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Semenov [28]
3 years ago
10

An Austin volleyball player bumps a 5 kg ball into the air. It reaches a height of 2.8 meters. How fast was the ball going as it

got bumped into the air?
O 137.2 m/s
O 7.4 m/s
O 19.6 m/s
O 14 m/s
Physics
1 answer:
KATRIN_1 [288]3 years ago
6 0

Answer:

v = 7.4 m/s

Explanation:

Given that,

Mass if a volleyball, m = 5 kg

The ball reaches a height of 2.8 m

We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 2.8} \\\\=7.4\ m/s

So, the required speed is 7.4 m/s. Hence, the correct option is (b).

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in this since your volume remains at a constant you'll need to use Gay-Lussacs law, p1/t1=p2/t2.

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victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

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Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

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