Work = force x distance
200 Newtons x 20 meters
= 4,000 Joules
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and
22.5 J
Explanation:
Given:
x = 3 m

The spring potential energy
is


Intensity of sunlight at given position is defined as power received per unit area
so here we can say

area on which photons are received is given as

now we can find the power received due to sunlight



now we can say this power is due to photons that strikes on surface of earth
so here we can say

given here that





so it will strike 2.47 * 10^18 photons on given area per second