528N
consider number of contestants= n
toatal weight of ccontestants=W=n(70 x 9.8)=n(686N)
since, styrofoam fully submerged into water, then bat force,
B=ρ(vs)g=1030 x 1.2 x 9.8 = 12112.8N
At equilibrium,
B - W - Ws = 0
12112.8 - 686n - 3528 = 0
[tex]n=\frac{12112.8-3528}{686}=12.5[/tex]
n=12 person(contestants)
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal.
At its highest point in this trajectory, the velocityof the projectile would be what?
Answer:
169.4 m/s
Explanation:
Given that the angle of projectile is θ_1 =450°
The speed of body at maximum height is U cosθ_1 = 150 m/s
The angle in second trail is θ_2 =37°
From the given data U cosθ_1 = 150 m/s
U = 150 m/s / cosθ_1
= 150m/s / cos45°
=212.13 m/s
The velocity of the projectile at maximum height in second trail= Ucos(θ_2)
=212.13 m/s×cos37°
=169.4 m/s
The velocity of a projectile at its highest point when fired at an angle to the horizontal remains the same if the initial speed is unchanged, regardless of the angle. Hence, even when changing the angle from 45 to 37 degrees, the velocity at the highest point would still be 150 m/s.
The student's question deals with the velocity of a projectile at the highest point in its trajectory. When a projectile is fired upward at an angle, its velocity at the highest point of its trajectory is only composed of the horizontal component because the vertical component of the velocity becomes zero at that point.
In the given scenario, when the projectile is shot at 45 degrees to the horizontal, the speed at the highest point is given as 150 m/s. This speed represents the horizontal component since there's no vertical component at the highest point. When the angle is changed to 37 degrees, the horizontal component of the initial velocity is calculated using the cosine component of the initial speed, hence the velocity at the highest point remains the same as when it was fired at 45 degrees, provided that the initial speed is unchanged.
Therefore, in the second trial, with the angle at 37 degrees, the velocity of the projectile at its highest point would still be 150 m/s, because the horizontal component of the velocity is not affected by the change in angle.
A cylinder is filled with 10.0 L of gas and a piston is put into it. The initial pressure of the gas is measured to be 163. kPa. The piston is now pushed down, compressing the gas, until the gas has a final volume of 4.40 L. Calculate the final pressure of the gas. Round your answer to 3 significant digits.
Answer:
P₂ = 370 kPa
Explanation:
Boyle's law: States that the volume of a given mass of gas is inversely proportional to its pressure provided the temperature remains constant. It can be expressed mathematically as,
P₁V₁ = P₂V₂................ Equation 1
Where P₁ = Initial Pressure, V₁ = Initial volume, P₂ = Final Pressure, V₂ = Final Volume.
Making P₂ The subject of the equation above,
P₂ = P₁V₁/V₂..................... equation 2
Substituting these vaues into equation 2,
Where P₁ = 163 kPa = 163000 pa, V₁ = 10 L, V₂ = 4.4 L.
P₂ = 163000(10)/4.4
P₂ = 370454.55 Pa
P₂ = 370000 Pa
P₂ = 370 kPa To 3 significant figure.
Therefore Final pressure = 370 kPa
What is the object’s velocity, in meters per second, at time t = 2.9? Calculate the object’s acceleration, in meters per second squared, at time t = 2.9. What is the magnitude of the object’s maximum acceleration, in meters per second squared? What is the magnitude of the object’s maximum velocity, in meters per second?
Answer:
the question is incomplete, below is the complete question
"An object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 5.5 cos(4.4t - 1.8), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.
a.What is the object's velocity, in meters per second, at time t = 2.9?
b.Calculate the object's acceleration, in meters per second squared, at time t = 2.9.
c. What is the magnitude of the object's maximum acceleration, in meters per second squared?
d.What is the magnitude of the object's maximum velocity, in meters per second?"
a.[tex]v(t)==24.1m/s[/tex]
b.[tex]a(t)=3.79m/s^{2}[/tex]
c.[tex]a_{max}=106.48m/s^{2}[/tex]
d.[tex]v_{max}=24.2m/s[/tex]
Explanation:
the gneral expression for the displacement of object in simple harmonic motion is represented by
[tex]x(t)=Acos(wt- \alpha)\\[/tex]
while the velocity is express as
[tex]v(t)=-Awcos(4.4t-1.8)\\[/tex]
and the acceleration is
[tex]a(t)=-aw^{2}cos(wt- \alpha )\\[/tex]
Note: the angle is in radians
The expression for the displacement from the question is [tex]x(t)=5.5cos(4.4t-1.8)\\[/tex]
comparing, A=5.5, w=4.4,α=1.8
a.To determine the object velocity at t=2.9secs,
we substitute for t in the velocity equation
[tex]v(t)=-5.5*4.4sin(4.4*2.9-1.8)\\v(t)=-24.2sin(10.96)\\[/tex]
[tex]v(t)=-24.2*(-0.9993)\\v(t)==24.1m/s[/tex]
b.To determine the object acceleration at t=2.9secs,
we substitute for t in the acceleration equation
[tex]a(t)=-5.5*4.4^{2} cos(4.4*2.9-1.8)\\a(t)=-106.48cos(10.96)\\[/tex]
[tex]a(t)=-106.48*0.0356\\a(t)=3.79m/s^{2}[/tex]
c. The acceleration is maximum when the displacement equals the amplitude. hence magnitude of the object acceleration is
[tex]a_{max}=-w^{2}A\\ a_{max}=-4.4^{2}*5.5\\ a_{max}=106.48m/s^{2}[/tex]
d.The maximum velocity is expressed as
[tex]v_{max}=wA\\v_{max}=4.4*5.5\\v_{max}=24.2m/s[/tex]
To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of 60.0°.What is the angular acceleration (in rad/s2) if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.250 kg · m2, and the net force she exerts is 550 N at an effective perpendicular lever arm of 2.10 cm? (Ignore gravity)How much work (in J) does she do?
To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:
[tex]\tau = Fr[/tex]
Where,
F = Force
r = Radius
Replacing we have that,
[tex]\tau = Fr[/tex]
[tex]\tau = 21cm (\frac{1m}{100cm})* 550N[/tex]
[tex]\tau = 11.55Nm[/tex]
The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore
[tex]I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2[/tex]
[tex]I = 0.394kg\cdot m^2[/tex]
Finally, angular acceleration is a result of the expression of torque by inertia, therefore
[tex]\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}[/tex]
[tex]\alpha = \frac{11.55}{0.394}[/tex]
[tex]\alpha = 29.3 rad/s^2[/tex]
PART B)
The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians [tex](\pi / 3)[/tex], therefore
[tex]W = \tau \theta[/tex]
[tex]W = 11.5* \frac{\pi}{3}[/tex]
[tex]W = 12.09J[/tex]
A projectile of mass m1 moving with speed v1 in the +x direction strikes a stationary target of mass m2 head-on. The collision is elastic. Use the Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target, making no approximations concerning the masses. (Use the following as necessary: m1, m2, and v1.)
Using Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target are respectively;
v1 = (m1•u1 - m2•u1)/(m1 + m2)
v1 = (m1•u1 - m2•u1)/(m1 + m2)v2 = (m1•u1 + m1•u2)/(m1 + m2)
Let us denote the following;
Mass of projectile; m1
Mass of target; m2
Initial velocity of projectile; u1
Final velocity of projectile; v1
Initial velocity of target; u2 = 0 m/s
Final velocity of target; v2
From conservation of momentum, we have;
(m1•u1) + (m2•u2) = (m1•v1) + (m2•v2)
Plugging in the relevant values;
(m1•u1) + (m2•0) = (m1•v1) + (m2•v2)
(m1•u1) = (m1•v1) + (m2•v2) - - - (eq 1)
From energy principle, we have;
u1 + v1 = u2 + v2
Thus;
Since u2 is zero, then we have;
u1 + v1 = v2 - - - (eq 2)
Put (u1 + v1) for v2 in eq 1 to get;
(m1•u1) = (m1•v1) + (m2(u1 + v1))
(m1•u1) = (m1•v1) + (m2•u1 + m2•v1) - - - (eq 3)
Let us make v1 the subject;
m1•u1 - m2•u1 = m1•v1 + m2•v1
v1(m1 + m2) = m1•u1 - m2•u1
v1 = (m1•u1 - m2•u1)/(m1 + m2)
Let us make v1 the subject in eq 2;
v1 = v2 - u2
Put v2 - u2 for v1 in eq 1 to get;
(m1•u1) = (m1(v2 - u2) + (m2•v2)
(m1•u1) = m1•v2 - m1•u2 + m2•v2
m1•u1 + m1•u2 = v2(m1 + m2)
v2 = (m1•u1 + m1•u2)/(m1 + m2)
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The final velocities of the projectile and the target after an elastic collision can be arrived at by solving the equations for Conservation of Momentum and Conservation of Kinetic Energy simultaneously. Here, the target's initial velocity is zero (it's at rest). By substituting this into the equations and solving them, we can find the final velocities.
Explanation:An elastic collision is a type of collision where both momentum and kinetic energy are conserved. For this problem, we can solve the final velocities using the conservation laws.
Conservation of Momentum gives us:
m1*v1 = m1*v1' + m2*v2' -------- (1)
V1' and V2' are the final velocities of the projectile and target, respectively.
Conservation of Kinetic Energy gives us:
1/2*m1*v1^2 = 1/2*m1*(v1')^2 + 1/2*m2*(v2')^2 -------- (2)
While solving these two equations, one should remember that m2 was initially at rest, so the momentum in the Y-direction is zero. The above-mentioned equations can be solved by substitution method or any other algebraic method to get the final velocities v1' and v2'.
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The free-fall acceleration on Mars is 3.7 m/s2.
(a) What length of pendulum has a period of 1.2 s on Earth? cm
(b) What length of pendulum would have a 1.2-s period on Mars? cm An object is suspended from a spring with force constant 10 N/m.
(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg
(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars.
Answer:
(a) The length of the pendulum on Earth is 36.8cm
(b) The length of the pendulum on Mars is 13.5cm
(c) Mass suspended from the spring on Earth is 0.37kg
(d) Mass suspended from the spring on Mars is 0.36kg
Explanation:
Period = 1.2s, free fall acceleration on Earth = 9.8m/s^2, free fall acceleration on Mars = 3.7m/s^2
( a) Length of pendulum on Earth = [( period ÷ 2π)^2] × acceleration = (1.2 ÷ 2×3.142)^2 × 9.8 = 0.0365×9.8 = 0.358m = 35.8cm
(b) Length of the pendulum on Mars = (1.2÷2×3.142)^2 × 3.7 = 0.0365×3.7 = 0.135cm = 13.5m
(c) Mass suspended from the spring on Earth = (force constant×length in meter) ÷ acceleration = (10×0.358) ÷ 9.8 = 0.37kg
(d) Mass suspended from the spring on Mars = (10×0.135)÷3.7 = 0.36kg
The length of a pendulum with a period of 1.2 s on Earth is approximately 36.95 cm, while on Mars it is around 16.99 cm. The mass suspended from a spring that would result in a period of 1.2 s on Earth is approximately 0.722 kg, and on Mars it is approximately 0.329 kg.
Explanation:(a) What length of pendulum has a period of 1.2 s on Earth? cm
Using the equation for the period of a pendulum, T = 2π √(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity, we can solve for L. Rearranging the equation, we have L = (T/2π)² * g.
Given that the free-fall acceleration on Earth is approximately 9.8 m/s², substituting the values into the equation, we have:
L = (1.2/2π)² * 9.8 = 0.3695 m = 36.95 cm
(b) What length of pendulum would have a 1.2-s period on Mars? cm
Using the same equation, L = (T/2π)² * g, we can substitute the values for the period and acceleration due to gravity on Mars:
L = (1.2/2π)² * 3.7 = 0.1699 m = 16.99 cm.
(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg
For a spring-mass system, the period is given by T = 2π √(m/k), where T is the period, m is the mass, and k is the spring constant. Rearranging the equation, we have m = (T/2π)² * k.
Given that the spring constant is 10 N/m, substituting the values into the equation, we have:
m = (1.2/2π)² * 10 = 0.722 kg.
(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars. kg
Using the same equation, m = (T/2π)² * k, we can substitute the values for the period and spring constant:
m = (1.2/2π)² * 10 = 0.329 kg.
The hydrofoil boat has an A-36 steel propeller shaft that is 100ft long. It is connected to an inline diesel engine that delivers a maximum power of 2500 hp and causes the shaft to rotate at 1700 rpm. If the outer diamater of the shaft is 8 in. and the wall thickness is 3/8 in., determine the maximum sheer stress developed in the shaft. Also, what is the "wind up," or angle of twist in the shaft at full power?
Final answer:
The maximum shear stress developed in the shaft is approximately 7.57 MPa. The angle of twist in the shaft at full power is approximately 3.00°.
Explanation:
To determine the maximum shear stress developed in the shaft, we can use the formula:
Shear stress (τ) = Torque (T) / Polar Moment of Inertia (J)
To find the torque, we can use the formula:
Torque (T) = Power (P) / Angular Velocity (ω)
We are given that the power is 2500 hp and the angular velocity is 1700 rpm. Converting these values to W and rad/s respectively, we have:
Power (P) = 2500 hp * 745.7 W/hp ≈ 1,864,250 W
Angular Velocity (ω) = 1700 rpm * 2π rad/minute ≈ 17897.37 rad/s
Substituting these values into the torque formula, we have:
Torque (T) = 1,864,250 W / 17897.37 rad/s ≈ 103.98 N*m
Next, we need to find the polar moment of inertia (J) of the shaft. The polar moment of inertia for a solid shaft can be calculated using the formula:
J = (π/2) * (outer diameter^4 - inner diameter^4)
Converting the diameter to meters, we have:
Diameter (d) = 8 in * 0.0254 m/in = 0.2032 m
Substituting this value into the polar moment of inertia formula, we have:
J = (π/2) * (0.2032^4 - (0.2032 - 2 * 3/8)^4)
= (π/2) * (0.2032^4 - 0.1926^4)
0.013743 m^4
Finally, we can calculate the maximum shear stress using the formula:
Shear stress (τ) = 103.98 N*m / 0.013743 m^4 ≈ 7.57 MPa
As for the angle of twist, we can use the formula:
Angle of twist (θ) = T * L / (G * Given that the length of the shaft is 100 ft and Young's modulus (G) for A-36 steel is approximately 200 GPa, we can calculate the angle of twist as follows:
Angle of twist (θ) = 103.98 N*m * 100 ft * 0.3048 m/ft / (200 GPa * 0.013743 m^4)
≈ 0.0524 radians or 3.00°
All Houston Methodist buildings system wide have an emergency power generator that turns on to supply emergency power after normal power shuts down within:
A. 5 seconds
B. 30 seconds
C. 60 seconds
D. 10 seconds
Answer:
(D) 10 seconds
Explanation:
The Houston Methodist Hospital automated the emergency power supply system (EPSS) testing processes in order to ensure the safety of patients during an outage
There is an emergency power generator that turns on to supply emergency power after normal power shuts down within 10 seconds.
Answer:
D. 10 seconds
Explanation:
The Houston Methodist Hospital has eleven diesen powered generators that turn on at most 10 seconds after the normal power shuts down.
Considering that it is a hospital, this system is really important.
So the correct answer is:
D. 10 seconds
A cart of mass 250 g is place on a frictionless horizontal air track. A spring having a spring constant of 9.5 N/m is attached between the cart and the left end o f the track. If the cart is displaced 4.5 cm from its equilibrium position, find: a) The period at which it oscillates.[1.0 s] b) Its maximum speed.[0.28 m/s] c) Its speed when it is located 2.0 cm from its equilibrium position.[0.25 m/s]
Answer:
a) [tex]T=1.02 s[/tex]
b) [tex]v=0.28\frac{m}{s}[/tex]
c) [tex]v=0.25\frac{m}{s}[/tex]
Explanation:
a) The period is defined as:
[tex]T=\frac{2\pi}{\omega}[/tex]
[tex]\omega[/tex] is the natural frequency of the system, in this case is given by:
[tex]\omega=\sqrt{\frac{k}{m}}\\\omega=\sqrt{\frac{9.5\frac{N}{m}}{250*10^{-3}kg}}\\\omega=6.16\frac{rad}{s}[/tex]
Now, we calculate the period:
[tex]T=\frac{2\pi}{6.16\frac{rad}{s}}\\T=1.02 s[/tex]
b) According to the law of conservation of energy, we have:
[tex]\frac{kx^2}{2}=\frac{mv^2}{2}\\v=\sqrt{\frac{kx^2}{m}}\\v=\sqrt{\frac{(9.5\frac{N}{m})(4.5*10^{-2}m)^2}{250*10^{-3}kg}}\\v=0.28\frac{m}{s}[/tex]
c) In this case, we have:
[tex]U_i=U_f+K_f\\\frac{kx_i^2}{2}=\frac{kx_f^2}{2}+\frac{mv^2}{2}\\v=\sqrt{\frac{k(x_i^2-x_f^2)}{m}}\\v=\sqrt{\frac{9.5\frac{N}{m}((4.5*10^{-2}m)^2-(2*10^{-2}m)^2)}{250*10^{-3}kg}}\\v=0.25\frac{m}{s}[/tex]
The maximum speed is; 0.28 m/s
The speed when it is located 2.0 cm from its equilibrium position is; 0.25 m/s
What is the speed from the equilibrium position?A) Formula for period is;
T = √(2π/ω)
where ω which is the natural angular frequency
ω = √(k/m)
we are given;
k = 9.5 N/m
m = 250 g = 0.25 kg
x = 4.5 cm = 0.045 m
Thus;
ω = √(9.5/0.25)
ω = 6.1 rad/s
Thus;
Period is; T = √(2π/6.1)
T = 1.02 s
B) From law of conservation of energy;
maximum speed is;
v = √(kx²/m)
v = √(9.5 * 0.045²/0.25)
v = 0.28 m/s
C) We are now given;
x₁ = 4.5cm = 0.045 m
x₂ = 2 cm = 0.02 m
Thus, speed is gotten from;
v = √((k(x₁ - x₂)/m)
v = √((9.5(0.045 - 0.02)/0.25)
v = 0.25 m/s
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The diagram shows models of both prokaryotic and eukaryotic cells. The diagram shows models of both prokaryotic and eukaryotic cells. Prokaryotic cells are considered to be the more primitive of the two. What is a similarity between eukaryotes and prokaryotes as illustrated by the models? A) Both contain ribosomes. B) Both contain mitochondria. C) Both are surrounded by a capsule. D) Both contain a membrane-bound nucleus.
Answer:
both have plasma membrane, ribosomes, cytoplasm, and DNA
Explanation:
Both eukaryotic and prokaryotic cells contain ribosomes, which are responsible for protein synthesis. Eukaryotic cells are more complex with a true nucleus, found in animals and plants, while prokaryotic cells, found in bacteria, are simpler and lack a true nucleus.
Explanation:Option A is the correct answer. Both eukaryotic and prokaryotic cells contain ribosomes. The ribosomes in both types of cells serve the same function: protein synthesis, or assembly of amino acids to form proteins. Eukaryotic cells are those with a true nucleus and complex structures, found in animals, plants, fungi, and protists, while prokaryotic cells are smaller, simpler cells without a true nucleus, found in bacteria and archaea.
Unlike eukaryotic cells, prokaryotic cells don't contain a membrane-bound nucleus or mitochondria. A capsule is found in some bacteria, not all prokaryotes and no eukaryotes.
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Which of the following statements concerning the nuclear force is false? O The nuclear force is attractive and not repulsive. O The nuclear is one of only four known types of forces in nature. O The nuclear force is very short-ranged. O The nuclear force acts on both protons and neutrons. O The nuclear force is very weak and much smaller in relative magnitude than the electrostatic and gravitational forces. Submit Request
Answer:
The nuclear force is attractive and not repulsive.The nuclear force is very weak and much smaller in relative magnitude than the electrostatic and gravitational forces.Explanation:
Nuclear force is the strongest existing force in the nature. It has the shortest range.Its main function is to hold the subatomic particles together in nature.The nuclear force is created by the exchange of pi mesons between the nucleons of an atom, but for this exchange to happen the particles must be close to one another of the order of few femtometer.At about 1 femtometer the nuclear force is very strongly attractive in nature but at distance greater than 2.5 femtometer it fades away. The force becomes repulsive in nature at distance less than 0.7 femtometer.This force holds the likely charged protons together in the nucleus.Question:
Jane, looking for Tarzan, is running at top speed (4.5 m/s) and grabs a vine hanging vertically from a tall tree in the jungle.
How high can she swing upward?
Energy Conservation
According to the conservation of the mechanical energy the total energy remains constant of the system, therefore the kinetic energy of the system will get converted into the potential energy of the system.
Answer:
She can swing 1.0 m high.
Explanation:
Hi there!
The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).
The kinetic energy is calculated as follows:
KE = 1/2 · m · v²
And the potential energy:
PE = m · g · h
Where:
m = mass of Jane.
v = velocity.
g = acceleration due to gravity (9.8 m/s²).
h = height.
Then:
ME = KE + PE
Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:
ME = KE + PE (PE = 0)
ME = KE
ME = 1/2 · m · (4.5 m/s)²
ME = m · 10.125 m²/s²
When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:
ME = KE + PE (KE = 0)
ME = PE
ME = m · 9.8 m/s² · h
Then, equallizing both expressions of ME and solving for h:
m · 10.125 m²/s² = m · 9.8 m/s² · h
10.125 m²/s² / 9.8 m/s² = h
h = 1.0 m
She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).
An object is 12 m long, 0.65 m wide, and 13 cm high. Calculate the volume of this object.
The volume corresponds to the measure of the space occupied by a body. From the given dimensions we can intuit that we are looking to find the Volume of an Cuboid, that is, an orthogonal rectangular prism, whose faces form straight dihedral angles.
Mathematically the volume of this body is given as
[tex]V = lWh[/tex]
Where,
L = Length
W = Width
H = High
[tex]V = (12)(0.65)(13*10^{-2})[/tex]
[tex]V = 1.014m^3[/tex]
Note: The value given for the height was in centimeters, so it was transformed to meters.