Answer:
this is a law because it is a constant fact of nature
Explanation:
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
Answer:
1470kgm²
Explanation:
The formula for expressing the moment of inertial is expressed as;
I = 1/3mr²
m is the mass of the body
r is the radius
Since there are three rotor blades, the moment of inertia will be;
I = 3(1/3mr²)
I = mr²
Given
m = 120kg
r = 3.50m
Required
Moment of inertia
Substitute the given values and get I
I = 120(3.50)²
I = 120(12.25)
I = 1470kgm²
Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²
