Answer:
The pressure of the remaining gas in the tank is 6.4 atm.
Explanation:
Given that,
Temperature T = 13+273=286 K
Pressure = 10.0 atm
We need to calculate the pressure of the remaining gas
Using equation of ideal gas
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
For a gas
![P_{1}V_{1}=nRT_{1}](https://tex.z-dn.net/?f=P_%7B1%7DV_%7B1%7D%3DnRT_%7B1%7D)
Where, P = pressure
V = volume
T = temperature
Put the value in the equation
....(I)
When the temperature of the gas is increased
Then,
....(II)
Divided equation (I) by equation (II)
![\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_%7B1%7DV%7D%7BP_%7B2%7DV%7D%3D%5Cdfrac%7BnRT_%7B1%7D%7D%7B%5Cdfrac%7Bn%7D%7B2%7DRT_%7B2%7D%7D)
![\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}](https://tex.z-dn.net/?f=%5Cdfrac%7B10%5Ctimes%20V%7D%7BP_%7B2%7DV%7D%3D%5Cdfrac%7BnR%5Ctimes286%7D%7B%5Cdfrac%7Bn%7D%7B2%7DR368%7D)
![P_{2}=\dfrac{10\times368}{2\times286}](https://tex.z-dn.net/?f=P_%7B2%7D%3D%5Cdfrac%7B10%5Ctimes368%7D%7B2%5Ctimes286%7D)
![P_{2}= 6.433\ atm](https://tex.z-dn.net/?f=P_%7B2%7D%3D%206.433%5C%20atm)
![P_{2}=6.4\ atm](https://tex.z-dn.net/?f=P_%7B2%7D%3D6.4%5C%20atm)
Hence, The pressure of the remaining gas in the tank is 6.4 atm.
Answer:
With an increase in temperature, there is typically an increase in the molecular interchange as molecules move faster in higher temperatures. The gas viscosity will increase with temperature. ... With high temperatures, viscosity increases in gases and decreases in liquids, the drag force will do the same.
Answer:
In-car temperature is controlled in a variable displacement compressor system by varying the capacity of the <em>refrigeration</em> not by cycling the <em>compressor </em>on and off.
Explanation:
A Variable Displacement Compressor (VDC) works in a separate and rather more efficient manner than the fixed displacement compressor (FDC). <em>The VDC automatically regulates the cooling effect of the refrigerator by changing its pumping capacity by means of a "wobbling piston" system. </em>
<em></em>
During times of peak load, the refrigeration pumping capacity is increased, delivering a large flow rate of refrigerant, thus giving a high refrigerating effect. The reverse happens during times of load refrigeration load.
<em>The fixed compressor tries to vary the temperature of the car by simply switching off the compressor</em>, and turning it on during times of low and peak loads.
Answer:
2.3687599 m/s
0.91106 m/s
0.617213012 J
Explanation:
f = Frequency = ![2.9\ Hz](https://tex.z-dn.net/?f=2.9%5C%20Hz)
A = Amplitude = 0.13 m
k = Spring constant
m = Mass of object = 0.22 kg
Angular velocity is given by
![\omega=2\pi f\\\Rightarrow \omega=2\pi 2.9\\\Rightarrow \omega=18.22123\ rad/s](https://tex.z-dn.net/?f=%5Comega%3D2%5Cpi%20f%5C%5C%5CRightarrow%20%5Comega%3D2%5Cpi%202.9%5C%5C%5CRightarrow%20%5Comega%3D18.22123%5C%20rad%2Fs)
Velocity is given by
![V=A\omega\\\Rightarrow V=0.13\times 18.22123\\\Rightarrow V=2.3687599\ m/s](https://tex.z-dn.net/?f=V%3DA%5Comega%5C%5C%5CRightarrow%20V%3D0.13%5Ctimes%2018.22123%5C%5C%5CRightarrow%20V%3D2.3687599%5C%20m%2Fs)
Speed when it passes the equilibrium point is 2.3687599 m/s
Frequency is given by
![f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow k=m4\pi^2f^2 \\\Rightarrow k=0.22\times 4\pi^2\times 2.9^2\\\Rightarrow k=73.04296\ N/m](https://tex.z-dn.net/?f=f%3D%5Cdfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%7D%5C%5C%5CRightarrow%20k%3Dm4%5Cpi%5E2f%5E2%20%5C%5C%5CRightarrow%20k%3D0.22%5Ctimes%204%5Cpi%5E2%5Ctimes%202.9%5E2%5C%5C%5CRightarrow%20k%3D73.04296%5C%20N%2Fm)
x = Displacement = 0.12 m
In this system the energies are conserved
![\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{73.04296(0.13^2-0.12^2)}{0.22}}\\\Rightarrow v=0.91106\ m/s](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7DkA%5E2%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%2B%5Cdfrac%7B1%7D%7B2%7Dkx%5E2%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B%5Cdfrac%7Bk%28A%5E2-x%5E2%29%7D%7Bm%7D%7D%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B%5Cdfrac%7B73.04296%280.13%5E2-0.12%5E2%29%7D%7B0.22%7D%7D%5C%5C%5CRightarrow%20v%3D0.91106%5C%20m%2Fs)
The speed when it is 0.12 m from equilibrium is 0.91106 m/s
The energy in the system is given by
![E=\dfrac{1}{2}kA^2\\\Rightarrow E=\dfrac{1}{2}\times 73.04296\times 0.13^2\\\Rightarrow E=0.617213012\ J](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B1%7D%7B2%7DkA%5E2%5C%5C%5CRightarrow%20E%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%2073.04296%5Ctimes%200.13%5E2%5C%5C%5CRightarrow%20E%3D0.617213012%5C%20J)
The total energy of the system is 0.617213012 J
Answer: -30kgm/s
Explanation:
Mass M = 5kg
initial velocity V1 = 8m/s
Final velocity V2= 2m/s
Change in momentum = ?
Recall that momentum is the product of mass of an object and the velocity by which it moves.
i.e Momentum = Mass x Change in velocity
Momentum = M x (V2 - V1)
= 5kg x (2m/s - 8m/s)
= 5kg x (-6m/s)
= -30kgm/s (The negative momentum mean the mass is travelling in a negative direction)
Thus, the change in momentum is -30kgm/s