Answer:
1. The gaming company installs games for an arcade in the set of a movie
2. moviemakers might give permission to make a game based on the movie.
3. Advertisements on a game the moviemakers might get a percentage of the money made on a game because there are characters from the movie on the box saying "I played this game and it was awesome,"
Hope this helped! :)
Explanation:
Given Information:
Inductance = L = 5 mH = 0.005 H
Time = t = 2 seconds
Required Information:
Current at t = 2 seconds = i(t) = ?
Energy at t = 2 seconds = W = ?
Answer:
Current at t = 2 seconds = i(t) = 735.75 A
Energy at t = 2 seconds = W = 1353.32 J
Explanation:
The voltage across an inductor is given as

The current flowing through the inductor is given by

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.
![i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\](https://tex.z-dn.net/?f=i%28t%29%20%3D%20%5Cfrac%7B1%7D%7B0.005%7D%20%5Cint_0%5Et%20%5Cmathrm%7B5%281-e%5E%7B-0.5t%7D%7D%29%20%5C%2C%5Cmathrm%7Bd%7Dt%20%5C%2C%2B%200%5C%5C%5C%5Ci%28t%29%20%3D%20200%20%5Cint_0%5Et%20%5Cmathrm%7B5%281-e%5E%7B-0.5t%7D%7D%29%20%5C%2C%5Cmathrm%7Bd%7Dt%20%5C%5C%5C%5Ci%28t%29%20%3D%20200%20%5C%3A%20%5B%20%7B5%5C%3A%20%28t%20%2B%20%5Cfrac%7Be%5E%7B-0.5t%7D%7D%7B0.5%7D%29%5D_0%5Et%20%5C%5Ci%28t%29%20%3D%20200%5Ctimes5%5C%3A%20%5C%3A%20%5B%20%7B%20%28t%20%2B%202e%5E%7B-0.5t%7D%20%2B%202%20%29%5D%20%5C%5C)

So the current at t = 2 seconds is

The energy stored in the inductor at t = 2 seconds is

Answer:
1) the final temperature is T2 = 876.76°C
2) the final volume is V2 = 24.14 cm³
Explanation:
We can model the gas behaviour as an ideal gas, then
P*V=n*R*T
since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:
P*V^k = constant = C, k= adiabatic coefficient for air = 1.4
then the work will be
W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)
W = (P1*V1/T1)*(T2-T1)/(1-k)
T2 = (1-k)W* T1/(P1*V1) +T1
replacing values (W=-450 J since it is the work done by the gas to the piston)
T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C
the final volume is
TV^(k-1)= constant
therefore
T2/T1= (V2/V1)^(1-k)
V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³