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Andrew [12]
3 years ago
14

Neglect air resistance in this question. If you drop a ball from a height of 4.9 m, it will hit the ground 1s later. If you fire

a bullet exactly horizontally from a height of 4.9 m, it will also hit the ground 1s later. Explain.
Physics
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

Explanation:

The time at which an object from from rest depends on air resistance, the height, gravity.

Since the question says we should neglect air resistance and both the ball and the bullet are from the same height they will get to the ground and the same time since they are both fired on earth which has the same gravity..

H=ut+1/2gt^2.

For an object dropping from a height the initial velocity is zero, therefore u=0m/s.

The formula becomes

t=√(2H/g)

So the time and object get to the ground depends on the height from the ground and gravity...

So since no air resistance they will get to the ground together.

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Explanation:

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Answer:

Velocity

Explanation:

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A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
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Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

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F_{Moon}=m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

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5 0
3 years ago
Chris threw a basketball a distance of 27.5 m to score and win his
salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

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g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}

<u>v₀ = 16.55 m/s</u>

8 0
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