3 years ago

Сс Explain the effect of Earth's gravity on objects on the surface of Earth.

1 answer:

6 0

Answer:

The effect of gravity on object makes objects stable and not floating in the air . It also makes any object thrown up to return back

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A ball was dropped from a height of 10 feet. Each time it hits the ground, it bounces 4/5 of its previous height. Find the total

Shtirlitz [24]

Answer:

d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

so we can say

$h_2 = \frac{4}{5}h$

so the distance covered is given as

$d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........$

here we know that

h = 10 feet

$d = h + 2(\frac{4}{5}h)(1 + \frac{4}{5} + (\frac{4}{5})^2 + ...........)$

$d = 10 + 2(\frac{4}{5}(10))(\frac{1}{1 - \frac{4}{5}})$

$d = 90 ft$

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3 years ago

What roles do the musk ox play in the tundra ecosystem? (Select all that apply.)

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Answer:

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3 years ago

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The velocity of a moving body increases from 10m/s to 15 m/s in 5 sec . calculate its acceleration

Gnoma [55]

Answer:

v=u+at

15=10+a x 5

15-10= 5a

5= 5a

a=1 m/sec^2

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3 years ago

Jana is observing a young autistic boy in a preschool classroom. Her job is to note each time the boy gets up from his chair wit

svp [43]

The answer is tallying

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4 years ago

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

v = R ω

using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

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3 years ago