Because of the costs involved in maintaining high temperatures and pressure, nuclear method for generating electrical energy. is
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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>
Learn more about Newton's Second Law of motion and force exerted water here:
Answer:
The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>
Explanation:
Given:
Time taken by the ball to reach maximum height is,
We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is,
Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is,
The negative sign is used as acceleration is a vector and it acts in the downward direction.
Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:
Where, 'u' is the initial velocity.
Plug in the given values and solve for 'u'. This gives,
Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s