' +4 m/s² ' means that the pigeon's speed is 4 m/s greater every second.
Starting from zero speed, after 10 seconds, its speed is
(10 x 4m/s) = 40 m/s.
We can't say anything about its velocity, because we have
no information regarding the direction of its flight.
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:
T is less than or equal to 19 N
Explanation:
It's one of your hands. Which one it is depends on how you sweep.
-- If you hold the top of the stick motionless and wave your bottom hand
back and forth, then your top hand is the fulcrum, and you're using the
broom as a Class-3 lever.
-- If you hold your bottom hand motionless and wiggle the top end of the
broom back and forth with your top hand, then your lower hand is the fulcrum,
and you're using the broom as a Class-1 lever.
