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e-lub
1 month ago
12

If Siobhan hits a 0.25 kg volleyball with 0.5 N of force, what is the acceleration of the ball?

Physics
2 answers:
Keith_Richards [3.2K]1 month ago
8 0

Answer:

2 m/s²

Explanation:

Maru [3.3K]1 month ago
7 0

Answer:2 m/s²

Explanation:

mass=0.25 kg

Force =0.5 N

Acceleration=force/mass

Acceleration=0.5/0.25

Acceleration=2

Acceleration =2 m/s²

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a stomp rocket takes 1.5 seconds to reach its maximum height what was the initial velocity and what was the maximum height ?
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1) Vf = Vo - gt; Setting Vf = 0 gives Vo = gt, resulting in Vo = 9.8 m/s^2 * 1.5 s = 14.7 m/s. 2) The displacement is calculated as d = Vo*t - gt^2 / 2 = 14.7 m/s * 1.5 - 9.8 m/s^2 * (1.5 s)^2 / 2 = 11.02 m.
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2 months ago
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1. A U.S. manufacturer wants to sell computer equipment to companies in Honduras. Because the U.S. government limits trade with
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2 months ago
A crow drops a 0.11kg clam onto a rocky beach from a height of 9.8m. What is the kinetic energy of the clam when it is 5.0m abov
kicyunya [3294]

Solution:

The kinetic energy of the clam at an elevation of 5.0 m is 5.19 J and the velocity of the clam at that height is 9.71 m/s.

Explanation:

Throughout its motion, mechanical energy remains constant. We understand that mechanical energy is the summation of potential energy and kinetic energy. Potential energy = m \times g \times h, Kinetic energy = \frac{1}{2} \times m \times v^{2} and Mechanical energy = m \times g \times h+\frac{1}{2} \times m \times v^{2} Initial kinetic energy is zero. At a height of 9.8 m, the mechanical energy of the clam with a mass of 0.11 kg and g=9.81\frac{m}{s^{2}} is calculated as follows: 0.11×9.81×9.8 = 10.58 J.

Mechanical energy of the clam at a height of 5.0 m = 0.11 \times 9.81 \times 5+\frac{1}{2} \times m \times v^{2} = 5.39+\frac{1}{2} \times m \times v^{2}. Given that mechanical energy is conserved, we can state that the mechanical energy of the clam at a height of 9.8 m is equal to that at 5.0 m. The representation is as follows:

10.58 = 5.39+\frac{1}{2} \times m \times v^{2} 10.58 – 5.39 = \frac{1}{2} \times m \times v^{2}  5.19 = \frac{1}{2} \times m \times v^{2} the clam's kinetic energy measures 5.19 J.

Lastly, the speed of the clam at 5.0 m is computed; thus, 5.19 = \frac{1}{2} \times 0.11 \times v^{2} \frac{5.19 \times 2}{0.11}=v^{2} 94.36 = v^{2} \sqrt{94.36}=v \quad v= 9.71 m/s. The clam's speed is determined to be 9.71 m/s.

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2 months ago
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