Answer:
a) 86 atm
b) 86 atm
c) 645 m/s
Explanation:
See attachment for calculations on how i arrived at the answer
Answer:
d. 332 V
Explanation:
Given;
number of turns in the wire, N = 40 turns
area of the coil, A = 0.06 m²
magnitude of the magnetic field, B = 0.4 T
frequency of the wave, f = 55 Hz
The maximum emf induced in the coil is given by;
E = NBAω
Where;
ω is angular velocity = 2πf
E = NBA(2πf)
E = 40 x 0.4 x 0.06 x (2 x π x 55)
E = 332 V
Therefore, the maximum induced emf in the coil is 332 V.
The correct option is "D"
d. 332 V
Answer:
595391.482946 m/s

Explanation:
E = Energy = 1.85 keV
I = Current = 5.15 mA
e = Charge of electron = 
t = Time taken = 1 second
m = Mass of proton = 
Velocity of proton is given by

The speed of the proton is 595391.482946 m/s
Current is given by

Number of protons is

The number of protons is 
"<span>The current is the same at all points" is the one among the following choices given in the question that answers the question correctly. The correct option among all the options that are given in the question is the fifth option or the last option. I hope that this is the answer that has come to your desired help.</span>
F = 750 N (Force)
d = 10 m (displacement
)
t = 25 s (time)
L = ? (Mechanical work
) = (Energy)
P = ? (Power)
Solve:
L = F × d = 750 × 10 = 7500 Joules
P = L / t = 7500 / 25 = 300 Watts