Answer:
Part a)

Part b)

Explanation:
Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

here we also use angular momentum conservation
so we have

also we know that the collision is elastic collision so we have

so we have

also we know

also we know

so we have


now we have


Part b)
Now we know that speed of the ball after collision is given as

so it is given as

Does this help?
When an object is
immersed in a fluid (in this case water, but may include both liquids and
gases) the fluid exerts an upward force on the object which is called buoyancy
force or <span>up-thrust. Archimedes’ Principle states that the buoyant
force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by
that object. Thus when an object is
placed in water the rise in the water level is dictated by the mass of that
object.</span>
<span>
</span>
<span>So for example if you fill a bucket with water and you drop a stone in that bucket, if you measure the weight of the water that overflows from the bucket due to the stone being dropped into the bucket is equivalent to the pushing force that the water has on the stone (as the stone drops to the bottom of the bucket the water is pushing it to stay afloat but the rock is more dense than water and as such its downthrust exceeds water's upthrust).</span>
Answer:
v = 384km/min
Explanation:
In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.
You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

hence, the speed of the Hubble is approximately 384km/min
Answer:
Block A
Explanation:
Block A will float higher in the water compared to the second Block.
The density of water is 1g/cm³.
According to the principle of floatation "an object that floats in a liquid will displace equal amount of fluid to the weight of the object".
A body will become more submerged in water if it has more density because density is the mass per volume of body.
An object with a higher density than another will sink in the liquid of the one with lesser density.
- Object A has lesser density and will float higher up and displace very little water.
- Object B has higher density and will be more submerged.