... then your weight is <em>25.2 lbf</em> on the moon.
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
Answer:
Approximately 
.
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant 
near the surface of the earth.
For an object that is accelerating constantly,
,
where
is the initial velocity of the object,
is the final velocity of the object.
is its acceleration, and
is its displacement.
In this case, is the same as the change in the ball's height: 
. By assumption, this ball was dropped with no initial velocity. As a result, 
. Since the ball is accelerating due to gravity, 
.
.
In this case, would be the velocity of the ball just before it hits the ground. Solve for
.
.
