i am right their with you i do believe it is A or D but i am more certain of D
Answer:
The average emf induced in the coil is 175 mV
Explanation:
Given;
number of turns of the coil, N = 1060 turns
diameter of the coil, d = 20.0 cm = 0.2 m
magnitude of the magnetic field, B = 5.25 x 10⁻⁵ T
duration of change in field, t = 10 ms = 10 x 10⁻³ s
The average emf induced in the coil is given by;

where;
A is the area of the coil
A = πr²
r is the radius of the coil = 0.2 /2 = 0.1 m
A = π(0.1)² = 0.03142 m²

Therefore, the average emf induced in the coil is 175 mV
Explanation:
Below is an attachment containing the solution.
Answer:0.967meV
Explanation:
-Find the difference in u, so 130.906118-130.90508= 0.001038u
- convert to meV
1 u = 931.494meV
multiply 0.001038 by 931.494
=0.001038 X 931.494
0.967 meV
Answer:
The work input during this process is -742 kJ
Explanation:
Given;
Initial temperature of nitrogen T₁ = 250 K
final temperature of nitrogen T₂ = 450 K
mass of nitrogen, m = 5 kg

The work input during the process is calculated as;

where;
R is gas constant = 0.2968 kJ/kgK
substitute given values in above equation.

Therefore, the work input during this process is -742 kJ