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VLD [36.1K]
3 years ago
10

Someone pls help mee

Physics
1 answer:
Alex17521 [72]3 years ago
3 0
This claim isn’t true. This claim is trying to say that once force is being applied to Newton’s Cradle then it will forever stay in motion. However, from the excerpt we learn that this isn’t possible. As one of the balls are pushed, it is set into kinetic energy, and then that ball will hit another and send it into kinetic energy as well. However, not all of the kinetic energy is kept through this process, some of the energy is lost and converted into different forms such as sound energy. Therefore, it isn’t possible for Newton’s Cradle to stay in motion forever.
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3 years ago
A baseball bat changes the momentum of a ball with an impulse of 13.8 Nᐧs. What is the average force that the bat exerts on the
vladimir2022 [97]

Answer:

13800 N

Explanation:

Impulse is the product of average force and time expressed as I=Ft where I is the impulse which results into change in momentum, F is the average force and t is the time of impact. Making F the subject of formula then

F=\frac {I}{t}

Substituting I with 13.8 N.s and time, t witg 0.001 s then the average force is calculated as

F=\frac {13.8 N.s}{0.001}=13800N

Therefore, the average force is equivalent to 13800 N

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3 years ago
A baseball has a mass of 0.15 kg and radius 3.7 cm. In a baseball game, a pitcher throws the ball with a substantial spin so tha
mylen [45]

Answer:

Rotational kinetic energy = 0.099 J

Translational kinetic energy = 200 J

The moment of inertia of a solid sphere is I = \frac{2}{5}mr^2.

Explanation:

Rotational kinetic energy is given by

\text{RKE} = \frac{1}{2}I\omega^2

where <em>I</em> is the moment of inertia and <em>ω</em> is the angular speed.

For a solid sphere,

I = \frac{2}{5}mr^2

where <em>m</em> is its mass and <em>r</em> is its radius.

From the question,

<em>ω</em> = 49 rad/s

<em>m</em> = 0.15 kg

<em>r</em> = 3.7 cm = 0.037 m

\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2

\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}

Translational kinetic energy is given by

\text{TKE} = \frac{1}{2} mv^2

where <em>v</em> is the linear speed.

\text{TKE} = \frac{1}{2} (0.15\ \text{kg})(52\ \text{m/s})^2 = 200\text{ J}

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