Answer:
a) 2250 J
b) 0 J
c) 2250 J
Explanation:
a) Since, the process is isochoric
the change in internal energy

Here, n = 0.2 moles
Cv = 12.5 J/mole.K
We have to find T_f so we can use gas equation as
![\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2 [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BP_2V_2%7D%20%3D%5Cfrac%7BT_i%7D%7BT_f%7D%5C%5CSince%2C%20V_1%3DV_2%20%20%20%20%5Bisochoric%2Fprocess%5D%5C%5C%5CRightarrow%20%5Cfrac%7BP_%7Batm%7D%7D%7B4P_%7Batm%7D%7D%20%3D%20%5Cfrac%7B300%7D%7BT_f%7D%20%5C%5C%5CRightarrow%20T_f%20%3D%201200%20K)
So, 
b) Since, the process is isochoric no work shall be done.
c) By first law of thermodynamics we have

Since, Q is positive 2250 J of heat will flow into the system.
Answer:
circuit sketched in first attached image.
Second attached image is for calculating the equivalent output resistance
Explanation:
For calculating the output voltage with regarding the first image.

![Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V](https://tex.z-dn.net/?f=Vout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5B%2F%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5DVout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5C%5CVout%20%3D%205%20%5Cfrac%7B2%7D%7B5%7D%20%3D%202%20V)
For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.
so.

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.
if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:




so.

(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)
</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.
Hope this helps :)