Answer:
Explanation:
For answer this we will use the law of the conservation of the angular momentum.
so:
where 
is the moment of inertia of the merry-go-round, 
is the initial angular velocity of the merry-go-round, 
is the moment of inertia of the merry-go-round and the child together and 
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where 
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:
Finally we replace all the data:
Solving for :
<span>Water is never added to earth system. Water forever remains in the water cycle on earth, so it goes from the ground, to the air, to the rain, to the sea, and round and round continuously. This cycle means that there does not need to be new water added to the earth, because it recycles any water that already exists of its own accord.</span>
Answer:
4.0 m/s
Explanation:
The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.
Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is
where here we have
d = 3.0 m is the horizontal distance covered
vx is the horizontal velocity
t = 1.3 s is the duration of the fall
Solving for vx,
Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by
where
h = 4.0 m is the initial height
vy is the initial vertical velocity
We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy
So now we can find the magnitude of the initial velocity:
Explanation:
Given formula:
ME=
mv²+mgh
To make height the subject of the formula, follow the following procedures;
Subtract mv² from both side of equation
M.E - mv² = mv² - mv²+mgh
This gives:
M.E - mv² = mgh
Multiply both sides of the expression by 
( M.E - mv² ) x = x mgh
h = ( M.E - mv² ) x
Learn more:
Kinetic energy brainly.com/question/6536722
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I suppose that you wanted write "uncharged". The particles without electrical charge present in the nucleus are called neutrons.
