In establishing standing waves with a Slinky®, your wrist oscillates with f frequency of 4.0 Hz, and the length between any two
1 answer:
You might be interested in
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 energy in its ground state is 10 meV.
Explanation:
It is given that,
The energy of the electron in its first excited state is 40 meV.
Energy of the electron in any state is given by :
For ground state, n = 1
.............(1)
For first excited state, n = 2
.............(2)
Dividing equation (1) and (2), we get :
So, the energy in its ground state is 10 meV. Hence, this is the required solution.