Just like mass, energy, linear momentum, and electric charge, angular momentum is also conserved.
The wheel has angular momentum. I don't remember whether it's
up or down (right-hand or left-hand rule), but it's consistent with
counterclockwise rotation as viewed from above.
When you grab the wheel and stop it from spinning (relative to you),
that angular momentum has to go somewhere.
As I see it, the angular momentum transfers through you as a temporary
axis of rotation, and eventually to the merry-go-round. Finally, all the mass
of (merry-go-round) + (you) + (wheel) is rotating around the big common
axis, counterclockwise as viewed from above, and with the magnitude
that was originally all concentrated in the wheel.
It's the natural tendency of things to keep going unless there's something trying to stop them.
It's usually called "inertia".
Don't get the idea from all of this that things stop unless there's something to keep them going. The truth is exactly the opposite: Things keep going unless there's something to make them stop.
An electric motor is an electrical machine that converts electrical energy into mechanical energy.
Electric motors work on the principal of the interaction between magnetic field electr-magnetism. A loop which is carrying the current is placed in a magnetic field. The loop will experience a torque. The torque starts rotating the coil and the propellers start to rotate when the current passes through the loop.
Answer:
a) 20s
b) 500m
Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.
To find time, we apply the UARM formula:
v final = (a x t) + v initial
Replacing the values gives us:
0 = (-10 x t) + 100
-100 = -10t
t = 10s
It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur
So 10s going up and another 10s going down:
10x2 = 20s
b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:
Δy = (1/2)(a)(t^2) + (v initial)(t)
Replacing the values gives us:
Δy = (1/2)(-10)(10^2) + (100)(10)
= (-5)(100) + 1000
= -500 + 1000
= 500 m
Hope this helps, brainliest would be appreciated :)
