Answer:
t = 1.16 s.
Explanation:
Given,
speed of conveyor belt, v = 3.2 m/s
coefficient of friction,f = 0.28
Using newton second law
f = ma
and we also know that frictional force
f = μ N
f = μ m g
equating both the force equation
a = μ g
a = 0.28 x 9.81
a = 2.75 m/s²
Using Kinematic equation
v = u + at
3.2 = 0 + 2.75 x t
t = 1.16 s.
Time taken by the box to move without slipping is 1.16 s.
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:
a dog walking or their phone rings or heard a neighbor talking to them
Answer:
Acceleration
Explanation:
Its speed or velocity change
