Idrk sorry... i’m doing this bc it’s asking me to. wish i could help you guys . i really do !
As we know by the first law of thermodynamics

here we know that
Q = heat given to the system

W = work done by the system
now here we can say


now we can say that heat will be given as

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy
So best describe Jin's Error is
<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>
Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:


DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation

according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero

solving for x we get

x = 0.200 m
b)electric field at half way mean x =0.25

E = 3.84*10^{-4} N/C
Answer:Given mass = 4.1kg
Radius = 0.0117m
Velocity V = 8.4m/s
Coefficient of friction = 0.25
Explanation: Below is an attached solution to the problem stated above.
1. The angular acceleration is equal to 524rad/s^2
2. The linear acceleration is equal to 2.45m/s^2
3.the time it takes the ball to begin rolling = 0.98s
4. The distance the ball slides before it begins to roll = 7.05m
Answer:
Explanation:
Given that,
We have two capacitors connected in series
C1=2.0-μF
C2=4.0-μF
Then the equivalent of their series connection
1/Ceq = ½ + ¼
1/Ceq= (2+1)/4
1/Ceq=¾
Taking the reciprocal
Ceq= 4/3 μF
The capacitors are connected to a battery of 1kv
V=1000Volts
We know that,
Q=CV
Where Q is charge
C is capacitance and
V is voltage
Then, Q=4/3 ×1000
Q=4000/3 -μC
Since the capacitors are in series, then the charge pass through them, so each charge on the capacitors are 4000/3 μF
After the capacitor has been charge, the capacitor are disconnect and reconnected in parallel to each other,
For parallel connection, they have the same voltage but different charges.
When connected in parallel, there is a charge redistribution,
And the total charge will be 2•4000/3=8000/3 -μF
Then, Q1 +Q2= 8000/3 μF
Now the charge on each capacitor will be, let them have a common voltage V
Q=CV
Then, Q1=C1V
Q1= 2×V=2V
Q2= 4×V=4V
Then, Q1+Q2=8000/3
4V+2V=8000/3
6V=8000/3
V=8000/(3×6)
V=4000/9
V=444.44Volts
Now, Q1=2V
Q1=2×4000/9
Q1=8000/9 μF
Also, Q2=4V
Q2=4×4000/9
Q2=16000/9 μF