Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J
Solution :
(1) The equation used is,

where,
= final internal energy
= initial internal energy
q = heat energy
w = work done
(2) The known variables are, q, w and 
initial internal energy =
= 2000 J
heat energy = q = 1000 J
work done = w = 500 J
(3) Now plug the numbers into the equation, we get

(4) By solving the terms, we get




(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J
No I don’t think so. But it worth a try tho. Try it out.
Answer:
(D) energy from one place to another
Answer:
T=1022.42 N
Explanation:
Given that
l = 32 cm ,μ = 1.5 g/cm
L =2 m ,V= 344 m/s
The pipe is closed so n= 3 ,for first over tone


f= 129 Hz
The tension in the string given as
T = f²(4l²) μ
Now by putting the values
T = f²(4l²) μ
T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100
T=1022.42 N
It i believe it would be 7.3 × 10∧3.
correct me if im wrong