Kinetic energy= 1/2 m v^2
so... 1/2(1350)(12^2)
kinetic energy = 97200Joules
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
:

The loss of matter is called the mass defect. The missing matter is converted into energy. You can actually calculate the amount of energy produced during a nuclear reaction with fairly simple equation developed by Albert Einstein; E = mc^2. In this equation, E is the amount of energy produced, m is the missing mass, or the mass defect, and c is the speed of light, which is a rather large number. The speed of light is squared, making that part of the equation a very large number that, even when multiplied by a small amount of mass, yields a large amount of energy.
The weight of the box is <em>w</em> = <em>mg</em>, where <em>m</em> is the mass. So
<em>m</em> = <em>w</em>/<em>g</em> = (3893.40 N) / (9.80 m/s²) ≈ 397 kg
Then the box has density
(397 kg)/(4.60 m³) ≈ 86.4 kg/m³
which is less than the density of the given liquid, so the box will float.
Answer:
The pendulum frequency is (c) the same, or very close to it
Explanation:
The simple pendulum corresponds to a simple harmonic movement, to reach this approximation in the expression of the force the sine of the angle (θ) approaches an angle value, this is only true for small angles, generally less than 15º
Sine (15th) = 0.2588
The angle in radians is 15º π / 180º = 0.26180.2588 / 0.2618
The difference between these two values is less than 1.2%
for smaller angle the difference is reduced more
Therefore, the period for both the 5º and 10º angles is almost the same