A cabinet weighing 100 N is placed on a floor. The amount of contact area between the cabinet and the floor is 0.5 m2. How much
1 answer:
Answer:
p = 200Pa
Explanation:
From its definition, the pressure depends on the force exerted on the floor by the cabinet, and the perpendicular area in which it is applied. This is expressed in the formula:
Since the force the cabinet exerts on the floor is its weight, we can use the given values to calculate the pressure exerted on the floor:
In words, the pressure the cabinet exerts on the floor is 200Pa.
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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:
Hooke's Law
k = spring constant
x = stretch
F = force
Input the value