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
What do you mean? Is there a picture ?
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.
Answer:
0.66c
Explanation:
Use length contraction equation:
L = L₀ √(1 − (v²/c²))
where L is the contracted length,
L₀ is the length at 0 velocity,
v is the velocity,
and c is the speed of light.
900 = 1200 √(1 − (v²/c²))
3/4 = √(1 − (v²/c²))
9/16 = 1 − (v²/c²)
v²/c² = 7/16
v = ¼√7 c
v ≈ 0.66 c
Answer:
The answer is A good luck :P
