Answer:
Induced emf, 
Explanation:
We have,
Number of turns in the coil, N = 40
Radius of coil, r = 3 cm = 0.03 m
The field increases from 0 to 0.75 T at a constant rate in a time interval of 225 s.
It is required to find the magnitude of the induced emf in the coil if the field is perpendicular to the plane of the coil. The induced emf is given by :
, is magnetic flux
So, the magnitude of induced emf is 
.
Answer:
V₁ = 5.6 m/s
V₂ = 7.2 m/s
V₃ = 8.8 m/s
Explanation:
Average velocity: Average velocity can be defined as the ratio of the total displacement to the total time taken. The S.I unit of Average velocity is m/s.
For the first 2 s,
V₁ = Δd₁/t
Where V₁ = Average velocity for the first 2 s
Where Δd₁= distance, t = time
Δd₁ = 25.6-14.4 = 11.2 m t = 2 s
V₁ = 11.2/2
V₁ = 5.6 m/s
For the second 2 s,
V₂ =Δd₂/t
Where V₂ = average velocity for the second 2 s.
Δd₂= 40-25.6 = 14.4 m, t= 2 s
V₂ = 14.4/2
V₂ = 7.2 m/s
For the last 2 seconds,
V₃ =Δd₃/t
Where V₃ = average velocity for the last 2 s
where Δd₃ = 57.6- 40 = 17.6 m, t = 2 s
V₃ = 17.6/2
V₃ = 8.8 m/s.
Newton’s second law states that force is the product of
mass and acceleration. This is expressed mathematically as:
F = m * a
Where F = force, m = mass, a = acceleration
Since in the given problem, the force is constant or same
force is acting upon two objects. Therefore we can simply equate the expression
m * a of the 2 objects.
m1 * a1 = m2 * a2
Where m1 = 2 kg, a1 = 38 m/s^2, m2 = 19 kg. Therefore
finding for a2:
2 kg * 38 m/s^2 = 19 kg * a2
<span>a2 = 4 m/s^2 (ANSWER)</span>
