Answer:
i) The average acceleration for the first time interval, from [tex]t_{0} = 0[/tex] to [tex]t_{1} =4\ s[/tex], is [tex]14 \frac{m}{s^{2} }[/tex] .
ii) The average acceleration for the second time interval, from [tex]t_{1} =4\ s[/tex] to [tex]t_{2} =10\ s[/tex], is zero.
iii) The instantaneous acceleration as a function of time a(t) between 0 s and 4 s is [tex]a(t)=-10\frac{m}{s^{2} } + 12\frac{m}{s^{3} } t[/tex] .
iv) The position as a function of time x(t) between 0 s and 4 s is [tex]x(t)=-5\frac{m}{s^{2} } t^{2} + 2\frac{m}{s^{3} } t^{3}[/tex] .
Explanation:
We are given as data that a car that is a rest, that means v₀=0, at the origin, x₀=0, accelerates in a straight line along the +x direction. We are told that from [tex]t_{0} = 0[/tex] to [tex]t_{1} =4\ s[/tex] the velocity is [tex]v(t)=-10\frac{m}{s^{2} } t + 6 \frac{m}{s^{3} } t^{2}[/tex] and that from [tex]t_{1} =4\ s[/tex] to [tex]t_{2} =10\ s[/tex] it maintains a constant velocity of [tex]v=56\frac{m}{s}[/tex] in the same direction.
i) The average acceleration is simply the rate of change of velocity, it can be expressed as:
[tex]\bar{a}=\frac{\vartriangle v}{\vartriangle t} =\frac{v_{f}-v_{0} }{t_{f}-t_{0}}[/tex]
we have to replace the values in this equation, the only value we need to calculate is [tex]v_{f}[/tex]
[tex]v(4 s)=-10\frac{m}{s^{2} } 4 s + 6 \frac{m}{s^{3} } (4 s)^{2}[/tex]
[tex]v(4 s)=-10\frac{m}{s^{2} } 4 s + 6 \frac{m}{s^{3} } 16 s^{2}[/tex]
[tex]v(4 s)= -40 \frac{m}{s} + 96 \frac{m}{s}= 56 \frac{m}{s}[/tex]
we get that [tex]v_{f}= 56 \frac{m}{s}[/tex] .
Then we pur this value in the expression of the average acceleration
[tex]\bar{a}=\frac{\vartriangle v}{\vartriangle t} =\frac{v_{f}-v_{0}}{t_{f}-t_{0}}=\frac{56\frac{m}{s}-0\frac{m}{s} }{4 s - 0 s}[/tex]
the average acceleration for the first time interval is
[tex]\bar{a}=14\frac{m}{s^{2} }[/tex]
ii) The average acceleration for the second time interval, from [tex]t_{1} =4\ s[/tex] to [tex]t_{2} =10\ s[/tex], is zero because the velocity is constant. The average acceleration is the rate of change of velocity, if this magnitude remains constant then it follows that the acceleration is zero.
iii) We calculate the instantaneous acceleration, which is the acceleration at a specific moment in time, as the derivative of the velocity function.
Mathematically [tex]a(t)=\frac{d v(t)}{d t }[/tex]
so we get
[tex]a(t)=-10 \frac{m}{s^{2} } + 12\frac{m}{s^{3} } t[/tex]
iv) To find the position in function of time x(t) we simply integrate the expression of v(t):
[tex]v(t)=\frac{d x(t)}{d t}= -10\frac{m}{s^{2} }t+ 6\frac{m}{s^{3} } t^{2}[/tex]
we get that
[tex]x(t)=-5\frac{m}{s^{2} } t^{2} + 2 \frac{m}{s^{3} } t^{2}[/tex]
A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(kg⋅K). The copper and water are initially at different temperatures. After a sufficiently long time, the copper and water come to a final equilibrium temperature. Part A Which of the following statements is correct concerning the temperature changes of both substances? (Ignore the signs of the temperature changes in your answer.) Which of the following statements is correct concerning the temperature changes of both substances? (Ignore the signs of the temperature changes in your answer.) The temperature change of the copper is equal to the temperature change of the water. The temperature change of the water is greater than the temperature change of the copper. The temperature change of the copper is greater than the temperature change of the water.
Answer:
The temperature change of the copper is greater than the temperature change of the water.
Explanation:
deltaQ = mc(deltaT)
Where,
delta T = change in the temperature
m =mass
c = heat capacity
[tex]\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}[/tex]
The temperature change in the copper is nearly 11 times the temperature change in the water.
So, the correct option is,
The temperature change of the copper is greater than the temperature change of the water.
Hope this helps!
The temperature change of the copper is equal to the temperature change of the water.
Explanation:The temperature change of the copper is equal to the temperature change of the water.
When two substances come into contact, heat is transferred between them until they reach thermal equilibrium, where their temperatures are equal. According to the Law of Conservation of Energy, the heat lost by one substance is equal to the heat gained by the other substance. In this case, since all heat transfer occurs between the copper and water, the temperature change of the copper is equal to the temperature change of the water.
Part A If the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity? If the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity? The accuracy in measuring its velocity also increases. The accuracy in measuring its velocity decreases. The accuracy in measuring its velocity remains the same. The accuracy in measuring its velocity becomes uncertain.
Answer:
The correct answer is the speed accuracy decreases
Explanation:
Before examining the final statements, let's review the uncertainty principle
Δx Δp> = h ’/ 2
h ’= h / 2π
This is because the process of measuring one quantity affects the measurement of the other.
Let's review the claims
The speed of the particle is proportional to the moment
p = mv
Therefore, if the position is measured more accurately (x) the accuracy of p must decrease
Δp = h ’/ 2 Δx
The correct answer is the speed accuracy decreases
According to the Heisenberg uncertainty principle in physics, if the accuracy in measuring a particle's position increases, the accuracy in measuring its velocity decreases. This is a fundamental characteristic of the quantum world due to the wave-particle duality of matter.
Explanation:The question refers to the Heisenberg uncertainty principle in physics, particularly quantum mechanics. This principle states that the more precisely the position of a particle is known, the less precisely its velocity (or momentum, to be more specific) can be known, and vice versa. Therefore, if the accuracy of measuring a particle's position increases, then the accuracy in measuring its velocity decreases.
This phenomenon isn't due to measurement techniques or technology. It's a fundamental limit defined by the nature of the quantum world. It derives from the wave-particle duality of matter, which means small particles like electrons behave both as particles and waves. If we measure the position very precisely, the particle acts more like a wave with an uncertain speed. Conversely, if we measure the speed very precisely, the particle acts more like a particle with an uncertain position.
For example, if we use an extremely short-wavelength electron probe to measure an electron's position, we'd be very accurate but massively disturb the electron's velocity in the process. Hence, increasing the precision in position measurement increases the uncertainty in velocity measurement.
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A counter attendant in a diner shoves a ketchup bottle with a mass 0.30 kg along a smooth, level lunch counter. The bottle leaves her hand with an initial velocity 2.8 m/s. As it slides, it slows down because of the horizontal friction force exerted on it by the countertop. The bottle slides a distance of 1.0 m before
coming to rest. What are the magnitude and direction of the friction force acting on it?
The magnitude of the friction force acting on the ketchup bottle is 0.84 N, and it is directed opposite to the initial velocity of the bottle.
Explanation:The friction force acting on the ketchup bottle can be determined using the equation:
Friction force = Mass x Acceleration
Since the bottle comes to rest, its final velocity is 0 m/s and the acceleration can be calculated using the equation:
Acceleration = (Final Velocity - Initial Velocity) / Time
By substituting the given values, we find that the friction force acting on the bottle is 0.30 kg x (-2.8 m/s) / (1.0 m/s^2) = -0.84 N. The negative sign indicates that the friction force is acting in the opposite direction of the initial velocity, which is in the direction of the attendant's hand.
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A water wave traveling in a straight line on a lake is described by the equation:y(x,t)=(2.75cm)cos(0.410rad/cm x+6.20rad/s t)Where y is the displacement perpendicular to the undisturbed surface of the lake. a. How much time does it take for one complete wave pattern to go past a fisherman in a boat at anchor, and what horizontal distance does the wave crest travel in that time? b. What are the wave number and the number of waves per second that pass the fisherman? c. How fast does a wave crest travel past the fisherman, and what is the maximum speed of his cork floater as the wave causes it to bob up and down?
Answer:
A) The wave equation is defined as
[tex]y(x,t) = A\cos(kx + \omega t)=0.0275\cos(0.0041x + 6.2t)\\[/tex]
Using the wave equation we can deduce the wave number and the angular velocity. k = 0.0041 and ω = 6.2.
The time it takes for one complete wave pattern to go past a fisherman is period.
[tex]\omega = 2\pi f\\ f = 1/ T[/tex]
T = 1.01 s.
The horizontal distance the wave crest traveled in one period is
[tex]\lambda = 2\pi / k = 2\pi / 0.0041 = 1.53\times 10^3~m[/tex]
[tex]y(x = \lambda,t = T) = 0.0275\cos(0.0041*1.53*\10^3 + 6.2*1.01) = 0.0275~m[/tex]
B) The wave number, k = 0.0041 . The number of waves per second is the frequency, so f = 0.987.
C) A wave crest travels past the fisherman with the following speed
[tex]v = \lambda f = 1.53\times 10^3 * 0.987 = 1.51\times 10^3~m/s[/tex]
The maximum speed of the cork floater can be calculated as follows.
The velocity of the wave crest is the derivative of the position with respect to time.
[tex]v(x,t) = \frac{dy(x,t)}{dt} = -(6.2\times 0.0275)\sin(0.0041x + 6.2t)[/tex]
The maximum velocity can be found by setting the derivative of the velocity to zero.
[tex]\frac{dv_y(x,t)}{dt} = -(6.2)^2(0.0275)\cos(0.0041*1.53\times 10^3 + 6.2t) = 0[/tex]
In order this to be zero, cosine term must be equal to zero.
[tex]0.0041*1.53\times 10^3 + 6.2t = 5\pi /2\\t = 0.255~s[/tex]
The reason that cosine term is set to be 5π/2 is that time cannot be zero. For π/2 and 3π/2, t<0.
[tex]v(x=\lambda, t = 0.255) = -(6.2\times0.0275)\sin(0.0041\times 1.53\times 10^3 + 6.2\times 0.255) = -0.17~m/s[/tex]
(a) The time taken "1.013 s".
(b) Number of waves "0.987 Hz".
(c) Maximum speed "0.1750 m/s".
A further explanation is below.
Given:
[tex]y(x,t) = (2.75 \ cm) Cos [(0.41 \ rad/cm)x+(6.20 \ rad/s)t][/tex](a)
The time taken will be:
→ [tex]T = \frac{2 \pi}{W}[/tex]
[tex]= \frac{2 \pi}{6.20}[/tex]
[tex]= 1.013 \ s[/tex]
The covered horizontal distance will be:
→ [tex]\lambda = \frac{2 \pi}{K}[/tex]
[tex]= \frac{2 \pi}{0.410}[/tex]
[tex]= 15.3 \ cm[/tex]
(b)
Wave number,
[tex]K = 0.410 \ rad/cm[/tex]The number of waves per second will be:
→ [tex]f = \frac{1}{T}[/tex]
[tex]= \frac{1}{1.013}[/tex]
[tex]= 0.987 \ Hz[/tex]
(c)
The speed in which the wave crest travel will be:
→ [tex]v = f \lambda[/tex]
[tex]= 15.3\times 0.987[/tex]
[tex]= 15.1 \ cm/s \ or \ 0.151 \ m/s[/tex]
and,
The maximum speed of the cork floater will be:
→ [tex]v_1 = AW[/tex]
[tex]=2.75\times 6.20[/tex]
[tex]= 0.1750 \ m/s[/tex]
Thus the above answers are correct.
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One of the greatest terrorism-related nuclear threats is from Select one: a. nuclear power plants. b. dirty bombs. c. nuclear warheads. d. None of these are correct.
Answer:
Dirty bomb
Explanation:
Among the nuclear bomb One type is a "dirty bomb." It combines a conventional explosive such as the dynamite with radioactive material which can spread when the system explodes. The explosion is releasing "dirty" bits of radioactive particles which are extremely harmful and can cause loss equivalent to a nuclear attack.
A sinusoidally-varying voltage V(t)=V0sin(2pift) with amplitude V0 = 10 V and frequency f = 100 Hz is impressed across the plates of a circular-shaped parallel plate air-gap capacitor of radius a = 1.0 cm and plate separation d = 0.01 mm. The amplitude of Maxwell's displacement current ID flowing across the gap between the plates of this capacitor is?
The amplitude of Maxwell's displacement current flowing across the gap between the plates of the capacitor is 1.753 x 10^-7 A.
Explanation:The amplitude of Maxwell's displacement current flowing across the gap between the plates of this capacitor can be determined using the formula for the displacement current, which is given by ID = ε0AdV/dt, where ε0 is the permittivity of free space, A is the area of the capacitor plates, and dV/dt is the rate of change of voltage with respect to time.
In this case, the area A of the circular plates is equal to πa^2, where a is the radius of the plates. Therefore, A = π(0.01 m)^2 = 0.000314 m^2. The rate of change of voltage can be obtained from the equation V(t) = V0sin(2πft), where V0 is the amplitude of the voltage, f is the frequency, and t is the time.
Substituting the given values V0 = 10 V and f = 100 Hz into the equation, we have V(t) = 10sin(2π(100)t). Differentiating this equation with respect to time gives dV/dt = 2000πcos(2π(100)t). Plugging in the value of t = 0 (for maximum displacement current) into the equation, we find that cos(2π(100)(0)) = cos(0) = 1.
Therefore, the amplitude of the Maxwell's displacement current ID is given by ID = (8.85 x 10^-12 F/m)(0.000314 m^2)(2000π)(1) = 1.753 x 10^-7 A.
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Stars form from clouds of gas and dust. As a protostar gravitationally contracts within its parent cloud, "conservation of angular momentum" says thatA) the protostar rotates more slowly.b) the protostar\'s rotation does not change.c) the protostar\'s axis of rotation will change direction.d) the protostar rotates more quickly.
Answer:
D) True. the protostar rotates more quickly.
Explanation:
If the system is isolated, the angular momentum must be retained.
Initial
L₀ = I w₀
Final
[tex]L_{f}[/tex] = [tex]I_{f}[/tex] [tex]w_{f}[/tex]
L₀ = [tex]L_{f}[/tex]
I w₀ = [tex]I_{f}[/tex][tex]w_{f}[/tex]
[tex]w_{f}[/tex] = I /[tex]I_{f}[/tex] w₀
In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase
Let's examine the answers
A) False. The opposite happens
B) False. Speed changes
C) False. For this there must be an external force, which does not exist
D) True. You agree with the above
Final answer:
The correct answer to the question of what occurs to a protostar as it contracts due to the 'conservation of angular momentum' is that the protostar rotates more quickly (d). This is analogous to a figure skater pulling their arms in to spin faster and is supported by observations of star-forming regions like the Orion Nebula.
Explanation:
Stars form from clouds of gas and dust. As these clouds collapse under their own gravity, they form protostars, which spin due to the conservation of angular momentum. Conservation of angular momentum dictates that as the radius of a spinning object decreases, its rotation speed must increase to conserve angular momentum. This concept is similar to a figure skater who spins faster when they pull their arms in. Therefore, when a protostar gravitationally contracts within its parent cloud, it rotates more quickly because as it shrinks, the protostar's rate of spin increases to conserve angular momentum.
This increase in rotation speed results in the formation of a spinning accretion disk around the equator, which is easier to observe in some regions, such as the Orion Nebula or the Taurus star-forming region. Observations from telescopes like the Hubble Space Telescope support this understanding of the role of angular momentum in star formation. In conclusion, the correct answer to the question is (d) the protostar rotates more quickly.
If two electrons in the same atom have the same four quantum numbers, then they must have the same energy.a. trueb. falsec. They cannot both have the same four quantum numbers.
Final answer:
The statement that two electrons in the same atom with the same four quantum numbers must have the same energy is false.
Explanation:
The statement that two electrons in the same atom with the same four quantum numbers must have the same energy is false. The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers. The four quantum numbers are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (m), and the spin quantum number (s).
Since the quantum numbers determine the energy and other properties of the electrons, if two electrons in the same atom have the same set of quantum numbers, they must have different spins in order to obey the Pauli exclusion principle. Therefore, they would have different energies.
A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.107 mm wide. At the point in the pattern which is an angular distance of 3.09° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad.
a) What is the wavelenght of the radiation?
b) What is the intensity at this point, if the intensity at the center of the central maximum is I/O?
Answer:
a. λ = 647.2 nm
b. I₀ 9.36 x 10⁻⁵
Explanation:
Given:
β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m
a.
The wavelength of the radiation can be find using
β = 2 π / γ * sin θ
λ = [ 2π * γ * sin θ ] / β
λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad
λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm
b.
The intensity of the central maximum I₀
I = I₀ (4 / β² ) * sin ( β / 2)²
I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²
I = I₀ 9.36 x 10⁻⁵
To find the wavelength of the radiation and the intensity at a specific point in a single-slit diffraction pattern, we can use formulas and calculations based on the given information.
Explanation:To find the wavelength of the radiation, we can use the formula for single-slit diffraction:
sin(θ) = mλ/w
Where θ is the angle from the center of the central maximum, m is the order of the bright fringe (m = 1 for the first bright fringe), λ is the wavelength, and w is the width of the slit. Rearranging the formula, we get:
λ = wsin(θ)/m
Plugging in the values, we have:
λ = (0.107 mm)(sin(3.09°))/(1)
Calculating this gives us the wavelength of the radiation.
To find the intensity at the given point, we can use the formula for intensity in single-slit diffraction:
I/I0 = (sin(θ)/θ)2
The given point is at an angular distance of 3.09°, so we can use this formula to calculate the intensity.
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For many years Colonel John P. Stapp, USAF, held the world's land speed record. He participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, he rode a rocket-propelled sled that moved down a track at a speed of 632 mi/h. He and the sled were safely brought to rest in 1.40s
(a) Determine the negative acceleration he experienced.
__ m/s2
(b) Determine the distance he traveled during this negative acceleration.
__ m.
Answer:
(a) -202 m/s²
(b) 198 m
Explanation:
Given data
Initial speed (v₀): 283 m/s[tex]\frac{632mi}{h} .\frac{1609.34m}{1mi} .\frac{1h}{3600s} =283m/s[/tex]
Final speed (vf): 0 (rest)Time (t): 1.40 s(a) The acceleration (a) is the change in the speed over the time elapsed.
a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²
(b) We can find the distance traveled (d) using the following kinematic expression.
y = v₀ × t + 1/2 × a × t²
y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²
y = 198 m
A cylinder with a piston contains 0.300 mol of oxygen at 2.50×105 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.Find the work done by the gas during the initial expansion.Find the heat added to the gas during the initial expansion.Find internal-energy change of the gas during the initial expansion.Find the work done during the final cooling;Find the heat added during the final cooling;Find the internal-energy change during the final cooling;Find the internal-energy change during the isothermal compression
Answer:
a) W = 900 J. b) Q = 3142.8 J . c) ΔU = 2242.8 J. d) W = 0. e) Q = 2244.78 J. g) Δ U = 0.
Explanation:
(a) Work done by the gas during the initial expansion:
The work done W for a thermodynamic constant pressure process is given as;
W = p Δ V
where
p is the pressure and Δ V is the change in volume.
Here, Given;
P 1 = i n i t i a l p r e s s u r e = 2.5 × 10^ 5 P a
T 1 = i n i t i a l t e m p e r a t u r e = 360 K
n = n u m b er o f m o l e s = 0.300 m o l
The ideal gas equation is given by
P V = nRT
where ,
p = absolute pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = universal gas constant = 8.314 K J / m o l K
T = absolute temperature of the gas
Now we will Calculate the initial volume of the gas using the above equation as follows;
PV = n R T
2.5 × 10 ^5 × V 1 = 0.3 × 8.314 × 360
V1 = 897.91 / 250000
V 1 = 0.0036 m ^3 = 3.6×10^-3 m^3
We are also given that
V 2 = 2× V 1
V2 = 2 × 0.0036
V2 = 0.0072 m^3
Thus, work done is calculated as;
W = p Δ V = p×(V2 - V1)
W = ( 2.5 × 10 ^5 ) ×( 0.0072 − 0.0036 )
W = 900 J.
(b) Heat added to the gas during the initial expansion:
For a diatomic gas,
C p = 7 /2 ×R
Cp = 7 /2 × 8.314
Cp = 29.1 J / mo l K
For a constant pressure process,
T 2 /T 1 = V 2 /V 1
T 2 = V 2 /V 1 × T 1
T 2 = 2 × T 1 = 2×360
T 2 = 720 K
Heat added (Q) can be calculated as;
Q = n C p Δ T = nC×(T2 - T1)
Q = 0.3 × 29.1 × ( 720 − 360 )
Q = 3142.8 J .
(c) Internal-energy change of the gas during the initial expansion:
From first law of thermodynamics ;
Q = Δ U + W
where ,
Q is the heat added or extracted,
Δ U is the change in internal energy,
W is the work done on or by the system.
Put the previously calculated values of Q and W in the above formula to calculate Δ U as;
Δ U = Q − W
ΔU = 3142.8 − 900
ΔU = 2242.8 J.
(d) The work done during the final cooling:
The final cooling is a constant volume or isochoric process. There is no change in volume and thus the work done is zero.
(e) Heat added during the final cooling:
The final process is a isochoric process and for this, the first law equation becomes ,
Q = Δ U
The molar specific heat at constant volume is given as;
C v = 5 /2 ×R
Cv = 5 /2 × 8.314
Cv = 20.785 J / m o l K
The change in internal energy and thus the heat added can be calculated as;
Q = Δ U = n C v Δ T
Q = 0.3 × 20.785 × ( 720 - 360 )
Q = 2244.78 J.
(f) Internal-energy change during the final cooling:
Internal-energy change during the final cooling is equal to the heat added during the final cooling Q = Δ U .
(g) The internal-energy change during the isothermal compression:
For isothermal compression,
Δ U = n C v Δ T
As their is no change in temperature for isothermal compression,
Δ T = 0 , then,
Δ U = 0.
During the initial expansion, the gas performed 900 J of work, absorbed 3142.8 J of heat, resulting in a change in internal energy of 2242.8 J. In subsequent processes, work and internal energy changes were determined, culminating in an isothermal compression with no internal energy change.
(a) Work done by the gas during the initial expansion:
The work done (W) for a thermodynamic constant pressure process is given by W = P ΔV, where P is the pressure and ΔV is the change in volume.
Given:
P₁ = initial pressure = 2.5 × 10^5 Pa
T₁ = initial temperature = 360 K
n = number of moles = 0.300 mol
The ideal gas equation is PV = nRT, where P is the absolute pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the universal gas constant (8.314 kJ/mol·K), and T is the absolute temperature of the gas.
Calculate the initial volume of the gas:
P₁V₁ = nRT₁
(2.5 × 10^5) × V₁ = 0.3 × 8.314 × 360
V₁ = (0.3 × 8.314 × 360) / (2.5 × 10^5)
V₁ = 0.0036 m³ = 3.6 × 10⁻³ m³
Given V₂ = 2 × V₁:
V₂ = 2 × 0.0036
V₂ = 0.0072 m³
Now, calculate the work done:
W = P ΔV = (2.5 × 10^5) × (0.0072 - 0.0036)
W = 900 J
(b) Heat added to the gas during the initial expansion:
For a diatomic gas, Cp = (7/2)R, where Cp is the molar heat capacity at constant pressure.
Cp = (7/2) × 8.314
Cp = 29.1 J/mol·K
For a constant pressure process, T₂/T₁ = V₂/V₁:
T₂ = (V₂/V₁) × T₁
T₂ = 2 × T₁ = 2 × 360 = 720 K
Heat added (Q) can be calculated as:
Q = nCpΔT = 0.3 × 29.1 × (720 - 360)
Q = 3142.8 J
(c) Internal-energy change of the gas during the initial expansion:
From the first law of thermodynamics Q = ΔU + W, where Q is the heat added or extracted, ΔU is the change in internal energy, and W is the work done on or by the system.
ΔU = Q - W = 3142.8 - 900
ΔU = 2242.8 J
(d) The work done during the final cooling:
The final cooling is a constant volume (isochoric) process, so there is no change in volume (ΔV = 0), and thus the work done is zero.
(e) Heat added during the final cooling:
For an isochoric process, Q = ΔU. The molar specific heat at constant volume is Cv = (5/2)R.
Cv = (5/2) × 8.314 = 20.785 J/mol·K
The change in internal energy and thus the heat added can be calculated as:
Q = ΔU = nCvΔT = 0.3 × 20.785 × (720 - 360)
Q = 2244.78 J
(f) Internal-energy change during the final cooling:
The internal-energy change during the final cooling is equal to the heat added during the final cooling, so ΔU = Q.
(g) The internal-energy change during the isothermal compression:
For isothermal compression, ΔU = 0 since there is no change in temperature (ΔT = 0).
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A flight attendant pulls her 70.0 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52 degrees above the horizontal. Find the following: a) the work she does on the flight bag b) the work done on the flight bag c) the coefficient of kinetic friction between the flight bag and the floor.
Answer:
6230.49413 J
0 J
0.63998
Explanation:
F = Force = 40 N
[tex]\theta[/tex] = Angle = 52°
Work done is given by
[tex]W=Fscos50\\\Rightarrow W=40\times 253\times cos52\\\Rightarrow W=6230.49413\ J[/tex]
The work she does on the flight bag is 6230.49413 J
The work done on the flight bag will be the opposite of the work done by the flight attendant
[tex]W=-6230.49413\ J[/tex]
So net work will be
[tex]W_n=6230.49413-6230.49413\\\Rightarrow W_n=0[/tex]
The net work done on the flight bag is 0 J
Coefficient of friction is given by
[tex]\mu=\dfrac{F_f}{F_N}\\\Rightarrow \mu=\dfrac{F_f}{F_g-F_{app}}\\\Rightarrow \mu=\dfrac{40cos52}{70-40sin52}\\\Rightarrow \mu=0.63998[/tex]
The coefficient of friction is 0.63998
(a) The work done by the attendant on the flight bag is 6230.49 J.
(b) The work on the flight bag is 0 J .
(c) The coefficient of kinetic friction between the flight bag and the floor is 0.35.
Given data:
The weight of flight bag is, W = 70.0 N.
The distance covered by the bag is, d = 253 m.
The magnitude of force exerted on bag is, F = 40.0 N.
The angle of inclination with horizontal is, [tex]\theta =52^{\circ}[/tex].
Work done defined as the product of force and distance covered due to applied force.
(a)
The work done on the flight bag is given as,
[tex]W'=F \times dcos\theta[/tex]
Solving as,
[tex]W'=40 \times 253cos52\\W'=6230.49 \;\rm J[/tex]
Thus, the work done by the attendant on the flight bag is 6230.49 J.
(b)
The work done on the flight bag will be the opposite of the work done by the flight attendant. So,
W'' = - W
W'' = - 6230.49 J
Then net work done on the flight bag is,
[tex]W_{net}=W'+W''\\W_{net}=6230.49 - 6230.49\\W_{net}=0[/tex]
Thus, the net work on the flight bag is 0 J .
(c)
The expression for the frictional force is given as,
[tex]f = \mu \times W[/tex]
[tex]\mu[/tex] is the coefficient of kinetic friction. And the frictional force is due to the horizontal component of applied force. Then,
[tex]f=Fcos\theta[/tex]
So,
[tex]Fcos\theta= \mu \times W\\\\40 \times cos52=\mu \times 70\\\\\mu=\dfrac{40 \times cos52}{70} \\\\\mu = 0.35[/tex]
Thus, the coefficient of kinetic friction between the flight bag and the floor is 0.35.
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A device plugged into a wall outlet experiences an alternating current. The RMS current value is 2 A. The peak current is ___.
Answer:
The peak value of current will be 2.828 A.
Explanation:
Given that
Value of RMS current I(rms) = 2 A
Lets take peak current = I(p)
As we know that relationship between RMS and peak current given as
[tex]I(rms)=\dfrac{I(p)}{\sqrt2}[/tex]
Now by putting the values in the above equation
[tex]I(rms)=\dfrac{I(p)}{\sqrt2}[/tex]
[tex]2\ A=\dfrac{I(p)}{\sqrt2}[/tex]
[tex]I(p)=2\sqrt2[/tex]
As we know that
[tex]\sqrt{2}=1.414[/tex]
Therefore
I(p)=2.828 A
The peak value of current will be 2.828 A.
will a flying bird have more kinetic energy than a sitting elephant
Answer:
Yes.
Explanation:
A sitting elephant has zero kinetic energy. A flying bird have some kinetic energy due to its motion. Regardless of their size, a moving object has always more kinetic energy than an object at rest.
In braking an automobile, the friction between the brake drums and brake shoes converts the car's kinetic energy into heat. If a 1 500-kg automobile traveling at 30 m/s brakes to a halt, how much does the temperature rise in each of the four 8.0-kg brake drums in °C? (The specific heat of each iron brake drum is 448 J/kg⋅°C).
To solve this problem it is necessary to apply the concepts related to energy conservation.
In this case the kinetic energy is given as
[tex]KE = \frac{1}{2} mv^2[/tex]
Where,
m = mass
v= Velocity
In the case of heat lost energy (for all 4 wheels) we have to
[tex]Q = mC_p \Delta T \rightarrow 4Q = 4mC_p \Delta T[/tex]
m = mass
[tex]C_p =[/tex] Specific Heat
[tex]\Delta T[/tex]= Change at temperature
For conservation we have to
[tex]KE = Q[/tex]
[tex]\frac{1}{2} mv^2 = 4mC_p \Delta T[/tex]
[tex]\Delta T = \frac{1}{2}\frac{mv^2}{4mC_p}[/tex]
[tex]\Delta T = \frac{1}{2}\frac{(1500)(30)^2}{4(8)(448)}[/tex]
[tex]\Delta T = 47.084\°C \approx 47\°C[/tex]
Therefore the temperature rises in each of the four brake drums around to 47°C
An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the observer manages to measure the time it.takes for the second cart to pass him to be 5.0 seconds How long will it take for the 10th car to pass him? Assume all cars to be of the same length.
a. 2.8 s
b. 2.4 s
c. 2.0 s
d. 1.5 s
e. 1.1 s
To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:
[tex]x = v_0 t \frac{1}{2} at^2[/tex]
Where,
x= Displacement
[tex]v_0[/tex] = Initial velocity
a = Acceleration
t = time
Since there is no initial velocity, the same equation can be transformed in terms of length and time as:
[tex]L = \frac{1}{2} a t_1 ^2[/tex]
For the second cart
[tex]2L \frac{1}{2} at_2^2[/tex]
When the tenth car is aligned the length will be 9 times the initial therefore:
[tex]9L = \frac{1}{2} at_3^2[/tex]
When the tenth car has passed the length will be 10 times the initial therefore:
[tex]10L = \frac{1}{2}at_4^2[/tex]
The difference in time taken from the second car to pass it is 5 seconds, therefore:
[tex]t_2-t_1 = 5s[/tex]
From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:
[tex]\frac{1}{2} = (\frac{t_1}{t_2})^2[/tex]
[tex]t_1 = \frac{t_2}{\sqrt{2}}[/tex]
From the relationship when the car has passed and the time difference we will have to:
[tex](t_2-\frac{t_2}{\sqrt{2}}) = 5[/tex]
[tex]t_2 (\sqrt{2}-1) = 3\sqrt{2}[/tex]
[tex]t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2[/tex]
Replacing the value found in the equation given for the second car equation we have to:
[tex]\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2[/tex]
Finally we will have the time when the cars are aligned is
[tex]18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2[/tex]
[tex]t_3 = 36.213s[/tex]
The time when you have passed it would be:
[tex]20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2[/tex]
[tex]t_4 = 38.172[/tex]
The difference between the two times would be:
[tex]t_4-t_3 = 38.172-36.213 \approx 2s[/tex]
Therefore the correct answer is C.
The passage time for the 10th car in a uniformly accelerating train would be less than 5.0 seconds but not as low as any of the options provided (a to e), as each car passes more quickly than the previous one due to constant acceleration.
Explanation:The question involves an observer timing how long it takes for individual cars of constant acceleration of a train to pass by. If it took 5.0 seconds for the second car to pass the observer, then the time it takes for the 10th car to pass him is merely the time it took for one car to pass because the train is accelerating at a constant rate, and all cars are of the same length. Considering the train is accelerating uniformly, the passage of each car happens more quickly over time.
Giving this knowledge, for the first car to pass, the train would have started at rest, so it would have taken longer than the second car which had the benefit of the train already moving at a certain velocity. Therefore, we can derive that the time for each subsequent car to pass will be less than 5.0 seconds, which questions the multiple-choice options. Based on this logic, we don't have to calculate the exact passing time for the 10th car, as there are no values given to calculate with, but we can infer that none of the times listed would be correct. Each car would be passing quicker than the last, but it would not be instantaneous; it would be a fraction of the 5 seconds but not as low as any of the times listed from a to e.
The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionless, using both the FLT system and the MLT system for basic dimensions, and determine its value for methane flowing at a velocity of 4 m/s through a 2-in-diameter pipe.
Answer:
Re = 1 10⁴
Explanation:
Reynolds number is
Re = ρ v D /μ
The units of each term are
ρ = [kg / m³]
v = [m / s]
D = [m]
μ = [Pa s]
The pressure
Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]
μ = [Pa s] = [kg / m s²] [s] = [kg / m s]
We substitute the units in the equation
Re = [kg / m³] [m / s] [m] / [kg / m s]
Re = [kg / m s] / [m s / kg]
RE = [ ]
Reynolds number is a scalar
Let's evaluate for the given point
Where the data for methane are:
viscosity μ = 11.2 10⁻⁶ Pa s
the density ρ = 0.656 kg / m³
D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m
Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶
Re = 1.19 10⁴
The Reynolds number is a key parameter in fluid mechanics that signals laminar or turbulent flow. It is a dimensionless quantity determined by fluid properties and flow characteristics.
The Reynolds number, a dimensionless parameter, indicates whether flow is laminar or turbulent in fluid mechanics. It is defined as Re = rho VD/mu, where rho is the fluid density, V is the velocity, D is the diameter, and mu is the fluid viscosity. The value of Reynolds number for methane flowing at 4 m/s through a 2-in-diameter pipe is calculated using this formula.
The size (radius) of an oxygen molecule is about 2.0 ×10−10 m. Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures (T= 300K ). Assume that deviatons would be noticeable when volume of the gas per molecule equals the volume of the molecule itself.
Express your answer using one significant figure.
P = ? Pa
Answer:
P = 1 x 10⁸ Pa
Explanation:
given,
radius = 2.0 ×10⁻¹⁰ m
Temperature
T = 300 K
Volume of gas molecule =
[tex]V = \dfrac{4}{3}\pi r^3[/tex]
[tex]V = \dfrac{4}{3}\pi (2\times 10^{-10})^3[/tex]
V = 33.51 x 10⁻³⁰ m³
we know,
P V = 1 . k T
k = 1.38 x 10⁻²³ J/K
P(33.51 x 10⁻³⁰) = 1 . (1.38 x 10⁻²³) x 300
P = 1.235 x 10⁸ Pa
for 1 significant figure
P = 1 x 10⁸ Pa
If a camera lens gives the proper exposure for a photograph at a shutter speed of 1/200 s at an f-stop of f/2.80, the proper shutter speed at f/7.92 is
Answer:
[tex]\dfrac{1}{28.15432}\ s[/tex]
Explanation:
The factor by which the shutter speed is increased is [tex]\dfrac{7.92}{2.8}[/tex]
Exposure time will be increased by
[tex]f=2^{\dfrac{7.92}{2.8}}=7.1037\ s[/tex]
The proper shutter speed is given by
[tex]\dfrac{1}{T}f=\dfrac{1}{200}\times 2^{\dfrac{7.92}{2.8}}=\dfrac{1}{\dfrac{200}{2^{\dfrac{7.92}{2.8}}}}\\ =\dfrac{1}{28.15432}\ s[/tex]
The proper shutter speed is [tex]\dfrac{1}{28.15432}\ s[/tex]
A 3.0-kg cylinder falls vertically from rest in a very tall, evacuated tube near the surface of the earth. What is its speed after the cylinder has fallen 6.0 m?
Answer:v=10.84 m/s
Explanation:
Given
mass of Cylinder [tex]m=3 kg[/tex]
height of cylinder [tex]h=6 m[/tex]
It is given that tube is evacuated so we can neglect air resistance so friction provided by the air is zero.
Since Energy cannot be destroyed but can be transformed from one form to another therefore Potential Energy of Cylinder is converted to Kinetic Energy of Cylinder
Potential Energy[tex]=mgh=3\times 9.8\times 6=176.4 J[/tex]
Kinetic Energy [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 3\times (v)^2[/tex]
[tex]176.4=\frac{1}{2}\times 3\times (v)^2[/tex]
[tex]v=\sqrt{117.6}[/tex]
[tex]v=10.84 m/s[/tex]
Speed of cylinder after the cylinder has fallen 6 m is 11 meter/second.
Energy conservation based problemWhat information do we have?
Mass of cylinder = 3 kg
Height = 6 m
Using energy conservation theroy
mgh = (1/2)mv²
gh = (1/2)v²
(9.8)(6) = (1/2)v²
Velocity = 11 m/s (Approx.)
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Two protons, with equal kinetic energy, collide head-on. What is the minimum kinetic energy Kp of one of these protons necessary to make a pion-antipion pair? The rest energy of a pion is 139.6MeV.
Answer:
[tex]K_p=139.6\ MeV[/tex]
Explanation:
It is given that,
The rest energy of a pion is 139.6 MeV. Here, two protons having equal kinetic energy collides elastically. We need to find the minimum kinetic energy of one of these protons necessary to make a pion-antipion pair.
It can be calculated using conservation of energy and momentum, the total energy of the particles gets converted into rest mass energy of new particles. So,
[tex]2K_p=2\times E_{\pi^+}[/tex]
[tex]K_p=\times E_{\pi^+}[/tex]
[tex]K_p=139.6\ MeV[/tex]
So, the minimum kinetic energy of one of these protons necessary to make a pion-antipion pair is 139.6 MeV. Hence, this is the required solution.
An object initially at rest falls from a height H until it reaches the ground. Two of the following energy bar charts represent the kinetic energy K and gravitational potential energy Ug of the object-Earth system at two positions. The first position is when the object is initially released, and the second position is when the object is halfway between its release point and the ground. Which two charts could represent the mechanical energy of the object-Earth system? Select two answers. by Ug K Ug к Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point d. c. к Ug Ug K Mechanical Energy at Release Point Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point
Answer:
Initial: bar power U₀
Final: bar power U = U₀ / 2
bar Kinetic energy K = U₀ / 2
These two bars are half the height of the initial bar
Explanation:
To know which graph is correct, let's discuss the solution to the problem
Initial mechanical energy
Em₀ = U₀ = m g H
The mechanical energy at the midpoint
Em₂ = K + U₂
As there is no friction, mechanical energy is conserved
Em₀ = Em₂
U₀ = K + U₂
K = U₀ - U₂
K = m g (H - y₂)
Indicates that position 2 corresponds to y₂ = H / 2
K = m g (H –H / 2)
K = ½ m g H
K = ½ Uo
Therefore the graph must be
Initial: bar power U₀
Final: bar power U = U₀ / 2
bar Kinetic energy K = U₀ / 2
These two bars are half the height of the initial bar
Answer: B&C
Explanation:
Release point: Ug bar graph only K none
Halfway point: Ug and K are equal bar graph
second option,
Release point: empty graph
Halfway point: Ug down half K up half
Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of 1.52\times 10^{11}~\text{V/m}1.52×10 11 V/m for a time of 1.00 ns. What energy does it deliver on a 1.00~\mathrm{mm^2}1.00 mm 2 area?
Answer:
30643 J
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]
t = Time taken = 1 ns
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
[tex]E_0[/tex] = Maximum electric field strength = [tex]1.52\times 10^{11}\ V/m[/tex]
A = Area = [tex]1\ mm^2[/tex]
Magnitude of magnetic field is given by
[tex]B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T[/tex]
Intensity is given by
[tex]I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2[/tex]
Power, intensity and time have the relation
[tex]E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J[/tex]
The energy it delivers is 30643 J
You place 100 grams of ice, with a temperature of −10∘C, in a styrofoam cup. Then you add an unknown mass of water, with a temperature of +10∘C, and allow the system to come to thermal equilibrium. For the calculations below, use the following approximations. The specific heat of solid water is 2 joules / gram, and the specific heat of liquid water is 4 joules / gram. The latent heat of fusion of water is 300 joules / gram. Assume no heat is exchanged with the surroundings. Express your answers in grams, using two significant digits.
If the final temperature of the system is -5∘C , how much water was added? ______________ grams
Answer:
Mass of water 2.9g
Explanation:
Ice
[tex]m_{ice}=100g[/tex]
[tex]c_{ice}=2J/g.K[/tex]
[tex]T_{ice,initial}=-10\°C[/tex]
[tex]T_{ice,final}=T_{equilibrium}=-5\°C[/tex]
Water
[tex]c_{water}=4J/g.K[/tex]
[tex]T_{water,initial}=10\°C[/tex]
[tex]T_{water,final}=0\°C[/tex]
[tex]T_{equilibrium}=-5\°C[/tex]
[tex]l_{water}=300J/g[/tex]
[tex]m_{water}=?g[/tex]
Step 1: Determine heat gained by ice
[tex]Q_{ice}=m_{ice}c_{ice}(T_{ice,final}-T_{ice,initial})[/tex]
[tex]Q_{ice}=100*2*(-5--10)[/tex]
[tex]Q_{ice}=1000J[/tex]
Step 2; Determine heat lost by water
[tex]Q_{water}=m_{water}c_{water}(T_{water,initial}-T_{water,final})+m_{water}l_{water}[/tex]
[tex]Q_{water}=m_{water}*4*(10-0)+m_{water}*300[/tex]
[tex]Q_{water}=40m_{water}+300m_{water}[/tex]
[tex]Q_{water}=340m_{water}[/tex]
Step 3: Heat gained by ice is equivalent to heat lost by water
[tex]Q_{ice}=Q_{water}[/tex]
[tex]1000=340m_{water}[/tex]
[tex]m_{water}=2.9g[/tex]
A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. The plates are separated by 3 mm and the magnitude of the magnetic field is 0.3 T. What voltage between the plates will allow particles of speed 5.0 x 10^5 m/sa. 1200 V b. 3800 V c. 7500 V d. 190 V e. 380 V
Answer:
Voltage, V = 450 volts
Explanation:
It is given that,
Separation between plates, d = 3 mm = 0.003 m
magnitude of magnetic field, B = 0.3 T
Speed of the particle, [tex]v=5\times 10^5\ m/s[/tex]
The relation between the magnetic field, electric field and the velocity of the particle is given by :
[tex]v=\dfrac{E}{B}[/tex]
Also, [tex]E=\dfrac{V}{d}[/tex]
[tex]v=\dfrac{V}{Bd}[/tex]
[tex]V=vBd[/tex]
[tex]V=5\times 10^5\times 0.3\times 0.003[/tex]
V = 450 volts
So, the voltage between the plates will be 450 V. Hence, this is the required solution.
The earth has radius R. A satellite of mass 100 kg is in orbit at an altitude of 3R above the earth's surface. What is the satellite's weight at the altitude of its orbit
Answer:
W= 61.3 N
Explanation:
The only force acting on the satellite is the one due to the attraction from Earth, which obeys the Newton's Universal Law of Gravitation, as follows:
Fg =G*ms*me / (res)²
This force, also obeys the Newton's 2nd Law, so we can write the following equation:
G*ms*me*/ (res)² = ms* a = ms*g
We call to the product of the mass times the acceleration caused by gravity (g), the weight of this mass, so we can write as follows:
G*ms*me / (res)² = ms*g = W (1)
where G = 6.67*10⁻11 N*m²/kg², ms= 100 kg, me= 5.97*10²⁴ kg, and
res= 4 *re = 4*6.37*10⁶ m.
Replacing all these known values in (1), we get the value of W:
W =(( 6.67*5.97/(4*6.37)²) *( 10⁻¹¹ * 10²⁴ /10¹²) )* 100 N = 61.3 N
The satellite's weight at the altitude of its orbit is;
61.31 N
Formula for gravitational force is;
F = GMm/R²
Where;
G is gravitational constant = 6.67 × 10⁻¹¹ N.m²/kg²
m is mass of earth = 5.97 × 10²⁴ kg
M is mass of satellite = 100 kg
Now, we are told that the altitude is 3R above the Earth's surface.
At the Earth's surface, the distance from the Earth's center is R where R is radius of earth.
Thus, total altitude from the Earth's center to the satellite it (3R + R) = 4R
Thus;
F = GMm/(4R)²
Where R is radius of earth = 6371 × 10⁶ m
Thus;
F = (6.67 × 10⁻¹¹ × 100 × 5.97 × 10²⁴)/(4 × 6371 × 10⁶)
F = 61.31 N
Now, from Newton's second law of motion, we know that the force is equal to the weight.
Thus;
Weight = 61.31 N
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One might be tempted to say that exothermic processes are always spontaneous since the system is emitting energy (heat) in order to reach a (preferred) lower energy state. However, as we have just investigated, the spontaneous process for polymers is endothermic. This reveals that we must consider entropy changes when determining the nature of spontaneity. The most probable configuration of a system and its surroundings, naturally, is the one that will be observed. The condition for spontaneity can be recast using the concept of the free energy of the system, where a change in free energy results both from changes in the enthalpy (which includes internal potential and kinetic energies) and the entropy (the number of states accessible to the system). Δ G = Δ H − T Δ S.
An unknown chemical reaction undergoes an enthalpy change of Δ H =17 kJ/mol while the entropy increases by Δ S =50 J/(mol * K).
Above what temperature (in Kelvin) does this reaction occur spontaneously?
Answer : This reaction occur spontaneously at temperature above in kelvins is, 340 K
Explanation : Given,
[tex]\Delta H[/tex] = 17 KJ/mole = 17000 J/mole
[tex]\Delta S[/tex] = 50 J/mole.K
Gibbs–Helmholtz equation is :
[tex]\Delta G=\Delta H-T\Delta S[/tex]
As per question the reaction is spontaneous that means the value of [tex]\Delta G[/tex] is negative or we can say that the value of [tex]\Delta G[/tex] is less than zero.
[tex]\Delta G<0[/tex]
The above expression will be:
[tex]0>\Delta H-T\Delta S[/tex]
[tex]T\Delta S>\Delta H[/tex]
[tex]T>\frac{\Delta H}{\Delta S}[/tex]
Now put all the given values in this expression, we get :
[tex]T>\frac{17000J/mole}{50J/mole.K}[/tex]
[tex]T>340K[/tex]
Therefore, this reaction occur spontaneously at temperature above in kelvins is, 340 K
Final answer:
The reaction occurs spontaneously above 340 Kelvin, determined by applying the Gibbs free energy formula (ΔG = ΔH - TΔS) and ensuring the values for ΔH and ΔS are in compatible units.
Explanation:
The question requires us to determine above what temperature a reaction occurs spontaneously, given an enthalpy change (ΔH) of 17 kJ/mol and an entropy change (ΔS) of 50 J/(mol·K). To find this, we use the formula for Gibbs free energy change (ΔG = ΔH - TΔS), where a reaction is spontaneous when ΔG < 0.
First, we need to ensure that both values are in the same units, so we convert ΔH from kJ to J: 17 kJ/mol = 17000 J/mol. Then, we solve for T in the equation ΔG < 0, substituting ΔH and ΔS into the equation:
0 > 17000 J/mol - T(50 J/(mol·K)),
implying T > 17000 J/mol / 50 J/(mol·K) = 340 K.
Therefore, the reaction occurs spontaneously above 340 K.
An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef- ficient based on area bl is CD = 0.06, estimate the power required to tow the banner. (b) For comparison, determine the power required if the airplane was instead able to tow a rigid flat plate of the same size. (c) Explain why one had a larger power requirement (and larger drag) than the other. (d) Finally, determine the power required if the airplane was towing a smooth spherical balloon with a diameter of 2 m.
Final answer:
We can estimate the power required to tow a banner by using the drag force formula. If the airplane was towing a rigid flat plate, the power required would be different due to a different drag coefficient. The power requirement and drag are larger for the banner due to its higher drag coefficient compared to a flat plate. No calculation can be made for a smooth spherical balloon without the drag coefficient.
Explanation:
(a) To estimate the power required to tow the banner, we can use the formula: Power = Drag force * velocity. The drag force can be calculated using the drag coefficient and the area of the banner. The area is given by multiplying the height (0.8 m) by the length (25 m). The drag force can then be multiplied by the velocity of the plane (150 km/hr converted to m/s) to obtain the power required.
(b) If the plane was towing a rigid flat plate of the same size, the drag coefficient would be different. The power required can be calculated using the new drag coefficient and the same formula as in part (a).
(c) The power requirement and drag are larger for the banner because the drag coefficient for the banner is higher compared to that of a rigid flat plate. This means that the banner experiences more air resistance, requiring more power to tow.
(d) To determine the power required to tow a smooth spherical balloon with a diameter of 2 m, we would need the drag coefficient associated with the balloon. Since it is not provided in the question, we cannot calculate the power required.
It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring’s effective force constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85-kg team?
Answer:
(a)spring effective force constant =1568N/cm
(b) Yes
Explanation:
Hooke's law is represented mathematically as, F=ke
where
F is force applied to elastic material,k= spring constante= extension(a).After the maximum load is exceeded, Hooke's law doesn't apply anymore. from the problem, its stated that a maximum load of 120kg will cause an extension of 0.75cm, we will use this to determine the spring constant.
F=mg, g=9.8[tex]m/s^{2}[/tex]
F= 120*9.8 =1176N
From Hooke's law, k=[tex]\frac{F}{e}[/tex]
k=[tex]\frac{1176}{0.75}[/tex]
k=1568N/cm
(b). The players who stands on the scale causes a 0.48cm extension.
e= 0.48cm, k= 1568N/cm
F=Ke
F= 1568*0.48
F= 752.64N
To calculate the mass of the player we divide this force by g=9.8[tex]m/s^{2}[/tex]
F=mg
m=F/g
[tex]m= \frac{ 752.64}{9.8} \\\\m=76.8kg[/tex]
since the rugby team is an under 85kg team, this player with 76.8kg mass is eligible
A resistor with R=300 ? and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 216 W. (a) What is the impedance Z of the circuit? (b) What is the amplitude of the voltage across the inductor? (c) What is the power factor?
Answer:
impedance Z = 416.66 ohm
voltage across inducance V = 346.99 V
power factor = 0.720
Explanation:
given data
resistor R = 300
voltage amplitude = 500 V
resistor = 216 W
to find out
impedance and amplitude of the voltage and power factor
solution
we apply here average power formula that is
average power = I²×R ............1
I = [tex]\frac{Vrms}{Z}[/tex]
so
average power = ([tex]\frac{Vrms}{Z}[/tex])²×R
Vrms = [tex]\frac{1}{\sqrt{2} }[/tex] × Vmax
Z = V × [tex]\sqrt{\frac{R}{2P} }[/tex]
Z = 500 × [tex]\sqrt{\frac{300}{2*216} }[/tex]
impedance Z = 416.66 ohm
and
we know voltage across inductor is here express as
V = I × X .............2
so here X will be by inductance
Z² = R² + (X)²
(X)² = 416.66² - 300²
X = 289.15 ohm
and I = [tex]\frac{V}{Z}[/tex]
I = [tex]\frac{500}{416.66}[/tex]
I = 1.20 A
so from equation 2
V = 1.20 × 289.15
voltage across inducance V = 346.99 V
and
average power = Vmax × Imax × cos∅
tan∅ = [tex]\frac{289.15}{300}[/tex]
tan∅ = 43.95°
so power factor is
power factor = cos43.95°
power factor = 0.720
The problem involves the use of Ohm's law, impedance, power, and a power factor in an AC circuit. The impedance of the circuit, the voltage across the inductor, and the power factor can be calculated using given values, the principle of impedance and the relationships among AC voltage, current, resistance and power.
Explanation:
The question involves an AC circuit composed of a resistor and inductor in series connected to an AC voltage source. We have a resistor with a resistance (R) of 300 Ω and a dissipated power (P) of 216W. The voltage amplitude (Vo) of the AC source is 500 V. It is important to remember that in this context, impedance (Z), which has a unit of ohms, represents the total resistance in an AC circuit and can be calculated using Ohm's law for AC circuits.
(a) To find the impedance Z of the circuit, we consider that the power P is given by the relation P = Vo^2/R, substituting for P, and R, we can solve for Vo, which will be sqrt(P*R). Then, the rms voltage (V) is given by Vo/sqrt(2). Our current I would be P/V. Finally, applying Ohm's law, Z=V/I would give us the impedance.
(b) The voltage across the inductor can be found by using Pythagoras' Theorem in the context of an AC circuit, VL = sqrt(Vo^2 - VR^2), where VR is the voltage across the resistor (equal to I* R).
(c) Lastly, the power factor can be found as the cosine of the phase angle θ, which can also be defined as R/Z. We'd first calculate θ = arccos(R/Z), and then find the power factor as cos(θ).
Learn more about AC Circuit here:https://brainly.com/question/31391159
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