Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Actually what the problem meant about the westward
component of the ball’s displacement is the horizontal component of the
displacement. To help us better understand the problem, I attached a figure of
the situation.
We can see from the figure that to solve for the value of
the horizontal component, we have to make use of the sin function. That is:
sin θ = side opposite to the angle / hypotenuse of the
triangle
sin 42 = x / 40 m
x = (40 m) sin 42
x = 26.77 m
Therefore the ball has a westward
displacement of about 26.77 m
Answer:
Time - taken = 2.5 s
deceleration= -8 m/s²
Solution:
Given:
speed, v = 8 m/s
distance, d = 20m
To Find:
deacceleration = ?
As we know speed is defined as
v = d/t
plugging in the values
t = 20/ 8
t = 2.5s
Now from deceleration formula
a = - v/ t
a = - 20/ 2.5
a = - 8 m/s²
Thus, the time taken and acceleration is 2.5 s and -8 m/s²
respectively.
Learn more about deceleration here:
brainly.com/question/13354629
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