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4 years ago

Describe the energy transformations that occur when you bounce a ball?

Physics

2 answers:

Gemiola [76]4 years ago

7 0

The transformation is kinetic energy

Semmy [17]4 years ago

4 0

The potential energy of the ball changes into kinetic energy.

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Two boys, Shrijan having weight 600N and Shrijesh having weight

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Answer:

to be balanced moments should be equal

Explanation:

$fd = fd \\ 600 \times 2 = 300 \times x \\ \frac{600 \times 2}{300} = x \\ 4m$

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3 years ago

Will a dead animal decay faster if it died in the hot summer or in the cold winter?

vivado [14]

Answer:

hot summer

Explanation:

it's common sense

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3 years ago

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If a ball of mass 0.30kg is dropped onto a flat surface, which it hits at 9m/s and rebounds to 3m. Find the rebound speed just a

VikaD [51]

Answer:

vinegar without v and i

Explanation:

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2 years ago

We observe the wheels of buses and trucks are heavier than the wheels of cars and scooters why.

scoray [572]

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Explanation:

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2 years ago

" A 4 kg ball is thrown at an angle of 40 degrees above the horizon from the top of a 150 m cliff. How far from the base of the

TEA [102]

Answer: 180.102m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the ball after it is thrown from the top of the cliff has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=30m/s$ is the ball's initial speed

$\theta=40\°$ is the angle above the horizon

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=150m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin with the calculations:

Firstly, we have to find the time the ball elapsed traveling. So, we will use equation (2) with the conditions given above:

$0=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (3)

Isolating $t$ from (3):

$t=\frac{V_{o}}{g}(sin\theta+\sqrt{{sin\theta}^{2}+2\frac{g.y_{o}}{V_{o}^{2}}})$ (4)

$t=7.83s$ (5)

Substituting (5) in (1):

$x_{max}=30m/s.cos(40\°) (7.83)$ (6)

Finally:

$x_{max}=180.102m$ (7)

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3 years ago