Answer:C
Explanation:
It is given that hand strap moves from the vertical in the backward direction.
The direction of strap depends upon the acceleration of bus i.e. if bus is accelerating in forward direction then strap will move in backward direction and vice-versa.
The reason for moving backwards is due to the psuedo acting on strap which bends the strap in backward direction
angle of inclination is given by 
where a=acceleration of bus
=inclination of strap from vertical
so we cannot conclude anything about the direction of the velocity of the bus
(a) The y-component or vertical velocity is calculated using:
Vy = Vsin(∅)
(b) The x-component or horizontal velocity is calculated using:
Vx = Vcos(∅)
Complete Question
For each of the following scenarios, describe the force providing the centripetal force for the motion:
a. a car making a turn
b. a child swinging around a pole
c. a person sitting on a bench facing the center of a carousel
d. a rock swinging on a string
e. the Earth orbiting the Sun.
Answer:
Considering a
The force providing the centripetal force is the frictional force on the tires \
i.e 
where 
is the coefficient of static friction
Considering b
The force providing the centripetal force is the force experienced by the boys hand on the pole
Considering c
The force providing the centripetal force is the normal from the bench due to the boys weight
Considering d
The force providing the centripetal force is the tension on the string
Considering e
The force providing the centripetal force is the force of gravity between the earth and the sun
Explanation:
F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer
Answer:
I = 16 kg*m²
Explanation:
Newton's second law for rotation
τ = I * α Formula (1)
where:
τ : It is the moment applied to the body. (Nxm)
I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)
α : It is angular acceleration. (rad/s²)
Kinematics of the wheel
Equation of circular motion uniformly accelerated :
ωf = ω₀+ α*t Formula (2)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (rad)
Data
ω₀ = 0
ωf = 1.2 rad/s
t = 2 s
Angular acceleration of the wheel
We replace data in the formula (2):
ωf = ω₀+ α*t
1.2= 0+ α*(2)
α*(2) = 1.2
α = 1.2 / 2
α = 0.6 rad/s²
Magnitude of the net torque (τ )
τ = F *R
Where:
F = tangential force (N)
R = radio (m)
τ = 80 N *0.12 m
τ = 9.6 N *m
Rotational inertia of the wheel
We replace data in the formula (1):
τ = I * α
9.6 = I *(0.6
)
I = 9.6 / (0.6
)
I = 16 kg*m²
