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seraphim [82]
3 years ago
9

What law states that, in any process, energy is neither created nor destroyed?

Physics
1 answer:
kupik [55]3 years ago
8 0
The law<span> of conservation of </span><span>energy.</span>
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The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time. According To a website on google 
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How fast would you have to jump off the ground in order to jump at least 10 meters in the air?
ivanzaharov [21]

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30.5

Explanation:

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What is the kinetic energy of a soccer ball which has a mass of 0.8 kg and is kicked at a velocity of 10 m/s
Lubov Fominskaja [6]

Answer:

Explanation:

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A diver can change his rotational inertia by drawing his arms andlegs close to his body in the tuck position. After he leaves th
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Answer:

3.14946 rad/s

Explanation:

I_i = Intial moment of inertia

I_f = Final moment of inertia

\omega_i = Initial angular velocity

\omega_f = Final angular velocity = \dfrac{2}{1.33}\times 2\pi\ rad/s

\dfrac{I_f}{I_i}=\dfrac{1}{3}

In this system the angular momentum is conserved

L_i=L_f\\\Rightarrow I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_i=\dfrac{I_f\omega_f}{I_i}\\\Rightarrow \omega_i=\dfrac{1\times \dfrac{2}{1.33}\times 2\pi}{3}\\\Rightarrow \omega_i=3.14946\ rad/s

The angular velocity when the diver left the board is 3.14946 rad/s

3 0
3 years ago
A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent
solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

Kinetic Energy K.E

Solution

From the moment of inertia

I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}

The angular acceleration is given as

a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}

The angular velocity is given as

w=at\\w=(0.905)(2.55)\\w=2.31rad/s

So the Kinetic Energy is given as

K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J

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