Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion

Here s = h,u = 450m/s a = -g and t = t+3
Substituting

Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting

Solving both equations

So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s

Passing of B occurs at 4108.31 height.
An arachnid has eight legs.
Answer:
60 kg m/s
Explanation:
Let
be the acceleration of the object.
As the acceleration of the object is constant, so

Given that applied force, F=6.00 N,
From Newton's second law, we have
,
[from equation (i)]


[given that time, t=10 s and F=6 N]

Here mv is the final momentum of the object and mu is the initial momentum of the object.
So, the change in the momentum of the object is mv-mu.
Hence, the change in the momentum of the object is 60 kg m/s.