When a stone is whirled at double the speed, the tension in the string increases to four times its original value, assuming the radius of the whirl remains the same.
Explanation:The tension in a string whirling a stone in a circle at a constant speed is directly proportional to the square of the speed. If the boy doubles the speed in the scenario you gave, keeping the radius of the circle unchanged, the tension in the string would increase as the square of that factor. So, between the options given, if the boy increases the speed of the stone so that it makes two complete revolutions every second instead of one, the magnitude of the tension in the string increases to four times its original value. Thus, the correct answer is (A) the magnitude of the tension increases to four times its original value, 4F.
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The tension in the string of a whirling stone increases by a factor of four when the speed of rotation doubles and the radius remains the same. It is because the tension is directly proportional to the square of the speed of the stone.
Explanation:The tension in the string of a whirling stone is related to the centripetal force, which is directly proportional to the square of the speed of rotation and the mass of the stone, and inversely proportional to the radius of the circle. If the speed of rotation doubles (from one revolution per second to two revolutions per second) and the radius of the circle remains the same, the resulting tension in the string (centripetal force) increases by a factor of four.
Hence, the answer is (A) The magnitude of the tension increases to four times its original value, 4F.
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An 80- kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15m/s . Sort the following quantities as known or unknown. Take the horizontal direction to be along the x axis. : the mass of the quarterback, : the mass of the football, : the horizontal velocity of quarterback before throwing the ball, : the horizontal velocity of football before being thrown, : the horizontal velocity of quarterback after throwing the ball, : the horizontal velocity of football after being thrown
Explanation:
Given that,
Mass of the quarterback, m = 80 kg
Mass of the football, m' = 0.43 kg
Speed of the football, v' = 15 m/s
We need to sort the following quantities as known or unknown.
1. The mass of the quarterback = known = 80 kg
2. The mass of the football = known = 0.43 kg
3. The horizontal velocity of quarterback before throwing the ball = known = 0
4. The horizontal velocity of football before being thrown = known = 0
5. The horizontal velocity of quarterback after throwing the ball, = unknown quantity and it can be calculated using the conservation of linear momentum.
6. The horizontal velocity of football after being thrown = known = 15 m/s
Hence, this is the required solution.
A solid plate, with a thickness of 15 cm and a thermal conductivity of 80 W/m·K, is being cooled at the upper surface by air. The air temperature is 10°C, while the temperatures at the upper and lower surfaces of the plate are 650 50 and 60°C, respectively. Determine the convection heat transfer coefficient of air at the upper surface and discuss whether the value is reasonable or not for force convection of air.
To determine the convection heat transfer coefficient of air at the upper surface, we need to know the surface area of the plate which is not provided in the given question.
Explanation:The convection heat transfer coefficient of air at the upper surface can be determined using Newton's law of cooling. According to Newton's law of cooling, the rate of heat transfer through convection is directly proportional to the temperature difference between the solid surface and the surrounding fluid and the surface area of the solid.
Therefore, the heat transfer rate can be calculated using the formula:
Q = h * A * (Ts - T∞)
Where:
In this case, we are given the following values:
To find the surface area, we need to know the dimensions of the plate. Once we have the surface area, we can solve for the convection heat transfer coefficient using the given formula. However, the surface area is not provided in the question, so we cannot determine the convection heat transfer coefficient without that information.
A child pulls a 15 kg sled containing a 5.0 kg dog along a straight path on a horizontal surface. He exerts a force of 55 N on the sled at an angle of 20 degrees above the horizontal. The coefficient of friction between the sled and the surface is 0.22.Calculate the word done by the child's pulling force as the system moves a distance of 7.0 m.
The work done by the child pulling the sled and dog, with a pulling force of 55 N, over a distance of 7.0 m, considering both pulling work and the work done against friction, is 59.42 J.
Explanation:The question concerns the work done by a child pulling a sled with a dog. 'Work done' in physics is calculated using the equation, work done = force x distance x cosine of the angle. The force exerted is 55 N, the distance is 7.0 m, and the angle is 20 degrees. Thus, the work done by the child's pulling force as the system moves a distance of 7.0 m, ignoring friction and because cos(20) is approximately 0.94, is calculated as: Work done = 55 N x 7.0 m x cos(20) = 55 N x 7.0 m x 0.94 = 361.3 J.
However, the total work done is reduced due to friction between the sled and the ground. The sled's total weight (15 kg sled + 5.0 kg dog = 20 kg) multiplied by gravity (9.8 m/s²) gives the normal force (20 kg * 9.8 m/s² = 196 N). Multiplying the normal force by the friction coefficient (0.22), gives the frictional force (196 N * 0.22 = 43.12 N). Hence, the work done against friction is: Work done against friction = frictional force x distance = 43.12 N x 7.0 m = 301.88 J. Therefore, the actual work done by the child equals the pull work minus the work against friction, which is 361.3 J - 301.88 J = 59.42 J.
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A liquid of density 1370 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.67 m/s and the pipe diameter d 1 is 12.9 cm . At Location 2, the pipe diameter d 2 is 16.7 cm . At Location 1, the pipe is Δ y = 9.85 m higher than it is at Location 2. Ignoring viscosity, calculate the difference Δ P between the fluid pressure at Location 2 and the fluid pressure at Location 1.
Answer:
ΔP = P₂ - P₁ = 132. 24 kPa
Explanation:
Given:
ρ = 1370 kg / m³ , v = 9.67 m / s , d₁ = 12.9 cm , d₂ = 16.7 cm , Δy = 9.85 m
Using Bernoulli's equations to determine the difference ΔP
P₂ + ρ * g * Z₂ + (ρ * V₂²) / 2 = P₁ + ρ * g * Z₁ + (ρ * V₁²) / 2
P₂ - P₁ = ρ * g * (Z₁ - Z₂) + [ ρ * (V₁² - V₂²) ] / 2
P₂ - P₁ = ρ * g * (Z₁ - Z₂) + ¹/₂ * ρ * V₁² * [ ( 1 - (d₁ / d₂) ⁴ ) ]
ΔP = 1370 kg / m³ * 9.8 m/s² * 9.85m + 0.5 * 1370 kg / m³ * ( 1 - (12.9 cm / 16.7 cm )⁴ )
ΔP = 132247.9364 Pa
ΔP = 132. 24 kPa
Answer:
P₂ - P₁=173.5kPa
Explanation:
The equation of continuity:
A₁v₁=A₂v₂
where A₁=πd₁²/4 and A₂=πd₂²/4
v₂=(A₁/A₂)v₁
v₂={(πd₁²/4)/(πd₂²/4)}v₁
v₂=(d₁²/d₂²)v₁
Use Bernoulli's equation
P₂+pgz₂+(pv₂²/2)=P₁+pgz₁+(pv₁²/2)
The difference between the fluid pressure at location 2 and the fluid pressure at location 1
P₂ - P₁=pg(z₁-z₂)+{p(v₁²-v₂²)}/2=pg(z₁-z₂)+1/2pv₁²(1-(d₁/d₂)⁴)
P₂ - P₁=(1.370×10³×9.8×9.85)+(1/2)(1.370×10³×(9.67)²){(1-(0.129m/0.167m)⁴}
P₂ - P₁=1.735×10⁵Pa
P₂ - P₁=173.5kPa
The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate the number of years required for the asteroid to make one revolution around the Sun. Number Enter your answer in accordance to the question statement Entry field with incorrect answer Units Choose the answer from the menu in accordance to the question statement Entry field with correct answer.
Answer:
The asteroid requires 5.14 years to make one revolution around the Sun.
Explanation:
Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:
[tex]T^{2} = a^{3}[/tex] (1)
Where T is the period of revolution and a is the semi-major axis.
In the other hand, the distance between the Earth and the Sun has a value of [tex]1.50x10^{8} Km[/tex]. That value can be known as well as an astronomical unit (1AU).
But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.
For the special case of the asteroid the distance will be:
[tex]a = 2.98(1.50x10^{8}Km)[/tex]
[tex]a = 4.47x10^{8}Km[/tex]
That distance will be expressed in terms of astronomical units:
[tex]4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km}[/tex] ⇒ [tex]2.98AU[/tex]
Finally, from equation 1 the period T can be isolated:
[tex]T = \sqrt{a^{3}}[/tex]
[tex]T = \sqrt{(2.98)^{3}}[/tex]
[tex]T = \sqrt{26.463592}[/tex]
[tex]T = 5.14AU[/tex]
Then, the period can be expressed in years:
[tex]5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr[/tex]
[tex]T = 5.14 yr[/tex]
Hence, the asteroid requires 5.14 years to make one revolution around the Sun.
Final answer:
Using Kepler's third law, the asteroid that is 2.98 times the mean distance of Earth from the Sun has an orbital period of approximately 5.14 years.
Explanation:
To calculate the orbital period of an asteroid using Kepler's third law of planetary motion, which relates the orbital period of a planet to its average distance from the Sun, we first need to understand that the asteroid's mean distance from the Sun is given as 2.98 times the Earth's mean distance from the Sun.
Since 1 astronomical unit (AU) is the mean distance of Earth from the Sun, we'll denote the asteroid's mean distance as 2.98 AU. Kepler's third law can be formulated as P² = a³, where P is the orbital period in Earth years, and a is the semimajor axis or mean distance from the Sun in astronomical units.
To find the period P, we take the cube of the mean distance (2.98³) and then take the square root of that number to find the orbital period in years.
Calculation:
a = 2.98 AU (mean distance to the Sun)
a³ = 2.98 x 2.98 x 2.98 = 26.45
P² = 26.45
P = √26.45
P ≈ 5.14 years
Therefore, the asteroid takes approximately 5.14 years to make one revolution around the Sun.
A propeller consists of two blades each 3.0 m in length and mass 120 kg each. The propeller can be approximated by a single rod rotating about its center of mass. The propeller starts from rest and rotates up to 1200 rpm in 30 seconds at a constant rate.
a) What is the angular momentum of the propeller at ????=10s;????=20s?
b) What is the torque on the propeller?
Answer:
Explanation:
a)
[tex]l[/tex] = length of each blade = 3 m
[tex]m[/tex] = mass of each blade = 120 kg
[tex]L[/tex] = length of the rod
[tex]M[/tex] = mass of the rod
Length of the rod is given as
[tex]L = 2 l = 2 (3) = 6 m[/tex]
[tex]w_{0}[/tex] = Angular velocity at t = 0, = 0 rad/s
[tex]w_{30}[/tex] = Angular velocity at t = 30, = 1200 rpm = 125.66 rad/s
[tex]\Delta t[/tex] = time interval = 30 - 0 = 30 s
angular acceleration is given as
[tex]\alpha = \frac{w_{30} - w_{0}}{\Delta t} = \frac{125.66 - 0}{30} = 4.2 rad/s^{2}[/tex]
Angular velocity at t = 10 s is given as
[tex]w_{10} = w_{0} + \alpha t\\w_{10} = 0 + (4.2) (10)\\w_{10} = 42 rad/s[/tex]
Angular velocity at t = 20 s is given as
[tex]w_{20} = w_{0} + \alpha t\\w_{20} = 0 + (4.2) (20)\\w_{10} = 84 rad/s[/tex]
Mass of the rod is given as
[tex]M = 2m = 2(120) = 240 kg[/tex]
Moment of inertia of the propeller is given as
[tex]I = \frac{ML^{2} }{12} = \frac{(240)(6)^{2} }{12} = 720 kgm^{2}[/tex]
Angular momentum at t = 10 s is given as
[tex]L_{10} = I w_{10} = (720) (42) = 30240 kgm^{2}/s[/tex]
Angular momentum at t = 20 s is given as
[tex]L_{20} = I w_{20} = (720) (84) = 60480 kgm^{2}/s[/tex]
b)
Torque on propeller is given as
[tex]\tau = I \alpha \\\tau = (720) (4.2)\\\tau = 3024 Nm[/tex]
Answer:
(a) 4800 kg m^2/s^2 , 9600 kg m^2/s^2
(b) 480 Nm
Explanation:
length of each blade = 3 m
total length, L = 2 x 3 = 6 m
mass of each blade = 120 kg
total mass, m = 2 x 120 = 240 kg
initial angular velocity, ωo = 0 rad/ s
final angular velocity, ω = 1200 rpm = 1200 / 60 = 20 rps = 125.6 rad/s
t = 30 s
Let the angular acceleration is α.
α = (ω - ωo)/t = 120 / 30 = 4 rad/s^2
(a) moment of inertia. I = mL^2 / 12
I = 240 x 6 / 12 = 120 kg m^2
Let ω' be the angular velocity at the end of 10 s
use first equation of motion
ω' = 0 + αt
ω' = 4 x 10 = 40 rad/s
Angular momentum at t = 10 s
L = I x ω = 120 x 40 = 4800 kg m^2/s^2
Let ω'' be the angular velocity at the end of 20 s
use first equation of motion
ω'' = 0 + αt
ω'' = 4 x 20 = 80 rad/s
Angular momentum at t = 20 s
L = I x ω = 120 x 80 = 9600 kg m^2/s^2
(b) Torque, τ = I x α
τ = 120 x 4 = 480 Nm
The temperature of the Sun’s atmosphere near the surface is about 6000 K, and the concentration of atoms is about 1015 particles/m3. What are the average pressure and density of its atmosphere? What assumptions did you make to solve the problem?
Answer:
average pressure = 8.314 ×[tex]10^{-5}[/tex] pascal
density = 1.66 ×[tex]10^{-12}[/tex] kg/m³
Explanation:
given data
temperature of sun = 6000 K
concentration of atoms = [tex]10^{15}[/tex] particles/m³
to find out
average pressure and density of its atmosphere
solution
first we get here average pressure of atmosphere that is express as
average pressure = [tex]\frac{nRT}{V}[/tex] .............1
put here value we get
average pressure = [tex]\frac{10^15*8.314*6000}{6*10^{23}}[/tex]
average pressure = 8.314 ×[tex]10^{-5}[/tex] pascal
and
density of atmosphere will be
density = [tex]\frac{nM}{V}[/tex] .............2
density = [tex]\frac{10^{15}*1.66*10^{-27}}{1}[/tex]
density = 1.66 ×[tex]10^{-12}[/tex] kg/m³
A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the speaker? What is the wavelength of this sound wave as it travels through air with a speed of about 341 m/s? What is the wavelength of the same sound wave as it enters some water, where it has a speed of about 1480 m/s?
Answer:
0.003034 s
1.035 m
4.5 m
Explanation:
[tex]f[/tex] = frequency of the tone = 329.6 Hz
[tex]T[/tex] = Time period of the sound wave
we know that, Time period and frequency are related as
[tex]T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s[/tex]
[tex]v[/tex] = speed of the sound in the air = 341 ms⁻¹
wavelength of the sound is given as
[tex]\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m[/tex]
[tex]v[/tex] = speed of the sound in the water = 1480 ms⁻¹
wavelength of the sound in water is given as
[tex]\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m[/tex]
A 539 g ball strikes a wall at 14.4 m/s and
rebounds at 13.8 m/s. The ball is in contact
with the wall for 0.033 s.
What is the magnitude of the average force
acting on the ball during the collision?
Answer in units of N.
Answer:
461 N
Explanation:
Impulse = change in momentum
F Δt = m Δv
F (0.033 s) = (0.539 kg) (14.4 m/s − (-13.8 m/s))
F = 461 N
A ladder whose length L is 12 m and mass m is 45 kg rests against a wall. Its upper end is a distance h of 9.3 m above the ground. The center of mass of the ladder is one-third of the way up the ladder. A fire fighter whose mass M is 72 kg climbs the ladder until his center of mass is halfway up. Assume the wall but not the ground is frictionless. What forces are exerted on the ladder by the wall and by the ground? Provide both the magnitude and direction of the ground force.
Answer:
FN₁ = 1146.6N : Force exerted on the ladder by the floor , vertical and upward
FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder
Explanation:
The equilibrium equation are:
∑Fx=0
∑Fy=0
∑M = 0
M = F*d
Where:
∑M : Algebraic sum of moments
M : moment ( N*m)
F : Force ( N)
d :Perpendicular distance of the force to the point ( meters )
Data
m =45 kg : mass of the ladder
M =72 kg : mass of the fire fighter
g = 9.8 m/s²: acceleration due to gravity
L = 12 m : ladder length
h = 9.3 m: ladder height
L/3 = 12/3 = 4m Location of the center of mass of the ladder of the way up
L/2 = 12/2 = 6m Location of the center of mass of the fire fighter
µ = 0 : coefficient of friction between the ladder and the wall
θ : angle that makes the ladder with the floor
sinθ = h/L = 9.3 m/12 m
θ =sin⁻¹( 9.3 / 12)
θ = 50.8°
Forces acting on the ladder
W₁ =m*g = 45 kg* 9.8 m/s² = 441 N: Weight of the ladder (vertical downward)
W₂ =M*g = 72 kg * 9.8 m/s² = 705.6 N : Weight of the fire fighter(vertical downward)
FN₁ :Normal force that the floor exerts on the ladder (vertical upward) (point A)
fs : friction force that the floor exerts on the ladder (horizontal and opposite the movement )(point A)
FN₂ : Normal Force that the wall exerts on the ladder ( horizontal and opposite to friction force between the floor and the ladder)
∑Fy=0
FN₁ -W₁ -W₂= 0
FN₁ = W₁ + W₂
FN₁ = 441N+ 705.6N
FN₁ = 1146.6N : Force exerted on the ladder by the wall (vertical and upward)
Calculation of the distances of the forces at the point A (contact point of the ladder on the floor)
d₁ = 4*cos 50.8° (m) = 2.53 m: Distance from W₁ to the point A
d₂ =6*cos 50.8° (m)= 3.79 m : Distance from W₂ to the point A
d₃ = 9.3 m : Distance from FN₂ to the point A
The equilibrium equation of the moments at the point A (contact point of the ladder with the floor)
∑MA = 0
FN₂(d₃) - W₁( d₁) - W₂(d₂) = 0
FN₂(d₃) = W₁(d₁) + W₂(d₂)
FN₂(9.3) = (441 )(2.53) + (705.6)( 3.79 )
FN₂(9.3) = 1115.73 + 2674.2
FN₂ = (3790) / (9.3)
FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder
Force excreted by floor on ladder is 1146.6 N (vertically upward) and force excreted by wall on ladder is 407.5 N (horizontally towards the normal force).
What is force?Force is the effect of pull or push due to which the object having a mass changes its velocity.
The force is of two types-
Push-When the force applied in the direction of motion of the object, then the force is called the push force. Pull- When the force applied in the opposite direction of motion of the object, then the force is called the pull force.The length of the ladder is 12 meter and the mass of the ladder is 45 kg.Its upper end is a distance h of 9.3 and the mass of fire fighter is 72 kg.
The sine angle for the ladder can be given as,
[tex]\sin\theta=\dfrac{9.3}{12}\\\theta=50.8^o[/tex]
The forces acting on the system is normal force and force due to the weight of the two bodies.
The summation of the vertical forces is equal to the zero to keep it at rest. Therefore,
[tex]F_n-45\times9.8-72\times9.8\\F_n=1146.6 \rm N[/tex]
By the equilibrium of the momentum for the system,
[tex]F_{N2}\times(9.3)+45(9.8)\times(4\cos50.8)+72(9.8)\times(6\cos50.8)\\F_{N2}=407.5\rm N[/tex]
Thus, the force excreted by the floor on ladder is 1146.6 N (vertically upward) and the force excreted by the wall on ladder is 407.5 N (horizontally towards the normal force).
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During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.146 kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 15.3 m/s , shatters the glass as it passes through, and leaves the window at 10.7 m/s with no change of direction. What is the direction of the impulse that the glass imparts to the baseball?
Calculate the magnitude of this impulse (a positive number).
The ball is in contact with the glass for 0.0106 s as it passes through. Find the magnitude of the average force of the glass on the ball (a positive number).
Answer:
Impulse, |J| = 0.6716 kg-m/s
Force, F = 63.35 N
Explanation:
It is given that,
Mass of the baseball, m = 0.146 kg
Initial speed of the ball, u = 15.3 m/s
Final speed of the ball, v = 10.7 m/s
To find,
(a) The magnitude of this impulse.
(b) The magnitude of the average force of the glass on the ball.
Solution,
(a) Impulse of an object is equal to the change in its momentum. It is given by :
[tex]J=m(v-u)[/tex]
[tex]J=0.146\ kg(10.7-15.3)\ m/s[/tex]
J = -0.6716 kg-m/s
or
|J| = 0.6716 kg-m/s
(b) Another definition of impulse is given by the product of force and time of contact.
t = 0.0106 s
[tex]J=F\times \Delta t[/tex]
[tex]F=\dfrac{J}{\Delta t}[/tex]
[tex]F=\dfrac{0.6716\ kg-m/s}{0.0106\ s}[/tex]
F = 63.35 N
Hence, this is the required solution.
The impulse direction is opposite to the baseball's initial direction, and its magnitude is 0.6716 kg·m/s. The average force magnitude exerted by the window on the baseball is 63.36 N.
The direction of the impulse imparted by the window to the baseball is opposite to the baseball's initial direction of motion. This is because impulse is equal to the change in momentum, and since the window slows the ball down, it is applying a force in the opposite direction of the ball's initial velocity.
To calculate the magnitude of the impulse (I), use the formula
I = change in momentum = m(vf - vi)
where m is the mass of the baseball, vf is the final velocity, and vi is the initial velocity. The mass m = 0.146 kg, vi = 15.3 m/s, and vf = 10.7 m/s.
I = (0.146 kg)(10.7 m/s - 15.3 m/s)
I = (0.146 kg)(-4.6 m/s)
I = -0.6716 kg·m/s.
The negative sign indicates the impulse is in the opposite direction of the ball's initial motion, but since the question asks for the magnitude, we take the absolute value: 0.6716 kg·m/s.
The magnitude of the average force (Favg) exerted on the ball can be found using the formula
Favg = I/t
where t is the contact time. For a contact time of 0.0106 s, we have:
Favg = 0.6716 kg·m/s / 0.0106 s
Favg = 63.36 N.
You are driving at 75 km/h. Your sister follows in the car behind at 75 km/h. When you honk your horn, your sister hears a frequency You are driving at 75 . Your sister follows in the car behind at 75 . When you honk your horn, your sister hears a frequency lower than the frequency you hear. higher than the frequency you hear. the same as the frequency you hear. You cannot tell without knowing the horn's frequency.
The ideal concept for solving this question is based on the Doppler effect, for which it is indicated that the source's listening frequency changes as the distance and the relative speed between the receiver and the transmitter are also changed. However, if the relative velocity between the two objects is zero as in the particular case presented (since both travel at 75km / h) we have that there will be no change in frequency.
Therefore the frequency that I hear and that my sister would listen would be the same.
19.55 The 8-kg uniform bar AB is hinged at C and is attached at A to a spring of constant k 5 500 N/m. If end A is given a small displace- ment and released, determine
(a) the frequency of small oscillations,
(b) the smallest value of the spring constant k for which oscillations will occur.
Answer:
Explanation:
C is center of mass of the bar ie middle point. Spring is attached with one end ie A of the bar . When this end is displaced by distance x ( small) a restoring force kx is produced which creates a torque
Torque created = kx . l /2 , which creates angular acceleration α
moment of inertia I = ml²/12
torque = I. x α
kx . L /2 = I. x α
α = L kx / 2I
a = linear acceleration
= α x L/2
= L² kx / 4I
= L²x 12 kx / 4 mL²
a = 3kx/m
This shows that motion is SHM as acceleration is proportional to displacement x .
angular frequency ω² = 3k / m
frequency f = 1/2π √ 3k/m
= (1/6.28) x √ 3x 500/8
= 2.18 Hz
Final answer:
The frequency of small oscillations is approximately 3.98 Hz, and the smallest value of the spring constant for which oscillations will occur is 0 N/m.
Explanation:
(a) To determine the frequency of small oscillations, we can use the formula: f = (1/2π) × (√(k/m)), where f is the frequency, k is the spring constant, and m is the mass of the bar. Plugging in the values, we get:
f = (1/2π) × (√(500/8)) ≈ 3.98 Hz
(b) To find the smallest value of the spring constant k for which oscillations will occur, we need to consider critical damping. Critical damping occurs when the damping force is equal to the force exerted by the spring. The formula for critical damping is c = 2√(mk), where c is the damping coefficient, m is the mass, and k is the spring constant. Since we know that the damping coefficient is 0, we can solve for k to get:
k = (c^2)/(4m). Substituting in the values, we get:
k = (0^2)/(4 x 8) = 0 N/m
A horizontal beam of unpolarized light is incident on a stack of three polarizing filters with their polarization axes oriented, in sequence, 30◦, 60◦ and 90◦ clockwise from the vertical. The intensity of the light emerging from the stack is measured to be 275 W/m2. What is the intensity of the emerging light (in W/m2) if the middle polarizing filter is removed?
Answer:
122.22 W/m²
Explanation:
Let the intensity of unpolarized light is Io.
from first polariser
I' = Io/2
From second polariser
I'' = I' Cos²30 = 3 Io/8
From third polariser
I''' = I'' Cos²30 = 9Io/32
According to the question
9Io/32 = 275
Io = 977.78 watt/m²
Now, from first polariser
I' = Io/2 = 977.78 / 2 = 488.89 W/m²
I'' = 488.89 x cos²60 = 122.22 W/m²
thus, the intensity of light is 122.22 W/m².
Answer:
Explanation:
Given
Intensity of light emerging out is [tex]I=275 W/m^2[/tex]
Polarizer axis are inclined at [tex]30^{\circ}] , [tex]60^{\circ}[/tex] , [tex]90^{\circ}[/tex]
If [tex]I_0[/tex] is the Intensity of Incoming light then
[tex]275=\frac{I_0}{2}\times \cos ^2{30}\times \cos^2 {30}[/tex]
as they are inclined to [tex]30^{\circ}[/tex]to each other
[tex]I_0=\frac{275}{9}\times 32[/tex]
[tex]I_0=977.77 W/m^2[/tex]
If middle Filter is removed then
[tex]I=0.5\cdot I_0\cos ^2{60}[/tex]
[tex]I=0.5\cdot 977.77\cdot \frac{1}{4}[/tex]
[tex]I=122.22 W/m^2[/tex]
A grocery shopper tosses a(n) 9.1 kg bag of
rice into a stationary 19.3 kg grocery cart.
The bag hits the cart with a horizontal speed
of 6.6 m/s toward the front of the cart.
What is the final speed of the cart and bag?
Answer in units of m/s.
Answer:
2.1 m/s
Explanation:
Momentum is conserved, so:
m₁ u₁ + m₂ u₂ = (m₁ + m₂) v
(9.1 kg) (6.6 m/s) = (9.1 kg + 19.3 kg) v
v = 2.1 m/s
Suppose that, at t = 9.00 × 10 − 4 s, the space coordinates of a particle are x = 135 m, y = 30.0 m, and z = 55.0 m according to coordinate system S . If reference frame S ′ moves at speed 1.50 × 10 5 m/s in the + x - direction relative to frame S , compute the corresponding coordinate values as measured in frame S ′ . The reference frames start together, with their origins coincident at t = 0 .
Answer:
[tex]x'=134.999983\ m[/tex]
[tex]y'=y=30\ m[/tex]
[tex]z'=z=55\ m[/tex]
Explanation:
Given:
time, [tex]t=9\times 10^{-4}\ s[/tex]x coordinates of a particle, [tex]x=135\ m[/tex]y coordinates of a particle, [tex]y=30\ m[/tex]z coordinates of a particle, [tex]x=55\ m[/tex]Relative Speed of frame of reference S' in the +x direction, [tex]v_x=1.5\times 10^{5} m.s^{-1}[/tex]Since the speed of the frame S' is comparable to the speed of light in vacuum therefore the observer from S' frame will observe a contracted length of the dimensions in the direction of motion.Now from the equation of length contraction:
[tex]x'=x\times \sqrt{1-\frac{v_x^2}{c^2} }[/tex]
[tex]x'=135\times \sqrt{1-\frac{(1.5\times 10^{5})^2}{(3\times 10^8)^2} }[/tex]
[tex]x'=134.999983\ m[/tex]
Rest other values will remain unaffected since they are along the axis of motion. So,
[tex]y'=y=30\ m[/tex]
[tex]z'=z=55\ m[/tex]
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12 cm2 is rotated in 0.040 s from a position where its plane is perpendicular to the earth’s magnetic field to a position where its plane is parallel to the field. The earth’s magnetic field at the lab location is 6.0×10−5 T. (a) What is the magnetic flux through each turn of the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?
To solve this problem it is necessary to apply the concepts related to the magnetic flow of a coil and take into account the angles for each case.
It is also necessary to delve into part C, the concept of electromotive force (emf) which is defined as the variation of the magnetic flux as a function of time.
By definition the magnetic flux is determined as:
[tex]\phi = NBA cos\theta[/tex]
Where
N = Number of loops [tex]\rightarrow[/tex] We will calculate the value for each of the spins
B = Magnetic Field
A = Cross-sectional Area
[tex]\theta =[/tex] Angle between the perpendicular cross-sectional area and the magnetic field.
PART A) The magnetic flux through the coil after it is rotated is as follows:
[tex]\phi_i = NBA cos\theta[/tex]
[tex]\phi_i = (1turns)(6*10^{-5}T)(12*10^{-4}m^2)cos(0)[/tex]
[tex]\phi_i = 7.2*10^{-8}T\cdot m^2[/tex]
PART B) For the second case the angle formed is perpendicular therefore:
[tex]\phi_f = NBA cos\theta[/tex]
[tex]\phi_f = (1turns)(6*10^{-5}T)(12*10^{-4}m^2)cos(90)[/tex]
[tex]\phi_f = 0[/tex]
PART C) The average induced emf of the coil is as follows:
[tex]\epsilon = - (\frac{\phi_f-\phi_i}{dt})[/tex]
[tex]\epsilon = -(\frac{0-7.2*10^{-8}}{0.04})[/tex]
[tex]\epsilon = 1.8*10^{-6}V[/tex]
A laser beam of wavelength λ=632.8 nm shines at normal incidence on the reflective side of a compact disc. The tracks of tiny pits in which information is coded onto the CD are 1.60 μm apart.For what nonzero angles of reflection (measured from the normal) will the intensity of light be maximum? Express your answer numerically. If there is more than one answer, enter each answer separated by a comma.
To solve this problem we will apply the concepts given by the principles of superposition, specifically those described by Bragg's law in constructive interference.
Mathematically this relationship is given as
[tex]dsin\theta = n\lambda[/tex]
Where,
d = Distance between slits
[tex]\lambda[/tex] = Wavelength
n = Any integer which represent the number of repetition of the spectrum
[tex]\theta = sin^{-1} (\frac{n\lambda}{d})[/tex]
Calculating the value for n, we have
n = 1
[tex]\theta_1 = sin^{-1} (\frac{\lambda}{d})\\\theta_1 = sin^{-1} (\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_1 = 23.3\°[/tex]
n=2
[tex]\theta_2 = sin^{-1} (\frac{2\lambda}{d})\\\theta_2 = sin^{-1} (2\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_2 = 52.28\°[/tex]
n =3
[tex]\theta_2 = sin^{-1} (\frac{2\lambda}{d})\\\theta_2 = sin^{-1} (3\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_2 = \text{not possible}[/tex]
Therefore the intensity of light be maximum for angles 23.3° and 52.28°
When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b) potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy? (Give the ratio of the answer to the amplitude)
Answer:
Explanation:
Given
Displacement is [tex]\frac{1}{3}[/tex] of Amplitude
i.e. [tex]x=\frac{A}{3}[/tex] , where A is maximum amplitude
Potential Energy is given by
[tex]U=\frac{1}{2}kx^2[/tex]
[tex]U=\frac{1}{2}k(\frac{A}{3})^2[/tex]
[tex]U=\frac{1}{18}kA^2[/tex]
Total Energy of SHM is given by
[tex]T.E.=\frac{1}{2}kA^2[/tex]
Total Energy=kinetic Energy+Potential Energy
[tex]K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2[/tex]
[tex]K.E.=\frac{8}{18}kA^2[/tex]
Potential Energy is [tex]\frac{1}{8}[/tex] th of Total Energy
Kinetic Energy is [tex]\frac{8}{9}[/tex] of Total Energy
(c)Kinetic Energy is [tex]0.5\times \frac{1}{2}kA^2[/tex]
[tex]P.E.=\frac{1}{4}kA^2[/tex]
[tex]\frac{1}{2}kx^2=\frac{1}{4}kA^2[/tex]
[tex]x=\frac{A}{\sqrt{2}}[/tex]
Answer:
Explanation:
Let the amplitude is A.
Displacement, x = one third of the amplitude = A/3
The total energy of the body executing Simple Harmonic Motion is given by
[tex]T=\frac{1}{2}KA^{2}[/tex]
(a) Kinetic energy of the particle executing SHM is given by
[tex]K=\frac{1}{2}K\left (A^{2} -x^{2} \right )[/tex]
[tex]K=\frac{1}{2}K\left (A^{2} -\frac{A}{9}^{2} \right )[/tex]
[tex]K=\frac{1}{2}\times \frac{8A^{2}}{9}[/tex]
So, the ratio of kinetic energy to the total energy is given by
K / T = 8 / 9
(b) Potential energy of a particle executing SHM is given by
[tex]U=\frac{1}{2}Kx^{2}[/tex]
[tex]T=\frac{1}{2}\times \frac{A^{2}}{9}[/tex]
So, the ratio of potential energy to the total energy is given by
U / T = 1 / 9
(c) Let at a displacement y the kinetic energy is equal to the potential energy
[tex]\frac{1}{2}\times K\times \left ( A^{2}-y^{2} \right )=\frac{1}{2}\times K\times y^{2}[/tex]
[tex]y=\frac{A}{\sqrt{2}}[/tex]
Suppose F is a radial force field, S1 is a sphere of radius 4 centered at the origin, and the flux integral ∫∫S1F⋅dS=8.Let S2 be a sphere of radius 16 centered at the origin, and consider the flux integral ∫∫S2F⋅dS.(A) If the magnitude of F is inversely proportional to the square of the distance from the origin,what is the value of ∫∫S2F⋅dS?(B) If the magnitude of F is inversely proportional to the cube of the distance from the origin, what is the value of ∫∫S2F⋅dS?
A. [tex]\( \iint_{S_2} F \cdot dS = 128 \)[/tex]
B. [tex]\( \iint_{S_2} F \cdot dS = 128 \)[/tex]
(A) If the magnitude of (F) is inversely proportional to the square of the distance from the origin:
We know that the flux integral over [tex]\( S_1 \)[/tex] is [tex]\( \iint_{S_1} F \cdot dS = 8 \)[/tex].
Since ( F ) is a radial force field, (F) and (dS) are parallel, and the dot product simplifies to [tex]\( |F| \cdot |dS| \)[/tex].
The surface area of a sphere is [tex]\( 4\pi r^2 \)[/tex], so the magnitude of [tex]\( dS \)[/tex] on [tex]\( S_1 \)[/tex] is [tex]\( |dS| = 16\pi \)[/tex].
Therefore, [tex]\( |F| \cdot |dS| = 8 \implies |F| \cdot 16\pi = 8 \)[/tex], which gives [tex]\( |F| = \frac{1}{2\pi} \)[/tex].
For [tex]\( S_2 \)[/tex], with radius 16, [tex]\( |dS| = 256\pi \)[/tex].
So, [tex]\( \iint_{S_2} F \cdot dS = |F| \cdot 256\pi = \frac{1}{2\pi} \cdot 256\pi = 128 \)[/tex].
(B) If the magnitude of (F) is inversely proportional to the cube of the distance from the origin:
For [tex]\( S_1 \)[/tex], [tex]\( |dS| = 16\pi \)[/tex] as before. So, [tex]\( |F| \cdot |dS| = 8 \implies |F| = \frac{1}{2} \)[/tex].
For [tex]\( S_2 \)[/tex], [tex]\( |dS| = 256\pi \)[/tex] as before.
So, [tex]\( \iint_{S_2} F \cdot dS = |F| \cdot 256\pi = \frac{1}{2} \cdot 256\pi = 128 \)[/tex].
When a battery, a resistor, a switch, and an inductor form a circuit and the switch is closed, the inductor acts to oppose the change in the current.
How is the time constant of the circuit affected by doubling the resistance in the circuit?
Answer:Time constant gets doubled
Explanation:
Given
L-R circuit is given and suppose R and L is the resistance and inductance of the circuit then current is given by
[tex]i=i_0\left [ 1-e^{-\frac{t}{\tau }}\right ][/tex]
where [tex]i_0[/tex] is maximum current
i=current at any time
[tex]\tau =\frac{L}{R}=time\ constant[/tex]
[tex]\tau '=\frac{2L}{R}=2\tau [/tex]
thus if inductance is doubled then time constant also gets doubled or twice to its original value.
As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.
In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.
The horizontal component of this force is given as
F_x = Fcos(6.7)
While the vertical component of this force would be
F_y = Fsin(6.7)
In the vertical component, the sum of Force indicates that:
[tex]\sum F_y= 0 [/tex]
The Normal Force would therefore be equivalent to the weight and vertical component of the applied force, therefore:
[tex]N = mg+Fsin(6.7)[/tex]
In the horizontal component we have that the Force of tension in its horizontal component is equivalent to the Force of friction:
[tex]\sum F_x = 0[/tex]
[tex]F_x = F_{friction}[/tex]
[tex]Fcos (6.7) = N\mu[/tex]
Using the previously found expression of the Normal Force and replacing it we have to,
[tex]Fcos(6.7)= \mu (mg+Fsin(6.7))[/tex]
Replacing,
[tex]Fcos(6.7)= (0.87) (mg+Fsin(6.7))[/tex]
[tex]Fcos(6.7) = (0.87)(mg) + (0.87)(Fsin(6.7))[/tex]
[tex]Fcos(6.7) -(0.87)(Fsin(6.7)) = 0.87 (mg)[/tex]
[tex]F(cos(6.7)-0.87sin(6.7)) = 0.87 (mg)[/tex]
[tex]F = \frac{0.87 (mg)}{(cos(6.7)-0.87sin(6.7))}[/tex]
[tex]F = \frac{0.87(128000*9.8)}{(cos(6.7)-0.87sin(6.7))}[/tex]
[tex]F = 1.95*10^6N[/tex]
Finally the acceleration would be by Newton's second law:
[tex]F = ma[/tex]
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{ 1.95*10^6}{128000}[/tex]
[tex]a = 15.234m/s^2[/tex]
Therefore the greatest acceleration the man can give the airplane is [tex]15.234m/s^2[/tex]
the greatest acceleration the man can give the airplane is approximately 7.38 m/s²
The greatest acceleration the man can give the airplane when pulling at an angle of 6.7° above the horizontal is given by the equation:
[tex]\[ a = \frac{\mu (m_m + m_a) g - m_a g \sin(\theta)}{m_m + m_a \cos(\theta)} \][/tex]
where:
- [tex]\( \mu \)[/tex] is the coefficient of static friction between the man's shoes and the runway,
- [tex]\( m_m \)[/tex] is the mass of the man,
- [tex]\( m_a \)[/tex] is the mass of the airplane,
- g is the acceleration due to gravity,
- [tex]\( \theta \)[/tex] is the angle at which the man pulls the cable.
Given:
[tex]- \( \mu = 0.87 \),[/tex]
[tex]- \( m_m = 76 \) kg,[/tex]
[tex]- \( m_a = 128000 \) kg,[/tex]
[tex]- \( g = 9.8 \) m/s²,[/tex]
[tex]- \( \theta = 6.7° \).[/tex]
First, we need to convert the angle from degrees to radians because the sine and cosine functions in trigonometry typically use radians:
[tex]\[ \theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{radians}} = 6.7° \times \frac{\pi}{180} \][/tex]
Now, we can plug in the values into the equation:
[tex]\[ a = \frac{0.87 (76 \text{ kg} + 128000 \text{ kg}) \times 9.8 \text{ m/s}^2 - 128000 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(6.7°)}{76 \text{ kg} + 128000 \text{ kg} \times \cos(6.7°)} \][/tex]
Calculating the sine and cosine of 6.7°:
[tex]\[ \sin(6.7)\approx 0.117 \][/tex]
[tex]\[ \cos(6.7) \approx 0.993 \][/tex]
Now, we substitute these values into the equation:
[tex]\[ a = \frac{0.87 (76 + 128000) \times 9.8 - 128000 \times 9.8 \times 0.117}{76 + 128000 \times 0.993} \][/tex]
Solving the numerator:
[tex]\[ (0.87 \times 128076 \times 9.8) - (128000 \times 9.8 \times 0.117) \][/tex]
[tex]\[ = (0.87 \times 128076 \times 9.8) - (128000 \times 1.1466) \][/tex]
[tex]\[ = 939608.16 \][/tex]
Solving the denominator:
[tex]\[ 76 + (128000 \times 0.993) \][/tex]
[tex]\[ = 127308 \][/tex]
Finally, we divide the numerator by the denominator to find the acceleration:
[tex]\[ a = \frac{939608.16}{127308} \][/tex]
[tex]\[ a \approx 7.38 \text{ m/s}^2 \][/tex]
The answer is: 7.38.
Learning Goal: To practice Problem-Solving Strategy 40.1 for quantum mechanics problems. Suppose a particle of mass m is confined to a one-dimensional box of length L. We can model this as an infinite square well in which the particle's potential energy inside the box is zero and the potential energy outside is infinite. For a particle in its first excited state, what is the probability Prob(center20%) of finding the particle within the center 20% of the box?
Answer
The answer and procedures of the exercise are attached in the following archives.
Explanation
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The probability of finding a particle in its first excited state within the center 20% of an infinite square well of length L is approximately 11.8%. This involves integrating the square of the wavefunction over the specific interval. The key steps are defining the potential energy and solving the Schrödinger equation.
To find the probability of locating a particle in its first excited state within the center 20% of a one-dimensional box of length L, we follow these steps:
Define the Potential Energy, V: Inside the box (0 ≤ x ≤ L), V(x) = 0. Outside the box, V(x) = ∞.
Solve the Schrödinger Equation: The normalized wavefunction for the first excited state (n=2) is ψ2(x) = √(2/L) * sin(2πx/L).
The center 20% of the box is the interval from 0.4L to 0.6L. We calculate the probability of finding the particle in this region by integrating the square of the wavefunction:
Prob(center20%) = ∫0.4L0.6L |ψ2(x)|² dx = ∫0.4L0.6L ">2/L * sin²(2πx/L) dx.
Using integration techniques, the result is:
Prob(center20%) = 2 * [0.1 - (sin(0.4π))/(π)]. This computes to approximately 0.118 or 11.8%.
Ten narrow slits are equally spaced 2.00 mm apart and illuminated with green light of wavelength 511 nm. The width of bright fringes can be calculated as the separation between the two adjacent dark fringes on either side. Find the angular widths (in rad) of the third- and fifth-order bright fringes. third-order bright fringe rad fifth-order bright fringe rad
Answer:
The width of third and fifth order bright fringe is 0.00076 rad and 0.00127 rad.
Explanation:
Given that,
Distance d = 2.00 mm
Wavelength = 511 nm
Order number = 3
Order number = 5
We need to calculate the width of third-order bright fringe
Using formula of width
[tex]d\sin\theta=m\lambda[/tex]
[tex]\theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]
Put the value into the formula
[tex]\theta=\sin^{-1}\dfrac{3\times511\times10^{-9}}{2.00\times10^{-3}}[/tex]
[tex]\theta=0.00076\ rad[/tex]
We need to calculate the width of fifth-order bright fringe
Using formula of width
[tex]d\sin\theta=m\lambda[/tex]
[tex]\theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]
Put the value into the formula
[tex]\theta=\sin^{-1}\dfrac{5\times511\times10^{-9}}{2.00\times10^{-3}}[/tex]
[tex]\theta=0.00127\ rad[/tex]
Hence, The width of third and fifth order bright fringe is 0.00076 rad and 0.00127 rad.
A welder using a tank of volume 7.20 x 10^−2 m^3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.20 x 10^5 Pa and temperature of 39.0°C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 20.9 °C, the gauge pressure of the oxygen in the tank is 1.85 x 10^5 Pa.
Part A) Find the initial mass of oxygen.
Part B) Find the mass of oxygen that has leaked out.
Answer:
a) the initial mass of O₂ is 373.92 gr
b) the mass leaked of O₂ is 104.26 gr
Explanation:
we can assume ideal gas behaviour of oxygen , then we can calculate the mass using the ideal gas equation
P*V = n*R*T ,
where P= absolute pressure , V= volume occupied by the gas , n = number of moles of gas , R= ideal gas constant = 8.314 J/mol K , T= absolute temperature
Initially
P = Pg + Pa ( 1 atm) = 3.20 *10⁵ Pa + 101325 Pa = 4.21*10⁵ Pa
where Pg= gauge pressure , Pa=atmospheric pressure
T = 39 °C= 312 K
V= 7.20 * 10⁻² m³
therefore
P*V = n*R*T → n = P*V/ (R*T)
replacing values
n = P*V/ (R*T) = 4.21*10⁵ Pa*7.20 * 10⁻² m³/(8.314 J/mol K*312 K) = 11.685 mol
since
m= n*M , where m= mass , n= number of moles , M= molecular weight of oxygen
then
m = n*M = 11.685 mol *32.0 g/mol = 373.92 gr of O₂
therefore the initial mass of O₂ is 373.92 gr
for the part B)
P₂= Pg₂ + Pa ( 1 atm) = 1.85*10⁵ Pa + 101325 Pa = 2.86*10⁵ Pa
T₂ = 20.9 °C= 293.9 K
V= 7.20 * 10⁻² m³
therefore
n₂ = P₂*V/ (R*T₂) = 2.86*10⁵ Pa*7.20 * 10⁻² m³/(8.314 J/mol K*293.9 K) = 8.427 mol
m₂ = n*M = 8.427 mol*32.0 g/mol = 269.66 gr of O₂
thus the mass leaked of oxygen is
m leaked = m - m₂ = 373.92 gr - 269.66 gr = 104.26 gr
If you drink cold water, it will soon warm up to body temperature (about 37 C) This takes energy. Therefore, you ought to be able to keep your weight down even if you eat fattening food provided you also drink lots of cold water, right?Estimate what volume of cold water you would need to drink to overcome the effect of eating one chocolate sundae. (A chocolate sundae contains about 500 Cal. Recall that 1 Cal, the "calorie" normally used in the US to measure the energy content of food, is 1000 cal, or 1kcal.)
Answer:
18.4 L
Explanation:
A chocolate sundae contains 500 Cal. Considering that 1 Cal = 1 kcal and 1 kcal = 4.184 kJ, a chocolate sundae contains:
500 Cal × (4.184 kJ / 1 Cal) = 2.09 × 10³ kJ
Let's suppose we drink water at 10°C that increases its temperature to 37°C by absorbing this heat (Q). We can find the mass of water using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity of the water
m: mass
ΔT: change in the temperature
Q = c × m × ΔT
2.09 × 10³ kJ = (4.19 kJ/kg.°C) × m × (37°C - 10°C)
m = 18.4 kg
Considering the density of water to be of 1 kg/L, the volume of water is 18.4L.
Final answer:
To counteract the 500 kcal from a chocolate sundae, you would need to drink approximately 14.3 liters of ice-cold water. However, this is neither practical nor safe, and weight management should involve a balance of diet and exercise instead.
Explanation:
Drinking cold water does indeed require energy from the body to warm it up to body temperature, thus expending calories in the process.
To estimate the volume of cold water needed to negate the caloric intake of one chocolate sundae, we can use the given information that 250 mL of ice-cold water takes approximately 8,750 calories (or 8.75 kcal) to be heated up to body temperature.
If a chocolate sundae contains about 500 Cal (500 kcal), we would need to drink a significantly larger volume of cold water to burn those calories. Specifically, we can calculate this using a simple ratio:
500 kcal (chocolate sundae) / 8.75 kcal (250 mL water) = x / 250 mL
This gives us:
x = 250 mL × (500/8.75)
x = 14,285.71 mL
So, you would need to drink approximately 14.3 liters of ice-cold water to counteract the caloric intake from one chocolate sundae, which is not practical or safe.
A small mirror is attached to a vertical wall, and it hangs a distance of 1.86 m above the floor. The mirror is facing due east, and a ray of sunlight strikes the mirror early in the morning and then again later in the morning. The incident and reflected rays lie in a plane that is perpendicular to both the wall and the floor. Early in the morning, the reflected ray strikes the floor at a distance of 3.80 m from the base of the wall. Later on in the morning, the ray is observed to strike the floor at a distance of 1.22 m from the wall. The earth rotates at a rate of 15.0˚ per hour. How much time (in hours) has elapsed between the two observations?
Answer:
[tex]t=2.044\ hr[/tex]
Explanation:
From the schematic we can visualize the situation and the position of the rays falling on the floor.
Considering the given data from the lowest edge of the mirror.
We get a triangle with height of 1.86 meters.In the first instance the base of the triangle is 3.8 meters.While in the second instance the base is 1.22 meters.Speed of rotation of earth, [tex]\omega=15^{\circ}\ hr^{-1}[/tex]Now applying the trigonometric ratio to known sides in the first instance:
[tex]tan\ \theta_1=\frac{1.86}{3.8}[/tex]
[tex]\theta_1=26.08^{\circ}[/tex]
Applying the trigonometric ratio to known sides in the second instance:
[tex]tan\ \theta_2=\frac{1.86}{1.22}[/tex]
[tex]\theta_2=56.74^{\circ}[/tex]
Now by the law of reflection we know that the angle of incidence is equal to the angle of reflection. So the sun would have been at the same angle on the opposite side of the normal.
Hence the change in angle of the sun with respect to the mirror (also the earth)
[tex]\Delta \theta=\theta_2-\theta_1[/tex]
[tex]\Delta \theta=56.74-26.08[/tex]
[tex]\Delta \theta=30.66^{\circ}[/tex]
Now the time past for this change:
[tex]t=\frac{\Delta \theta}{\omega}[/tex]
[tex]t=\frac{30.66}{15}[/tex]
[tex]t=2.044\ hr[/tex]
How much is the moment due to force P about point B. P(unit=N) vector is equal to (150i+260j ) and vector BA (unit=meter) is equal to (0.23i+0.04j)?
Answer:
The moment (torque) is given by the following equation:
[tex]\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)[/tex]
Explanation:
The cross-product between the distance and the force can be calculated using the method of determinant. Since the z-components are zero, it is easy to calculate.
A He-Ne laser produces light of 633 nm wavelength, 1.5 mW power, with a cylindrical beam of 0.64 mm in diameter.
(a) What is the intensity of this laser beam?
(b) The damage threshold of the retina is 100 W/m2 . Is this laser safe to view head-on?
(c) What are the maximum values of the electric and magnetic fields?
(d) What is the average energy density in the laser beam?
Final answer:
The intensity of the laser beam can be calculated from the given power and beam area, which is needed to assess safety for retinal exposure. The max electric and magnetic fields depend on the intensity and constants from electromagnetic wave equations. The average energy density is the ratio of beam intensity to the speed of light.
Explanation:
Calculating Intensity and Assessing Laser Safety:
(a) To find the intensity of the laser beam, we use the formula: I = P / A, where P is the power and A is the area of the beam. The power is given as 1.5 mW, which is 1.5 x 10-3 W. The area, given the diameter of the beam is 0.64 mm, can be calculated using the area of a circle, πr2 (where r is the radius in meters). So, the intensity is:
I = (1.5 x 10-3 W) / (π x (0.64 x 10-3 m / 2)2) = Intensity Value in W/m2.
(b) To determine if the laser is safe for the retina, we compare its intensity to the damage threshold of the retina, which is 100 W/m2. If the calculated intensity is less than this value, then the laser is considered safe for direct viewing.
(c) The maximum values of the electric (E) and magnetic (B) fields can be found using the following relationships, which are derived from the electromagnetic wave equations: E = cB, where c is the speed of light, and B = √(2I/ε0c), where ε0 is the vacuum permittivity.
(d) The average energy density of the beam can be found using: u = E / c, where u is the energy density and E is the intensity of the beam.
A beam of electrons, each with the same kinetic energy, illuminates a pair of slits separated by a distance of 51 nm. The beam forms bright and dark fringes on a screen located a distance 2.6 m beyond the two slits. The arrangement is otherwise identical to that used in the optical two-slit interference experiment. The bright fringes are found to be separated by a distance of 0.4 mm. What is the kinetic energy of the electrons in the beam? Planck’s constant is 6.63 × 10−34 J · s. Answer in units of keV.
Answer:
K = 24.5 keV
Explanation:
The interference phenomenon is described by the equation
.d sin θ = m λ m = 1,2,3,…
The pattern is observed on a screen at a distance L = 2.6 m
tan θ = y / L
As these experiments the angle is very small we can approximate the tangent
tan θ = sin θ / cos θ
For small angles
tan θ = sin θ
Let's replace
d y / L = m λ
λ = d y / m L
Let's reduce the units to the SI system
d = 51 nm = 51 10⁻⁹ m
y = 0.4 mm = 0.4 10⁻³ m
Let's calculate the wavelength
Let's use m = 1 for the first interference line
λ = 51 10⁻⁹ 0.4 10⁻³ / 2.6
λ = 7.846 10⁻¹² m
Let's look for kinetic energy
K = ½ m v²
p = mv
K = ½ m p² / m
K = p² / 2m
Let's use the wave-particle duality relationship
p = h /λ
K = h² / 2m λ²
Let's calculate
K = (6.63 10⁻³⁴)² / (2 9.1 10⁻³¹ (7.846 10⁻¹²)²)
K = 3,923 10⁻¹⁵-15 J
K = 3.923 10⁻¹⁵ J ( 1 eV / 1.6 10⁻¹⁹ J) =2.452 10⁴ eV
K = 24.5 keV