Diameter = 0.170 meter
Circumference = 0.170 π meters
530 rpm = 530 circumferences / minute
= (530 x 0.170 π meters) / minute
= 283.06 meter.minute
= 4.72 meters/second
Sorry don’t know this one
Answer:
Explanation:
Given
Each student exert a force of 
Let mass of car be m
there are 18 students who lifts the car
Total force by 18 students 
therefore weight of car 
mass of car 

(b)
Answer:
A)
= 1.44 kg m², B) moment of inertia must increase
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is
I = ½ m R²
A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is
I =
+ m D²
Let's apply these equations to our case
The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms
=
+ 2
= ½ M R²
The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body
M = 7/8 m total
M = 7/8 64
M = 56 kg
The mass of the arms is
m’= 1/8 m total
m’= 1/8 64
m’= 8 kg
As it has two arms the mass of each arm is half
m = ½ m ’
m = 4 kg
The arms are very thin, we will approximate them as a particle
= M D²
Let's write the equation
= ½ M R² + 2 (m D²)
Let's calculate
= ½ 56 0.20² + 2 4 0.20²
= 1.12 + 0.32
= 1.44 kg m²
b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase
Answer:
Explanation:
Let the volume below water be v . Then
buoyant force = v d g where d is density of water , g is acceleration due to gravity
= v x 1000 x g
weight of wood piece = volume x density of wood x g
= .6 x 600 x g
for equilibrium while floating
buoyant force = weight
= v x 1000 x g = .6 x 600 x g
v = .36 m²
volume above water or volume exposed = .6 - .36
= .24 m²
When immersed completely ,
buoyant force = .6 x 1000 x 9.8
= 5880 N
weight of wood
= .6 x 600 x g
= 3528 N
buoyant force is more than the weight . In order to equalise them for floating with full volume in water
weight required = 5880 - 3528
= 2352 N.