by angular momentum conservation we will have
angular momentum of child + angular momentum of merry go round = 0
angular momentum of child = mvR
m = mass of child
R = radius of child
v = speed = 2 m/s
now let's say moment of inertia of merry go round is I
so we will have


so merry go round will turn in opposite direction with above speed
Answer:
The answer <em><u>is C. Mars</u></em>. Mars and Mercury are both smaller than Earth's core. Hope this helps you :)
Answer:
4960 N
Explanation:
First, find the acceleration.
Given:
v₀ = 6.33 m/s
v = 2.38 m/s
Δx = 4.20 m
Find: a
v² = v₀² + 2aΔx
(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)
a = -4.10 m/s²
Next, find the force.
F = ma
F = (1210 kg) (-4.10 m/s²)
F = -4960 N
The magnitude of the force is 4960 N.
Answer:
Deltoid Force, 
Additional Information:
Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.
Explanation:
The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.
1. The torque about the point in the shoulder for the deltoid muscle,
2. The torque of the arm,
Assuming the arm is just being stretched and there is no rotation going on,
= 0
= 0
⇒ 

Where,
is radius of the deltoid
is the force of the deltiod
is the angle of the deltiod
is the radius of the arm
is the force of the arm ,
which is the mass of the arm and acceleration due to gravity
is the angle of the arm
The force of the deltoid muscle is,

but
,
∴ 
Answer:
1. 2.5s
Explanation:
1. For time, divide Distance / speed
25m / 10
=2.5s