Answer: 7.5 rev/s
Explanation:
We are given the angular velocity
a helicopter's main rotor blades:

However, we are asked to express this
in the International Systrm (SI) units. In this sense, the SI unit for time is second (
):


The addition of any numbers of vector provide the magnitude as well as the direction of the resultant vector, hence the mentioned first option is not true.
The addition of vector required to connect the head of the one vector with the tail of the other vector and any vector can be moved in the plane parallet to the previous location, so, the mentioned second and third options are true.
Answer:
D) The ball exerts a force on the wall and the wall exerts a force back.
Explanation:
Newton's third law of motion states that:
"When an object A exerts a force on another object B, then object B exerts an equal and opposite force on object A"
In this problem, we can identify (for instance) object A with tha ball and object B with the wall. Therefore, if we apply Newton's third law, we get:
The ball (object A) exerts a force on the wall (object B), therefore the wall (object B) exerts an equal and opposite force on the ball (object A). So, option D is the correct one.
The correct answer is D) The closet point in the Moon's orbit to Earth
This does not refer to the Moon only. It refers to any satellite and to its closest point to Earth.
Answer:
M
Explanation:
To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after
As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved
⇒
-
is the moment of interia of earth before impact -
is the angular velocity of earth about an axis passing through the center of earth before impact
is moment of interia of earth and asteroid system
is the angular velocity of earth and asteroid system about the same axis
let 
since 

⇒ if time period is to increase by 25%, which is
times, the angular velocity decreases 25% which is
times
therefore

(moment of inertia of solid sphere)
where M is mass of earth
R is radius of earth

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)
where
is mass of asteroid
⇒ 

=
+ 

⇒
