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
<em>The answer is </em>Ninth <em>and </em>Tenth <em>grade so the answer would be</em> B
<em>I hope this helps you </em>
Answer:
C). Take your foot off the gas pedal. Then brake lightly until you are moving at low speed.
Explanation:
While driving on roads of rural areas, if our right wheel moves off the pavement, we should always hold the steering wheel firmly and then take our foot off the gas pedal, then apply brake lightly until we are moving at a low speed.
When our wheels drift off the pavement area, we should not panic and yank. And instead of turning the wheel back in the left direction towards the road, it is always safer to take off our foot from the gas pedal and then apply brakes slowly. When our vehicle slows down check the incoming traffic behind us and then we should slowly move back on to the pavement.
Answer:
- 5436 J
Explanation:
mass of car, m = 120 kg
radius of loop, r = 12 m
velocity at the bottom (A) = Va = 25 m/s
Velocity at the top(B) = Vb = 8 m/s
Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m
by use of Work energy theorem
Work done by all the forces = change in kinetic energy of the body
Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A
- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)
- 120 x 9.8 x 24 + Work done by friction = 60 x (64 - 625)
- 28224 + Work done by friction = - 33660
Work done by friction = -33660 + 28224 = - 5436 J
Answer: 
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas at STP = 1 atm
= final pressure of gas = 2.67 atm
= initial volume of gas = 
= final volume of gas = ?
= initial temperature of gas at STP = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
Thus the final volume will be
