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3 years ago

Calculate the kinetic energy of the following object.

Physics

1 answer:

Arte-miy333 [17]3 years ago

5 0

(m.v^2)/2

(5,97x10^24 x 461^2)/2
= 6,34x10^29

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olga_2 [115]

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2 years ago

What is the kinetic energy of a 7.26 kg bowling ball that is rolling at a speed of 2m/s

vlada-n [284]

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 7.26 \times {2}^{2} \\ = 2 \times 7.26$

We have the final answer as

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2 years ago

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TiliK225 [7]

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3 years ago

A car accelerates from 10.0 m/s to 30.0 m/s at a rate of 3.00 m/s2. how far does the car travel while accelerating?

Snezhnost [94]

Answer is in attachment.

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3 years ago

A runner moves 2.88 m/s north. She accelerates at 0.350 m/s^2 at a -52.0 angle. At the point in the motion where she is running

MrRissso [65]

The runner has initial velocity vector

$\vec v_0=\left(2.88\dfrac{\rm m}{\rm s}\right)\,\vec\jmath$

and acceleration vector

$\vec a=\left(0.350\dfrac{\rm m}{\mathrm s^2}\right)(\cos(-52.0^\circ)\,\vec\imath+\sin(-52.0^\circ)\,\vec\jmath)$

so that her velocity at time $t$ is

$\vec v=\vec v_0+\vec at$

She runs directly east when the vertical component of $\vec v$ is 0:

$2.88\dfrac{\rm m}{\rm s}+\left(0.350\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-52.0^\circ)\,t=0\implies t=10.4\,\rm s$

It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time $t$ would be

$\vec x=\vec v_0t+\dfrac12\vec at^2$

so that after 10.4 s, her position would be

$\vec x=(10.1\,\mathrm m)\,\vec\imath+(17.2\,\mathrm m)\,\vec\jmath$

which is 19.9 m away from her starting position.

8 0

3 years ago

Read 2 more answers