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ycow [4]
3 years ago
10

What is the horizontal component of a ball thrown at a 27 degree angle at 16 m/s?

Physics
1 answer:
Airida [17]3 years ago
6 0

Answer:

14.25 m/s

Explanation:

In this problem, we need to find the horizontal component of a ball thrown at a 27 degree angle at 16 m/s.

It can be given by :

v_x=v\cos\theta\\\\=16\times \cos(27)\\\\=14.25\ m/s

So, the horizontal component of the ball is 14.25 m/s.

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steposvetlana [31]

Answer: KE = 62.5J

Explanation:

Given that

Mass of object = 5kg

kinetic energy KE = ?

velocity of object = 5m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.

i.e K.E = 1/2mv^2

KE = 1/2 x 5kg x (5m/s)^2

KE = 0.5 x 5 x 25

KE = 62.5J

Thus, the object has 62.5 joules of kinetic energy.

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A boy is using a rope to pull a sled to the left on a horizontal surface. The rope is parallel to the surface. What are the dire
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4 years ago
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Answer:

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c. Using the law of momentum conservation:

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