With what magnitude of force does a ball of mass 0.75 kilograms need to be hit so that it accelerates at the rate of 25 meters/
1 answer:
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The sound power the person generated is .
The area of the sound wave is calculated as follows;
The sound power the person generated is calculated as follows;
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Answer:
|F| = 393750 N
Explanation:
Given that,
Total mass of the train, m = 750000 kg
Initial speed, u = 84 m/s
Final speed, v = 42 m/s
Time, t = 80 s
We need to find the net force acting on the train. The formula for force is given by :
F = ma
So, the magnitude of net force is 393750 N.
The average force of the water droplets is the force given by the impact
per second of the droplets on the limestone floor.
- The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>
Reasons:
The given parameters are;
Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³
Height from which the water falls, <em>h </em>= 5 meters
Rate at which the water falls = 10 per minute
Required:
The average force exerted on the floor by the water droplets.
Solution:
According to Newton's Second Law of motion, we have;
Force = Rate of change of momentum
Momentum = Mass × Velocity
Mass of a droplet of water = Volume × Density
Density of water = 997 kg/m³
Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg
The velocity just before the droplet reaches the ground, v = √(2·g·h)
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Which gives;
v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s
The rate of change in momentum per minute = 1
Therefore;
ΔMomentum = Mass × ΔVelocity
Considering the 10 drops per minute, we have;
ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s
ΔTime = 1 minute = 60 seconds
Therefore;
- The average force exerted on the limestone floor by the droplets of water is
≈ <u>1.6013 × 10⁻² N</u>
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