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scZoUnD [109]
4 years ago
15

A super ball is dropped from a height of 100 feet. Each time it bounces, it rebounds half the distance it falls. How many feet w

ill the ball have traveled when it hits the ground for the fourth time
Physics
1 answer:
Crank4 years ago
4 0

The total distance travelled by the ball after the fourth impact is 275 feet.

<u>Explanation:</u>

Given-

Height, h = 100 feet

Rebounds half the distance

Distance in feet for the fourth time, x = ?

For the first time, the distance travelled by the ball is, x = 100 feet

For the second time, it will bounce up to 50 feet and fall upto 50 feet( half of 100 feet)

So, the distance travelled after the second impact, x = 100 + 50 + 50 = 200 feet

For the third time, it will bounce up to 25 feet and fall upto 25 feet( half of 50 feet)

So, the distance travelled after the third impact, x = 200 + 25 + 25 = 250 feet

For the fourth time, it will bounce up to 12.5 feet and fall upto 12.5 feet( half of 25 feet)

So, the distance travelled after the fourth impact, x = 250 + 12.5 + 12.5 = 275 feet

Therefore, total distance travelled by the ball after the fourth impact is 275 feet.

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Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite
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Answer:

The required work done is 6.5\times10^{9}\ J

Explanation:

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U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}

U_{A}=-3.44\times10^{10}\ J

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Using formula of potential

U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}

Put the value into the formula

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U_{B}=-2.14\times10^{10}\ J

We need to calculate the value of k_{A}

Using formula of k_{A}

k_{A}=-\dfrac{1}{2}U_{A}

Put the value into the formula

k_{A}=\dfrac{1}{2}\times3.44\times10^{10}

k_{A}=1.72\times10^{10}\ J

We need to calculate the value of k_{B}

Using formula of k_{B}

k_{B}=-\dfrac{1}{2}U_{B}

Put the value into the formula

k_{B}=\dfrac{1}{2}\times2.14\times10^{10}

k_{B}=1.07\times10^{10}\ J

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W=\Delta K+\Delta U

W=(k_{B}-k_{A})+(U_{B}-U_{A})

W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})

W=\dfrac{1}{2}(U_{B}-U_{A})

Put the value into the formula

W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})

W=6.5\times10^{9}\ J

Hence, The required work done is 6.5\times10^{9}\ J

7 0
3 years ago
A block with a weight of 3.0 N is at rest on a horizontal surface.A 1.0 N upward force is applied to the block by means of an at
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Answer:

Explanation:

Given

weight of block W=3\ N

A force is of F=1 N is applied on the block

As 1 N is less than weight of block so block exert  Force less than its weight

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8 0
4 years ago
!!Help!!
Ira Lisetskai [31]

Answer:

s = 37.81 m

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Mass of an apple, m = 1.5 kg

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s=ut+\dfrac{1}{2}gt^2

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s=\dfrac{1}{2}gt^2\\\\s=\dfrac{1}{2}\times 10\times (2.75)^2\\\\=37.81\ m

So, Apple dropped 37.81 m before hitting the ground.

4 0
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