Answer:
No..
Explanation:
As the bird releases the drop there is no internal force which will drive it into circular path but it will fall on tangent of the arc at the point of release because it has a tangential velocity same as bird. Path will be parabola in vertical plane.
As the person is on circular arc constantly moving it will never meet that drop.
Answer:
Explanation:
The momentum of the 25 kg mass is


If this whole momentum of the object is transferred to the 5.0 kg object then according to the law of conservation of momentum, the momentum of the 25.0 kg object must be transferred to the 5.0 kg object:



Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:


Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:


Where:
d: is the distance traveled = 46.1 m
: is the final speed of the truck = 0 (it stops)
: is the initial speed of the truck = 17.9 m/s
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!
Answer:
James Chadwick
Explanation:
In May 1932 James Chadwick announced that the core also contained a new uncharged particle, which he called the neutron
Answer: 27 joules
Explanation:
Work is done when force is applied on the bench over a distance. it is measured in joules.
Workdone = force x distance
= 45 N x 0.6 metres
= 27 joules
Thus, 27 joules of work is done on the bench.