Answer:
The correct answer is B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block.
Explanation:
Hi there!
The work done on each block is calculated as follows:
W = F · d
Since the two blocks were pushed the same distance with the same force, the work done on each object is the same.
Using the work-energy theorem, we know that the work done on an object is equal to its change in kinetic energy (KE):
W = ΔKE
W = final KE - initial KE
Since the objects are at rest, initial KE = 0, then:
W = final KE
Since the work done on each block is the same, so will be its final kinetic energy.
The correct answer is B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block.
Final answer:
When an equal force is applied over the same distance to two blocks of ice, with one being four times heavier than the other, the kinetic energy gained by both blocks is equal, because the work done on them is the same.
Explanation:
The question involves two blocks of ice on a frozen lake, with one block being four times as heavy as the other. When an equal force F is exerted on each block, pushing them the same distance d, we are asked which statement is true about the kinetic energy of the heavier block after the push compared to the kinetic energy of the lighter block. Since the work done (work = force × distance) on both blocks is the same and work done on an object is equal to the change in its kinetic energy (Work-Energy Principle), both blocks will have the same increase in kinetic energy. Therefore, without considering initial kinetic energies (since both start from rest), the correct answer is that it is equal to the kinetic energy of the lighter block (Option B). This is because the amount of work done on both blocks is the same, leading to an equal increase in kinetic energy.
While jogging, a 70.0-kg student generates thermal energy at a rate of 1200 W. To maintain a constant body temperature of 37.0∘C, this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, irreversible body damage could occur. Protein structures in the body are irreversibly damaged if body temperature rises to 44.0∘C or above. The specific heat of a typical human body is 3480J/(kg⋅K), slightly less than that of water. (The difference is due to the presence of protein, fat, and minerals, which have lower specific heat capacities.)
Answer:
1421 seconds or 23.67 minutes
Explanation:
Q = Heat = 1200 W
c = Specific heat of human body = 3480 J/kgK
[tex]\Delta T[/tex] = Change in temperature = (44-37)°C
t = Time taken
[tex]Q=\dfrac{mc\Delta T}{t}\\\Rightarrow t=\dfrac{mc\Delta T}{Q}\\\Rightarrow t=\dfrac{70\times 3480\times (44-37)}{1200}\\\Rightarrow t=1421\ s[/tex]
The time student jogs before irreversible body damage occurs is 1421 seconds or 23.67 minutes
The body maintains a stable temperature through thermoregulation, with mechanisms like sweating playing an important role in removing heat. During physical activity, energy production increases and this heat needs to be dissipated to maintain body temperature. Failure of these mechanisms can lead to dangerous overheating.
Explanation:The phenomenon mentioned in the question deals with the concept of thermoregulation, which is the process through which the body maintains its core temperature within certain boundaries. When a person exercises, the body produces more heat as a byproduct of the energy utilised during physical activity. This heat is then removed by perspiration and other mechanisms to ensure body temperature remains within a safe range.
The energy generated (in this case, 1200 W) can be converted into heat and then removed from the body to maintain a constant body temperature. Sweat, as part of the body's thermoregulation system, plays a crucial role in this process. As sweat (which is water) evaporates from the skin, it takes a significant amount of heat energy with it, effectively cooling the body.
If these cooling mechanisms were ineffective and the body could not release this heat, it could cause a dangerous increase in body temperature, potentially leading to irreversible damage to proteins in the body if the temperature reaches 44°C or above. That's why adequate fluid intake is essential to replace the liquid lost through sweat and to prevent dehydration.
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On a bet, you try to remove water from a glass by blowing across the top of a vertical straw immersed in the water. What is the minimum speed you must give the air at the top of the straw to draw water upward through a height of 1.6cm?
Answer:
v₂ = 0.56 m / s
Explanation:
This exercise can be done using Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Where points 1 and 2 are on the surface of the glass and the top of the straw
The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of the straw the velocity of the surface of the vessel is almost zero v₁ = 0
The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m
We substitute in the equation
[tex]P_{atm}[/tex] + ρ g y₁ = [tex]P_{atm}[/tex] + ½ ρ v₂² + ρ g y₂
½ v₂² = g (y₂-y₁)
v₂ = √ 2 g (y₂-y₁)
Let's calculate
v₂ = √ (2 9.8 1.6 10⁻²)
v₂ = 0.56 m / s
Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and place it in a circular low earth orbit--that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE = 6370 km.) Ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation.
To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.
Mathematically the conservation of these two energies can be given through
[tex]W = U_f - U_i[/tex]
Where,
W = Work
[tex]U_f =[/tex] Final gravitational Potential energy
[tex]U_i =[/tex] Initial gravitational Potential energy
When the spacecraft of mass m is on the surface of the earth then the energy possessed by it
[tex]U_i = \frac{-GMm}{R}[/tex]
Where
M = mass of earth
m = Mass of spacecraft
R = Radius of earth
Let the spacecraft is now in an orbit whose attitude is [tex]R_{orbit} \approx R[/tex] then the energy possessed by the spacecraft is
[tex]U_f = \frac{-GMm}{2R}[/tex]
Work needed to put it in orbit is the difference between the above two
[tex]W = U_f - U_i[/tex]
[tex]W = -GMm (\frac{1}{2R}-\frac{1}{R})[/tex]
Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is
[tex]W = \frac{GMm}{2R}[/tex]
A 4-kg block is sliding down a plane as pictured, with the plane forming a 3-4-5 right triangle. Initially, the block is not moving. (However, we will assume only kinetic friction applies.) The coefficient of static friction between the block and the plane is 0.25. How long does it take for the block to reach the bottom of the plane? (Pick the answer closest to the true value.)A. 3.2 secondsB. 1.6 secondsC. 2.5 secondsD. 1.0 secondsE. 0.5 seconds
Answer:b
Explanation:
Given
mass of block [tex]m=4 kg[/tex]
coefficient of static friction [tex]\mu =0.25 [/tex]
height of triangle is [tex]h=3 m[/tex]
[tex]F_{net}=mg\sin \theta -\mu _kmg\cos \theta [/tex]
[tex]a_{net}=g\sin \theta -\mu _kg\cos \theta [/tex]
[tex]a_{net}=9.8\sin 37-0.25\times 9.8\times \cos 37[/tex]
[tex]a_{net}=5.897-1.956=3.94 m/s^2[/tex]
here [tex]s=5 m[/tex]
[tex]v^2-u^2=2 a_{net}s[/tex]
[tex]v=\sqrt{2\times 3.94\times 5}[/tex]
[tex]v=6.27 m/s[/tex]
time taken to reach bottom of plane
[tex]v=u+at[/tex]
[tex]6.27=0+3.94\times t[/tex]
[tex]t=1.59 s\approx 1.6 s[/tex]
A student stands on the edge of a merry-go-round which rotates five times a minute and has a radius of two meters one evening as the sun is setting. The student produces a shadow on the nearby building. (a) Write an equation for the position of the shadow. (b) Write an equation for the velocity of the shadow.
To solve this problem it is necessary to apply the concepts related to the simple harmonic movement, to the speed in terms of displacement and the timpo, as well as the angular frequency and the period of frequency.
PART A) According to the description given, 5 revolutions are made in one minute (or 60 seconds) that is to say that the frequency would be given by
[tex]f = \frac{1}{12s^{-1}}[/tex]
Therefore the angular velocity can be found as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 0.52rad/s[/tex]
The displacement that determines the maximum displacement based on the angular velocity time and the time in the simple harmonic movement, is equal to the radius of the circle, in other words
[tex]x = Acos(\omega t)[/tex]
Where,
A = Amplitude
[tex]\omega =[/tex] Angular velocity
t = time
If for our given values the value of the amplitude is 2m and the value of the angular velocity is 0.52rad/s
[tex]x = 2cos(0.52t)[/tex]
So the equation for the position of the shadow is of
[tex]x(t) = 2cos(0.52t)[/tex]
PART B) The equation for the velocity of the shadow is calculated as a expression of the displacement against the time, if we differenciate the previous value found, we have that,
[tex]\frac{dx}{dt} = \frac{d(2cos(0.52t))}{dt}[/tex]
[tex]v(t) = 2(-0.52)sin(0.523t)[/tex]
[tex]v(t) = -1.05sin(0.52t)[/tex]
The position of the shadow can be described using the equation x=2cos(10πt+φ) and the velocity of the shadow can be described using the equation v=-2*10π*sin(10πt+φ), using principles of simple harmonic motion.
Explanation:The subject of this question involves physics and in particular, kinematics and rotational motion, as it relates to the shadow produced by a student standing on a rotating merry-go-round.
The position and velocity of the shadow will vary as the merry-go-round rotates, and these can be represented using mathematical equations. Let's denote the angular speed of the merry-go-round as 'w', which is given by the formula w = 2πn, where n is the number of rotations per minute. For the given question, n = 5 rotations per minute. Therefore, w = 10π rad/min.
(a) Position of the shadow: We can describe the position of the shadow in terms of simple harmonic motion - as the merry-go-round rotates, the shadow produced on the nearby building would be oscillating back and forth. This kind of motion can be mathematically represented as x=Acos(wt+φ), where x is the position of the shadow, A is the amplitude (which is the radius of the merry-go-round, hence 2 meters), w is the angular frequency, t is time, and φ is the phase constant.
(b) Velocity of the shadow: The velocity of the shadow is the derivative of the position with respect to time, which can be represented as v=-Aw*sin(wt+φ). In our case, the angular speed w = 10π rad/min, A is the amplitude (radius of the merry-go-round, hence 2 meters), and 't' is the time.
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We've seen that bees develop a positive charge as they fly through the air. When a bee lands on a flower, charge is transferred, and an opposite charge is induced in the earth below the flower. The flower and the ground together make a capacitor; a typical value is 0.60 pF. If a flower is charged to 30 V relative to the ground, a bee can reliably detect the added charge and then avoids the flower in favor of flowers that have not been recently visited Approximately how much charge must a bee transfer to the flower to create a 30 V potential difference?
To solve the problem it is necessary to apply the concept of Load on capacitors. The charge Q on the plates is proportional to the potential difference V across the two plates.
It can be mathematically defined as:
Q= CV
Where,
C = Capacitance
V = Voltage
Our values are given as,
[tex]C = 0.60pF\\V = 30V[/tex]
Substituting values in the above formula, we get
[tex]Q=CV\\Q = 0.6*30\\Q = 18pC[/tex]
Where
[tex]1pC = 10^{-12}Coulomb[/tex]
Therefore the charge must be 18pC to create a 30V Poential difference.
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 2.4 s on Earth. 0.142 Incorrect: Your answer is incorrect. Use the expression for the period of oscillation for a mass attached to a spring to find the mass of the object. kg (d) Find the mass suspended from this spring that would result in a period of 2.4 s on Mars. 0.142 Incorrect: Your answer is incorrect.
To solve this problem we must use the perioricity equations given as a function of the mass and spring constant. Mathematically this can be expressed as:
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
m = mass
k = Spring constant
Re-arrange to find the mass we have
[tex]m = \frac{T^2k}{4\pi^2 }[/tex]
Replacing with our values we have that
[tex]m = \frac{2.4^2*10}{4\pi^2}[/tex]
[tex]m = 1.459kg[/tex]
D) Mass is independent of acceleration due to gravity (as you can see at the equation previously given) for this reason the mass suspended on mars is given as the same found. Therefore the mass would be
m = 1.459kg
Final answer:
The mass suspended from the spring that would result in a period of 2.4 s is 0.142 kg.
Explanation:
To find the mass suspended from the spring that would result in a period of 2.4 s, we can use the formula for the period of oscillation:
T = 2π √(m/k)
Where T is the period, m is the mass, and k is the force constant of the spring.
Let's rearrange the formula to solve for m:
m = (T^2 · k) / (4π^2)
Substituting the given values:
m = (2.4^2 · 10) / (4π^2)
m = 0.142 kg
A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 = 4ft/s. Knowing that the tape does not slip on the drums, determine (a) the angular acceleration of drum B, and (b) the number of revolutions executed by drum B during the 4-s interval. 5) A rocket weighs 2600 lb, including 2200 lb of fuel, which is consumed at the rate of 25 lb/s and ejected with a relative velocity of 13000 ft/s. Knowing that the rocket is fired vertically from the ground, determine its acceleration (a) as it is fired, and (b) as the last particle of fuel is being consumed.
Answer:
Question 1)
a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.
[tex]v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2[/tex]
This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,
[tex]v = \omega R\\a = \alpha R[/tex]
where α is the angular acceleration.
In order to continue this question, the radius of the drums should be given.
Let us denote the radius of the drums as R, the angular acceleration of drum B is
α = 0.5/R.
b) The distance travelled by the drums can be found by the following kinematics formula:
[tex]v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft[/tex]
One revolution is equal to the circumference of the drum. So, total number of revolutions is
[tex]x / (2\pi R) = 6/(\pi R)[/tex]
Question 2)
a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:
[tex]a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2[/tex]
b) [tex]a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2[/tex]
The angular acceleration of drum B is 0.5 ft/s² and the number of revolutions executed by drum B during the 4-s interval is 6 ft.
Explanation:To determine the angular acceleration of drum B, we can use the formula:
α = (v1 - v0) / t
Where α is the angular acceleration, v1 and v0 are the final and initial speeds respectively, and t is the time interval.
Substituting the given values, we have:
α = (4 ft/s - 2 ft/s) / 4 s = 0.5 ft/s²
To find the number of revolutions executed by drum B during the 4 s interval, we can use the formula:
θ = (ω0 + ω1) / 2 * t
Where θ is the angle in radians, ω0 and ω1 are the initial and final angular velocities respectively, and t is the time interval.
Since the tape does not slip on the drums, the angular velocity of drum B is the same as the linear velocity of the tape. Thus, ω0 = v0 and ω1 = v1.
Substituting the given values, we have:
θ = (2 ft/s + 4 ft/s) / 2 * 4 s = 6 ft
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Which of the following experiments could be used to determine the inertial mass of a block? A. Place the block on a rough horizontal surface. Lift one end of the surface up and measure the angle the surface makes with the horizontal at the moment the block begins to slide. B. A Drop the block from different heights and measure the time of fall from each height. C. Place the block on a rough horizontal surface. Give the block an initial velocity and then let it come to rest. Measure the initial velocity and the distance the block moves in coming to rest. D. Use a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force.
Answer:
D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force
Explanation:
For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.
The experiment that could be used to determine the inertial mass of a block is ; ( D ) Using a spring scale to exert a force on the block.Measure the acceleration of the block and the applied force.
Inertial mass is the mass of a body that poses an inertial resistance to the acceleration of a body when forces are applied to it. before you inertial mass can be determined, force must be applied to the body at rest .
The experiment that is suitable to determine the inertial mass is using a spring scale to apply a force on the block, then take measurement of the acceleration experienced by the block and amount of force applied.
Inertial mass = acceleration / net force applied
Hence we can conclude that The experiment that could be used to determine the inertial mass of a block is Using a spring scale to exert a force on the block.Measure the acceleration of the block and the applied force.
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In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way in which this is done is to seat them in a chair of known mass attached to a spring of known force constant and measure the period of the oscillations of this system. The 36.2kg chair alone oscillates with a period of 1.15s , and the period with the astronaut sitting in the chair is 2.20s
Part A: Find the force constant of the spring.
Part B: Find the mass of the astronaut.
Answer:
a. K = 1080.61 N/m
b. mₐ = 96.28 kg
Explanation:
T = 2π * √ m / K
T₁ = 2π * √m₁ / K , T₂ = 2π * √m₂ / K
m₁ = 36.2 kg , m₂ = m₁ + mₐ
T₁ = 1.15 s , T₂ = 2.2 s
Equal period to determine the force constant of the spring
a.
K = 4π²* m₁ / T₁²
K = 4π² * 36.2 kg / 1.15² s
K = 1080.61 N/m
So replacing and solve can find m₂
b.
m₂ = T₂² * m₁ / T₁² ⇒ m₂ = 2.20² s * 36.2 kg / 1.15 ² s
m₂ = 132.48
mₐ = m₂ - m₁
mₐ = ( 132.48 - 36.2 ) kg = 96.28 kg
A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.Part AFind the work done by the gas during the initial expansion.W initial = JPart BFind the heat added to the gas during the initial expansion.Q initial = JPart CFind internal-energy change of the gas during the initial expansion.ΔU initial = JPart DFind the work done during the final cooling;W final = J
Answer:
A. [tex]W=600\ J[/tex]
B. [tex]Q=2112\ J[/tex]
C. [tex]\Delta U=1512\ J[/tex]
D. [tex]W=0\ J[/tex]
Explanation:
Given:
no. of moles of oxygen in the cylinder, [tex]n=0.2[/tex]initial pressure in the cylinder, [tex]P_i=2\times 10^5\ Pa[/tex]initial temperature of the gas in the cylinder, [tex]T_i=360\ K[/tex]According to the question the final volume becomes twice of the initial volume.
Using ideal gas law:
[tex]P.V=n.R.T[/tex]
[tex]2\times 10^5\times V_i=0.2\times 8.314\times 360[/tex]
[tex]V_i=0.003\ m^3[/tex]
A.
Work done by the gas during the initial isobaric expansion:
[tex]W=P.dV[/tex]
[tex]W=P_i\times (V_f-V_i)[/tex]
[tex]W=2\times 10^5\times (0.006-0.003)[/tex]
[tex]W=600\ J[/tex]
C.
we have the specific heat capacity of oxygen at constant pressure as:
[tex]c_v=21\ J.mol^{-1}.K^{-1}[/tex]
Now we apply Charles Law:
[tex]\frac{V_i}{T_i} =\frac{V_f}{T_f}[/tex]
[tex]\frac{0.003}{360} =\frac{0.006}{T_f}[/tex]
[tex]T_f=720\ K[/tex]
Now change in internal energy:
[tex]\Delta U=n.c_p.(T_f-T_i)[/tex]
[tex]\Delta U=0.2\times 21\times (720-360)[/tex]
[tex]\Delta U=1512\ J[/tex]
B.
Now heat added to the system:
[tex]Q=W+\Delta U[/tex]
[tex]Q=600+1512[/tex]
[tex]Q=2112\ J[/tex]
D.
Since during final cooling the process is isochoric (i.e. the volume does not changes). So,
[tex]W=0\ J[/tex]
Wire resistor A has twice the length and twice the cross sectional area of wire resistor B. Which of the following accurately compares the resistances of wire resistors A and B?a) Wire A has twice the resistance of wire B.b) Wire A has half the resistance of wire B.c) Wire A has the same resistance as wire B.d) None of the above
Answer: Option (c) is the correct answer.
Explanation:
It is known that the relation between resistance, length and cross-sectional area is as follows.
R = [tex]\rho \frac{l}{A}[/tex]
Let the resistance of resistor A is denoted by R and the resistance of resistor B is denoted by R'.
Hence, for resistor A the expression for resistance according to the given data is as follows.
R = [tex]\rho \frac{2l}{2A}[/tex]
On cancelling the common terms we get the expression as follows.
R = [tex]\rho \frac{l}{A}[/tex]
Now, the resistance for resistor B is as follows.
R' = [tex]\rho \frac{l'}{A'}[/tex]
Thus, we can conclude that the statement, Wire A has the same resistance as wire B, accurately compares the resistances of wire resistors A and B.
A glass tea kettle containing 500 g of water is on the stove. The portion of the tea kettle that is in contact with the heating element has an area of 0.090 m2 and is 1.5 mm thick.
At a certain moment, the temperature of the water is 75°C, and it is rising at the rate of 3°C per minute.
What is the temperature of the outside surface of the bottom of the tea kettle?
Neglect the heat capacity of the kettle, and assume that the inner surface of the kettle is at the same temperature as the water inside.
Answer:
77.08 C
Explanation:
[tex]m[/tex] = mass of the water = 500 g = 0.5 kg
[tex]c[/tex] = specific heat of water = 4186 J/(kg °C)
[tex]\Delta T[/tex] = Rate of change of temperature = 3 °C /min = (3/60 ) °C /s = 0.05 °C /s
[tex]k[/tex] = thermal conductivity of glass = 0.84
[tex]A[/tex] = Area of the element = 0.090 m²
[tex]t[/tex] = thickness of the element = 1.5 mm = 0.0015 m
[tex]T_{i}[/tex] = Temperature inside = 75 °C
[tex]T_{o}[/tex] = Temperature outside = ?
Using conservation of energy
Heat gained by water = Heat transferred through glass
[tex]m c \Delta T = \frac{kA(T_{o} - T_{i})}{t} \\(0.5) (4186 (0.05) = \frac{(0.84)(0.090)(T_{o} - 75)}{0.0015} \\104.65 = (50.4)(T_{o} - 75)\\T_{o} = 77.08 C[/tex]
A rigid rod is rotating at a constant angular speed about an axis that is perpendicular to one end of the rod. A small ball is fixed to the rod at a distance that is 12 cm from the axis. The centripetal force acting on this ball is 1.7 N. What would be the centripetal force that acts on the ball if the ball were fixed to the rod at a distance of 33 cm from the axis?
Answer:
Fc₂ = 4.675 N
Explanation:
r₁ = 12 cm = 0.12 m
Fc= 1.7 N
Fc = mV²/r = m w² r ( ∴ V= r w )
Fc = m w² r ----------------------- (1)
As angular speed (w) and mass are constant.
let m * w² = k ------------------(2)
Put equation (2) in equation (1).
⇒ Fc = m w² r = k * r ( ∴ m w² = k)
⇒ Fc =k * r ----------------------( 3)
⇒ Fc₁ =k * r₁ = 1.7 N
⇒ k = 1.7/ r = 1.7 / 0.12 N ( as r₁=0.12 m )
⇒ k = 14.17
Centripetal force hen the rod is fixed at 33 cm from the axis:
From equation ( 3 )
Fc₂ = k * r₂
Fc₂ = 14.17 * 0.33 N ( ∴ r₂ = 0.33 m)
Fc₂ = 4.675 N
Consider the four quantum numbers of an electron in an atom, n, l, ml, and ms. The energy of an electron in an isolated atom depends on:______a. l, ml, and msonly.b. n only.c. n and l only.d. n, l, and mlonly.e. all four quantum numbers.
Answer:
The energy of an electron in an isolated atom depends on b. n only.
Explanation:
The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.
The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.
The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates [tex]E_n[/tex], and for the case of an hydrogen atom we have:
[tex]E_n=-\cfrac{13.6}{n^2}\, eV[/tex]
Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.
What force is required to push a block (mass m) up an inclined plane that makes an angle of θ with the horizon at a constant velocity, if the coefficient of friction between the plane and the block is μ? Group of answer choices
a. mg μ cosθ
b. mg (μ cosθ + sinθ)
c. mg (μ sinθ + cosθ)
d. mg (μ cosθ + μ sinθ)
e. mg (μ cosθ + m sinθ)
The force needed to push a block up an incline at a constant velocity is the sum of the gravitational force parallel to the incline and the frictional force, which is mg sin(θ) + μmg cos(θ). So the correct option is e.
Explanation:The force required to push a block of mass m up an inclined plane at an angle θ with the horizon at a constant velocity, given the coefficient of friction between the plane and the block is μ, can be determined by analyzing the forces acting on the block. We need to consider both the component of the block's weight parallel to the incline and the frictional force.
Since the block is moving at a constant velocity, the net force along the incline must be zero. This implies that the applied force must counteract the combined forces of gravity pulling the block down the incline and the frictional force opposing the motion. The gravitational component parallel to the incline is mg sin(θ) and the frictional force is μmg cos(θ). Therefore, the required force F is:
F = mg sin(θ) + μmg cos(θ)
This corresponds to answer choice e.
If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 55 MN/m , determine the displacement of end A when the 60-kN force is applied. Take E = 200 GPa.
The question pertains to calculating the displacement at end A of a steel rod with a spring attached when subject to a specific force, using concepts of mechanical engineering such as Hooke's law and elastic deformation.
Explanation:The student's question involves calculating the displacement of end A when a 60-kN force is applied to a 20-mm-diameter rod made of A-36 steel which has a spring attached with a stiffness k = 55 MN/m and taking into account that the modulus of elasticity E = 200 GPa. This problem is a straightforward application of Hooke's law combined with the elastic deformation formula (stress = force/area, strain = change in length/original length, and Hooke's law: F = kx for springs), which are part of mechanical engineering and physics topics on material strength and deformation.
To find the displacement, we should consider the deformation of the rod under the applied force and the compression of the spring separately. The rod's deformation can be found using the modulus of elasticity E and the cross-sectional area derived from the diameter, while the spring's compression is directly related to the force applied and the spring constant k.
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How much heat energy is required to vaporize a 1.0-g ice cube at 0°C? The heat of fusion of ice is 80 cal/g.
The heat of vaporization of water is 540 cal/g, and cwater = 1.00 cal/g⋅°C.
a. 620 cal
b. 720 cal
c. 820 cal
d. 1 kcal
The heat required to convert 1.0-g of ice cube at 0°C to vapor is 720 cal.
We know that when heat is supplied to a substance, change of state occurs as the object moves from a particular state of matter to another. The heat required to convert 1.0-g of ice cube at 0°C to vapor is obtained from;
H = mLfus + mcdT + mLvap
H = (1 g * 80 cal/g) + (1 g * 1.00 cal/g⋅°C * (100 - 0)) + (1 g * 540 cal/g)
H = 80 cal + 100 cal + 540 cal
H = 720 cal
The heat required to convert 1.0-g of ice cube at 0°C to vapor is 720 cal.
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The total amount of heat required to vaporize a 1-g ice cube at 0°C considering the stages of heat of fusion, heating up water, and heat of vaporization is 720 cal.
Explanation:To calculate the total amount of heat energy required to vaporize a 1.0-g ice cube at 0°C, we need to consider the heat added to turn the ice into water (heat of fusion), heat to raise the temperature of that water from 0°C to 100°C, and finally heat to turn the water into steam (heat of vaporization).
First, we use the heat of fusion of ice, which is 80 cal/g, thus the heat needed to turn 1g of ice into water at 0°C is 1g * 80 cal/g = 80 cal.
Secondly, to raise the water temperature from 0 to 100°C, we use cwater = 1.00 cal/g⋅°C, thus the required heat is 1g * 100°C * 1.00 cal/g⋅°C = 100 cal.
Finally, we use the heat of vaporization of water which is 540 cal/g, for 1g of water, the heat needed is 1g * 540 cal/g = 540 cal.
Adding all those up give us a total heat of 80 cal + 100 cal + 540 cal = 720 cal.
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A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm , y = 0, and then in another straight line from x = 40.0 mm , y = 0 to x = 40.0 mm , y = 55.0 mm . The wire is in an external uniform 0.300-T magnetic field in the +z direction, and the current through the wire is 5.10 A , directed from the origin into the wire.
A/Determine the magnitude of the magnetic force exerted by the external field on the wire.
B/Determine the direction of the magnetic force exerted by the external field on the wire measured clockwise from the positive x axis.
C/The wire is removed and replaced by one that runs directly from the origin to the locationx = 20.0mm , y = 35.0mm . If the current through this wire is also 4.90A , what is the magnitude of the magnetic force exerted by the external field on the wire?
D/The wire is removed and replaced by one that runs directly from the origin to the locationx = 20.0mm , y = 35.0mm . If the current through this wire is also 4.90A , what is the magnitude of the direction force exerted by the external field on the wire measured clockwise from the positive x axis.?
Answer:
0.1040512455 N
[tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]
0.05925 N
[tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]
Explanation:
I = Current
B = Magnetic field
Separation between end points is
[tex]l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm[/tex]
Effective force is given by
[tex]F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N[/tex]
The force is 0.1040512455 N
[tex]tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}[/tex]
The angle the force makes is given by
[tex]\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]
The direction is [tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]
[tex]F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N[/tex]
The force is 0.05925 N
[tex]tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}[/tex]
[tex]\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]
The direction is [tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]
A. Magnitude of the force: [tex]\( 0.104 \, \text{N} \)[/tex]
B. Direction of the force: [tex]\( 323.8^\circ \)[/tex]
C. Magnitude of the force: [tex]\( 0.0593 \, \text{N} \)[/tex]
D. Direction of the force: [tex]\( 330^\circ \)[/tex]
Part A: Determine the magnitude of the magnetic force exerted by the external field on the wire
Given: Length of wire segments: [tex]\( L_1 = 40.0 \, \text{mm} = 0.040 \, \text{m} \) and \( L_2 = 55.0 \, \text{mm} = 0.055 \, \text{m} \)[/tex]
- Magnetic field, [tex]\( B = 0.300 \, \text{T} \)[/tex]
- Current, [tex]\( I = 5.10 \, \text{A} \)[/tex]
The force on a current-carrying wire in a magnetic field is given by [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \).[/tex]
For the first segment along the x-axis:
- [tex]\( \mathbf{L}_1 = 0.040 \, \text{m} \, \hat{i} \)[/tex]
[tex]\( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]
[tex]\[ \mathbf{F}_1 = I (\mathbf{L}_1 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.040 \, \hat{i} \times 0.300 \, \hat{k}) \][/tex]
[tex]\[ \mathbf{F}_1 = 5.10 \times 0.040 \times 0.300 \, \hat{j} \][/tex]
[tex]\[ \mathbf{F}_1 = 0.0612 \, \text{N} \, \hat{j} \][/tex]
For the second segment along the y-axis:
[tex]- \( \mathbf{L}_2 = 0.055 \, \text{m} \, \hat{j} \)[/tex]
[tex]- \( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]
[tex]\[ \mathbf{F}_2 = I (\mathbf{L}_2 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.055 \, \hat{j} \times 0.300 \, \hat{k}) \][/tex]
[tex]\[ \mathbf{F}_2 = 5.10 \times 0.055 \times (-0.300) \, \hat{i} \][/tex]
[tex]\[ \mathbf{F}_2 = -0.08415 \, \text{N} \, \hat{i} \][/tex]
The net force [tex]\(\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2\)[/tex]:
[tex]\[ \mathbf{F} = -0.08415 \, \hat{i} + 0.0612 \, \hat{j} \][/tex]
The magnitude of the force is:
[tex]\[ |\mathbf{F}| = \sqrt{(-0.08415)^2 + (0.0612)^2} \][/tex]
[tex]\[ |\mathbf{F}| = \sqrt{0.00708 + 0.00375} \][/tex]
[tex]\[ |\mathbf{F}| = \sqrt{0.01083} \][/tex]
[tex]\[ |\mathbf{F}| \approx 0.104 \, \text{N} \][/tex]
Part B: The direction of the force can be found using the angle [tex]\( \theta \)[/tex] with respect to the positive x-axis:
[tex]\[ \theta = \tan^{-1} \left( \frac{F_y}{F_x} \right) \][/tex]
[tex]\[ \theta = \tan^{-1} \left( \frac{0.0612}{-0.08415} \right) \][/tex]
[tex]\[ \theta = \tan^{-1} \left( -0.727 \right) \][/tex]
[tex]\[ \theta \approx -36.2^\circ \][/tex]
Since this angle is measured counterclockwise from the positive x-axis, the direction clockwise is:
[tex]\[ 360^\circ - 36.2^\circ = 323.8^\circ \][/tex]
Part C: Determine the magnitude of the magnetic force exerted by the external field on the wire
For the wire directly from the origin to [tex]\( x = 20.0 \, \text{mm}, y = 35.0 \, \text{mm} \):[/tex]
- Length [tex]\( L = \sqrt{(20.0 \, \text{mm})^2 + (35.0 \, \text{mm})^2} = \sqrt{(0.020 \, \text{m})^2 + (0.035 \, \text{m})^2} \)[/tex]
- Length [tex]\( L = \sqrt{0.0004 + 0.001225} = \sqrt{0.001625} \)[/tex]- Length [tex]\( L \approx 0.0403 \, \text{m} \)[/tex]
[tex]\[ \mathbf{L} = 0.0403 \, \text{m} \][/tex]
Given current [tex]\( I = 4.90 \, \text{A} \)[/tex]:
[tex]\[ \mathbf{F} = I L B \sin(\theta) \][/tex]
[tex]\[ \theta = \angle \text{between } \mathbf{L} \text{ and } \mathbf{B} = 90^\circ \][/tex]
[tex]\[ \mathbf{F} = 4.90 \times 0.0403 \times 0.300 \][/tex]
[tex]\[ \mathbf{F} \approx 0.0593 \, \text{N} \][/tex]
Part D: - Current: [tex]\( 4.90 \, \text{A} \)[/tex]
- Magnetic field: [tex]\( 0.300 \, \text{T} \, \hat{k} \)[/tex]
- Displacement vector: [tex]\( \mathbf{L} = 0.020 \, \hat{i} + 0.035 \, \hat{j} \)[/tex]
- Force vector: [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \)[/tex]
[tex]\[ \mathbf{L} \times \mathbf{B} = (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]
[tex]\[ \mathbf{F} = 4.90 (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]
[tex]\[ \mathbf{F} = 4.90 \times (0.020 \times 0.300 \, \hat{j} - 0.035 \times 0.300 \, \hat{i}) \][/tex]
[tex]\[ \mathbf{F} = 4.90 \times (0.006 \, \hat{j} - 0.0105 \, \hat{i}) \][/tex]
[tex]\[ \mathbf{F} = 0.0294 \, \hat{j} - 0.05145 \, \hat{i} \][/tex]
The magnitude of the force:
[tex]\[ |\mathbf{F}| = \sqrt{(0.0294)^2 + (-0.05145)^2} \approx 0.0593 \, \text{N} \][/tex]
The angle with respect to the x-axis:
[tex]\[ \theta = \tan^{-1} \left( \frac{0.0294}{-0.05145} \right) = \tan^{-1} \left( -0.571 \right) \approx -30^\circ \][/tex]
The direction clockwise is:
[tex]\[ 360^\circ - 30^\circ = 330^\circ \][/tex]
A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put on the launcher, the launcher spring compresses by 10.0 cm. The watermelon is then pushed down by an additional 30.0 cm and it’s ready to go. Just before the launch, how much energy is stored in the spring?
To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.
Since the forces are balanced the Spring force is equal to the force of the weight that is
[tex]F_s = F_g[/tex]
[tex]kx = mg[/tex]
Where,
k = Spring constant
x = Displacement
m = Mass
g = Gravitational Acceleration
Re-arrange to find the spring constant
[tex]k = \frac{mg}{x}[/tex]
[tex]k = \frac{8*9.8}{0.1}[/tex]
[tex]k = 784N/m[/tex]
Just before launch the compression is 40cm, then from Potential Elastic Energy definition
[tex]PE = \frac{1}{2} kx^2[/tex]
[tex]PE =\frac{1}{2} 784*0.4^2[/tex]
[tex]PE = 63.72J[/tex]
Therefore the energy stored in the spring is 63.72J
Manufacturers of headache remedies routinely claim that their own brands are more potent pain relievers than the competing brands. Their way of making the comparison is to compare the number of molecules in the standard dosage. Tylenol uses 325 mg of acetaminophen (C8H9NO2) as the standard dose, while Advil uses 2.00 x 102 mg of ibuprofen (C13H18O2). Find the number of molecules of pain reliever in the standard doses of
(a) Tylenol and
(b) Advil.
a) 12.95*10^20 molecules
b) 5.84*10^20 molecules
Explanation:
a)Tylenol
Mass of Tylenol = 325 mg
(m)=325*10^-3 g
Molecular weight of Tylenol(M) = 151.16256 g/mol
the number of moles present
n = m/M
= 325*10^-3 g /151.1656 g/mol
= 2.15 *10^-3 mol
the number of molecules present
N= nNa
=(2.15*10^ -3 mol)(6.023*10^23 molecules/mol)
N=12.95*10 ^20 molecules
b)Advil
Mass of Advil = 200 mg
m=200*10^-3 g
Molecular weight of Advil(M) =206.28082 g/mol
the number of moles
n=m/M
=200*10^ -3 g/ 206.28082 g/mol
=0.96955*10^ -3 mol
the number of molecules resent
N=nNa
=(0.96955*10^ -3)(6.023*10^ 23 molecules/ mol)
N=5.84*10^20 molecules
Tylenol contains more molecules of active ingredient hence it is a more effective drug.
We have to find the number of molecules in each of the drugs. In the drug, tylenol, the molar mass of the compound C8H9NO2 is;
8(12) + 9(1) + 14 + 2(16) = 96 + 9 + 14 + 32 =151 g/mol
Number of moles = 325 × 10^-3 g/151 g/mol = 0.0022 moles
1 mole contains 6.02 × 10^23 molecules
0.0022 moles contains 0.0022 moles × 6.02 × 10^23 molecules /1 mole = 1.32 × 10^21 molecules
For Advil;
Molecular mass= 13(12) + 18(1) + 2(16) = 156 + 18 + 32 = 206 g/mol
Number of moles = 0.2 g/ 206 g/mol = 0.00097 moles
1 mole contains 6.02 × 10^23 molecules of advil
0.00097 moles contains 0.00097 moles × 6.02 × 10^23 molecules /1 mole = 5.8 × 10^20
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Long Snorkel: Inhalation of a breath occurs when the muscles surrounding the human lungs move to increase the volume of the lungs, thereby reducing the air pressure in the lungs. The difference between the reduced pressure in the lungs and outside atmospheric pressure causes a flow of air into the lungs. The maximum reduction in air pressure that muscles of the chest can produce in the lungs against the surrounding air pressure on the chest and body is about 3740 Pa. Consider the longest snorkel that a human can operate. The minimum pressure difference needed to take in a breath is about 188 Pa. Find the depth h that a person could swim to and still breathe with this snorkel
Answer:
h = 0.362 m
Explanation:
The pressure equation with depth is
P₂ = [tex]P_{atm}[/tex] +ρ g h
The gauge pressure is
P2 - [tex]P_{atm}[/tex] = ρ g h
This is the pressure that muscles can create
P₂ - [tex]P_{atm}[/tex]= 3740 Pa
But still the person needs a small pressure for the transfer of gases, so
P₂ - [tex]P_{atm}[/tex] = 3740 - 188 = 3552 Pa
This is the maximum pressure difference, where the person can still breathe,
Let's clear the height
h = 3552 / ρ g
h = 3552 / (1000 9.8)
h = 0.362 m
This is the maximum depth where the person can still breathe normally.
Final answer:
The maximum depth that a person could swim to and still breathe with the snorkel is approximately 1.9 cm.
Explanation:
To find the depth that a person could swim to and still breathe with the snorkel, we need to consider the pressure difference needed for inhalation. The maximum reduction in air pressure that the chest muscles can produce in the lungs is 3740 Pa. The minimum pressure difference needed for inhalation is 188 Pa. Therefore, the maximum depth h that a person could swim to and still breathe with the snorkel is determined by the difference in atmospheric pressure at the surface and the pressure at that depth.
Using the equation for gauge pressure, we can calculate the maximum depth:
Pgauge = ρgh
Where:
Pgauge is the gauge pressure
ρ is the density of the fluid (water)
g is the acceleration due to gravity
h is the depth
Given that the minimum pressure difference is 188 Pa, we can rearrange the equation to solve for h:
h = Pgauge / (ρg)
Using the values for the density of water (1000 kg/m³) and acceleration due to gravity (9.8 m/s²), we can calculate the maximum depth:
h = 188 Pa / (1000 kg/m³ * 9.8 m/s²) = 0.019 m = 1.9 cm
Therefore, a person could swim to a maximum depth of approximately 1.9 cm and still breathe with the snorkel.
A container of water, diameter 12 cm, has a small opening near the bottom that can be unplugged so that the water can run out. If the top of the tank is open to the atmosphere, what is the exit speed of the water leaving through the hole. The water level is 15 cm above the bottom of the container. The center of the 3.0 diameter hole is 4.0 cm from the bottom. How long does it take the water to hit the ground? How far from the container will the water land?
Answer:
1.32 m
Explanation:
from the given figure we can find the velocity at the hole in the container
let it be v and we know that
then [tex]v= \sqrt{2gh}[/tex]
h= 15-4= 11 cm , g=9.81
[tex]v= \sqrt{2(9.81)(11)}[/tex]
v=14.69 m/s
now using
s=ut+0.5at^2
t= is the time required by water to reach bottom of the container
u = velocity in vertical direction is zero.
therefore,
[tex]0.04= 0\times t+\frac{1}{2}\times9.81\times t^2[/tex]
t= 0.09030 sec
let x be the distance far from the container will the water land
x=vt
x=14.69×0.09030 = 1.32 m
A light ray enters a glass enclosed fish tank. From air it enters the glass at 20.° with respect to the surface, then emerges into the water. The index for glass is 1.50 and for water 1.33.
(a) What is the angle of refraction in the glass?
(b) What is the angle of refraction in the water?
(c) Is there any incident angle in air for which the ray will not enter the water due to total internal reflection?
Answer:
a) [tex]\angle r_{ag}=38.79^{\circ}[/tex]
b) [tex]\angle r_{gw}=44.95^{\circ}[/tex]
c) not possible
Explanation:
Given:
angle of incidence on the air-glass interface, [tex]\angle i_{ag}=90-20=70^{\circ}[/tex]
refractive index of glass with respect to air, [tex]n_g=1.5[/tex]
refractive index of water with respect to air, [tex]n_a=1.33[/tex]
a)
For angle of refraction in glass we use Snell's law:
[tex]n_g=\frac{sin\ i_{ag}}{sin\ r{ag}}[/tex]
[tex]1.5=\frac{sin\ 70}{sin\ r_{ag}}[/tex]
[tex]\angle r_{ag}=38.79^{\circ}[/tex]
b)
Now we have angle of incident for glass-water interface, [tex]\angle i_{gw}=\angle r_{ag}=38.79^{\circ}[/tex]
And the refractive index of water with respect to glass:
[tex]n_{gw}=\frac{n_w}{n_g}[/tex]
[tex]n_{gw}=\frac{1.33}{1.5}[/tex]
[tex]n_{gw}=0.8867 [/tex]
Therefore, angle of refraction in the water:
[tex]n_{gw}=\frac{sin\ i_{gw}}{sin\ r_{gw}}[/tex]
[tex]0.8867 =\frac{sin\ 38.79}{sin\ r_{gw}}[/tex]
[tex]\angle r_{gw}=44.95^{\circ}[/tex]
c)
For total internal reflection through water the light must enter the glass-water interface at an angle greater than the critical angle.
So,
[tex]n_{gw}=\frac{sin\ i_{(gw)_c}}{sin\ 90}[/tex]
[tex]0.8867 =\frac{sin\ i_{(gw)_c}}{1}[/tex]
[tex]i_{(gw)_c}=62.46^{\circ}[/tex]
Now this angle will become angle of refraction for the air-glass interface.
Hence,
[tex]n_g=\frac{sin\ i_{ag}}{sin\ i_{(gw)_c}}[/tex]
[tex]1.5=\frac{sin\ i_{ag}}{0.8867 }[/tex]
[tex]sin\ i_{ag}=1.33005[/tex]
Since we do not have any angle for sine for which the value exceeds 1, therefore it is not possible that the light will reflect from the water.
Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.
If the temperature difference is 20° C from the inside of the house to the outside air, what is the rate of heat flow through this window?
(Thermal conductivity for glass is 0.84 J/s⋅m⋅°C and for air 0.023 4 J/s⋅m⋅°C.)
a. 7 700 Wb. 1 900 Wc. 547 Wd. 180 W
To solve this problem it is necessary to apply the concepts related to rate of thermal conduction
[tex]\frac{Q}{t} = \frac{kA\Delta T}{d}[/tex]
The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, [tex]\Delta T[/tex], T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.
The change made between glass and air would be determined by:
[tex](\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}[/tex]
[tex] k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}[/tex]
[tex]\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}[/tex]
[tex]\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}[/tex]
[tex]\Delta T_{air} = 44.9 \Delta T_{glass}[/tex]
There are two layers of Glass and one layer of Air so the total temperature would be given as,
[tex]\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}[/tex]
[tex]\Delta T = 2\Delta T_{glass} +\Delta T_{air}[/tex]
[tex]20\°C = 46.9\Delta T_{glass}[/tex]
[tex]\Delta T_{glass} = 0.426\°C[/tex]
Finally the rate of heat flow through this windows is given as,
[tex]\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}[/tex]
[tex]\Delta {Q}{t} = 0.84*24*10 -3*0.426[/tex]
[tex]\Delta {Q}{t} = 179W[/tex]
Therefore the correct answer is D. 180W.
A woman of mass 44 kg jumps off the bow of
a 49 kg canoe that is intially at rest.
If her velocity is 2.5 m/s to the right, what
is the velocity of the canoe after she jumps?
Answer in units of m/sˆı.
Answer:
2.2 m/s to the left
Explanation:
Momentum is conserved, so:
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
0 = (44 kg) (2.5 m/s) + (49 kg) v
v = -2.2 m/s
The canoe will move in the opposite direction of the woman with a velocity of approximately 2.24 m/s.
The velocity of the canoe after the woman jumps off can be determined by using the principle of conservation of momentum, which states that the total momentum before an event must equal the total momentum after the event when no external forces act on the system. In this scenario, the system consists of the woman and the canoe.
Before the woman jumps, the system is at rest, so its initial momentum is zero. When the woman jumps off the canoe to the right with a velocity of 2.5 m/s, by conservation of momentum, the canoe must move in the opposite direction to maintain the total momentum at zero.
The momentum possessed by the woman is given by the product of her mass and velocity (mw×vw).
Similarly, the momentum of the canoe is the product of its mass and its velocity in the opposite direction (mc×vc).
Momenta are equal in magnitude and opposite in direction:
mw ×vw = mc ×vc
Therefore, vc = (mw ×vw) / mc
Substituting the given values:
vc = (44 kg ×2.5 m/s) / 49 kg
= 2.24 m/s
The canoe will move to the left (opposite to the woman's direction) with a velocity of approximately 2.24 m/s.
PART ONE
A jet aircraft is traveling at 281 m/s in horizontal flight. The engine takes in air at a
rate of 107 kg/s and burns fuel at a rate of
4.23 kg/s. The exhaust gases are ejected at
679 m/s relative to the aircraft.
Find the thrust of the jet engine.
Answer in units of N.
PART TWO
Find the delivered power.
Answer in units of W
Answer:
1. F = 45,458.17 N
2. P = 12,800,000 W
Explanation:
Part 1. The thrust force is the sum of the forces on the air and on the fuel.
For the air, 107 kg of air is accelerated from 281 m/s to 679 m/s in 1 second.
F = ma
F = (107 kg) (679 m/s − 281 m/s) / (1 s)
F = 42,586 N
For the fuel, 4.23 kg of fuel is accelerated from 0 m/s to 679 m/s in 1 second.
F = ma
F = (4.23 kg) (679 m/s − 0 m/s) / (1 s)
F = 2,872.17 N
So the thrust on the jet is:
F = 42,586 N + 2,872.17 N
F = 45,458.17 N
Rounded to three significant figures, the force is 45,500 N.
Part 2. Power = work / time, and work = force × distance, so:
Power = force × distance / time
Power = force × velocity
P = (45,458.17 N) (281 m/s)
P = 12,773,745.77 W
Rounded to three significant figures, the power is 12,800,000 W.
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is empty, aside from containing air. The bucket is then inverted so that its open end is down and, being careful not to lose any of the air trapped inside, the bucket is lowered below the surface of a fresh-water lake so the water-air interface in the bucket is 20.0 m below the surface of the lake.
To keep the calculations simple, use g = 10 m/s2, atmospheric pressure = 1. 105 Pa, and the density of water as 1000 kg/m3.a. If the temperature is the same at the surface of the lake and at a depth of 20.0 m below the surface, what is the height of the cylinder of air in the bucket when the bucket is at a depth of 20.0 m below the surface of the lake?b. If, instead, the temperature changes from 300 K at the surface of the lake to 275 K at a depth of 20.0 m below the surface of the lake, what is the height of the cylinder of air in the bucket?
Answer:
12 cm
Explanation:
[tex]P_1[/tex] = Initial pressure = [tex]P_a=1\times 10^5\ Pa[/tex]
[tex]P_2[/tex] = Final pressure = [tex]P_a+\rho_w gh[/tex]
h = Depth of cylinder = 36 cm
g = Acceleration due to gravity = 10 m/s²
[tex]\rho_w[/tex] = Density of water = 1000 kg/m³
[tex]h_1[/tex] = Depth of lake = 20 m
From the ideal gas relation we have
[tex]P_1V_1=P_2V_2\\\Rightarrow P_a(\pi r^2h)=(P_a+\rho_w gh_1)\pi r^2h'\\\Rightarrow 1\times 10^5\times 36=(1\times 10^5+1000\times 10\times 20)h'\\\Rightarrow h'=\dfrac{1\times 10^5\times 36}{1\times 10^5+1000\times 10\times 20}\\\Rightarrow h'=12\ cm[/tex]
The height of the cylinder of air in the bucket when the bucket is at the given depth is 12 cm
Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C.
The specific heat of the solid:
a. is equal to that of the fluid.
b. is less than that of the fluid.
c. is more than that of the fluid.
d. cannot be compared to that of a material in a different phase.
Answer:
c. is more than that of the fluid.
Explanation:
This problem is based on the conservation of energy and the concept of thermal equilibrium
[tex]heat= m s \Delta T [/tex]
m= mass
s= specific heat
\DeltaT=change in temperature
let s1= specific heat of solid and s2= specific heat of liquid
then
Heat lost by solid= [tex]20(s_1)(70-30)=800s_1 [/tex]
Heat gained by fluid=[tex]100(s_2)(30-20)=1000s_2 [/tex]
Now heat gained = heat lost
therefore,
1000 S_2=800 S_1
S_1=1.25 S_2
so the specific heat of solid is more than that of the fluid.
The correct option is:
(c) is more than that of the fluid.
Conservation of energy:
The conservation of energy suggests that the heat energy lost by the solid must be equal to the heat energy gained by the liquid.
Let the specific heat of the solid be [tex]s_s[/tex] and the specific heat of the liquid be [tex]s_l[/tex].
Heat energy lost by the solid is given by:
[tex]\Delta Q_s=ms_s\Delta T\\\\ \Delta Q_s=20s_s(70-30)\\\\ \Delta Q_s=800s_s[/tex]
Heat energy gained by the liquid:
[tex]\Delta Q_l=ms_l\Delta T\\\\ \Delta Q_l=100s_l(30-20)\\\\ \Delta Q_l=1000s_l[/tex]
According to the conservation of energy:
[tex]\Delta Q_s=\Delta Q_l\\\\800s_s=1000s_l\\\\\frac{s_s}{s_l}=\frac{1000}{800}\\\\\frac{s_s}{s_l}=\frac{5}{4}[/tex]
Hence, the specific heat of the solid is more than that of the fluid
Learn more about specific heat:
https://brainly.com/question/2516072?referrer=searchResults
A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of the surroundings). Given these two temperatures, what is the maximum possible efficiency the car can have? (Note that actual car engine efficiencies are in the 20-25% range.) _______ %.
One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.
From the temperature the value is given as
[tex]\eta = 1-\frac{T_L}{T_H}[/tex]
Where,
T_L = Cold focus temperature
T_H = Hot spot temperature
Our values are given as,
T_L = 20\° C = (20+273) K = 293 K
T_H = 440\° C = (440+273) K = 713 K
Replacing we have,
[tex]\eta = 1-\frac{T_L}{T_H}[/tex]
[tex]\eta = 1-\frac{293}{713}[/tex]
[tex]\eta = 0.589[/tex]
Therefore the maximum possible efficiency the car can have is 58.9%