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3 years ago

A 0.0100-kilogram grape hangs from a vine 1.50 meters from the ground. What is the potential energy of the grape?

2 answers:

8 0

The potential energy is defined as Ep=m*g*h where m is the mass of the body, g=9.81 m/s² and h is the height of the body. In our case m=0.01 kg and h=1.5 m. So when we input the values into the equation:

Ep=0.01*9.81*1.5= 0.14715 J.

So the potential energy of a grape is Ep=0.14715 J.

FromTheMoon [43]3 years ago

3 0

Answer:

0.147

Explanation:

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11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

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A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H

Aleks04 [339]

Answer:

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Original KE = (1/2) (1430 kg) (7.5 m/s)² = 40,218.75 joules

Final KE = (1/2) (1430 kg) (11.0 m/s)² = 86,515 joules

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