initially coin is at rest and then it drop for total time t = 1.5 s
so here the speed of the coin at which it will hit the floor is to be find

here we know that

a = 9.8 m/s^2
t = 1.5 s
now from above equation


so it will hit the floor with speed 14.7 m/s
B-things that can attract iron.
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
Wave speed = frequency * wavelength
Wave speed = 4 * 25
Wave speed = 100 m/s
Answer:
d = 68.18 m
Explanation:
Given that,
Initial velocity, u = 15 m/s
Finally it comes to stop, v = 0
Acceleration, a = -1.65 m/s²
Time, t = 2.5 s
We need to find the distance covered by the hayride before coming to a stop. Let d is the distance covered. Using third equation of motion to find it :

So, the hayride will cover a distance of 68.18 m.