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.
Answer:
Part a)
Part b)
Explanation:
Part a)
Level of sound = 75 dB
now we know that
here we know that
now we have
Part b)
Intensity of sound wave is given as
here we know that
so we have
now we know
now we have
Fill in the fraction: 3,600/90 = 40; turn it into a unit fraction.
40 mi/min
From A to B its 5 ohm.
above shown 6 and 12 ohm resistors are in parallel to short circuit hence their equivalent resistance is zero.
(Current doesnt flow through a resisstor if there is a Short circuit alternate.
Answer: Diagram B
Explanation:
A free body diagram shows the forces acting on an object in a certain scenario.
In this scenario there are two forces acting on the carrot: the Tension force (Ft) from the rope that the carrot is hanging from and Gravitational force(Fg) which is pulling the carrot to the Earth.
The diagram depicting this is diagram B.
