Answer:
Explanation:
From newton's equation of motion of uniform acceleration
v = u + at
where v is final velocity , u is initial velocity , a is acceleration and time is t .
putting the values
v = 0 + .5 x 3 x 60 ( time in second = 3 x 60 s )
= 90 m /s
So , final velocity is 90 m /s .
Answer:

It will take 10 seconds to travel 200m at a speed of 10m/s
Explanation:
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Complete Question
A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.
Answer:
The value 
Explanation:
From the question we are told that
The charge on the first sphere is 
The charge on the second sphere is 
The mass of the second charge is 
The distance apart is 
The speed of the second sphere is 
Generally the total energy possessed by when
and
are separated by
is mathematically represented

Here KE is the kinetic energy which is mathematically represented as

substituting value


And U is the potential energy which is mathematically represented as

substituting values


So


Generally the total energy possessed by when
and
are separated by
is mathematically represented

Here
is the kinetic energy which is mathematically represented as

substituting value


And
is the potential energy which is mathematically represented as

substituting values


From the law of energy conservation

So


Answer:
The depth is 5.15 m.
Explanation:
Lets take the depth of the pool = h m
The atmospheric pressure ,P = 101235 N/m²
The area of the top = A m²
The area of the bottom = a m²
Given that A= 1.5 a
The force on the top of the pool = P A
The total pressure on the bottom = P + ρ g h
ρ =Density of the water = 1000 kg/m³
The total pressure at the bottom of the pool = (P + ρ g h) a
The bottom and the top force is same
(P + ρ g h) a = P A
P a +ρ g h a = P A
ρ g h a = P A - P a




h=5.15 m
The depth is 5.15 m.
A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:
- From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
- From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.
We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:
- The speed increases from 0 m/s to 1341 m/s.
- The time elpased is 2.0 min.
- 1 min = 60 s.
The acceleration of the ship during the first 2.0 minutes is:

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
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