Answer:
True
Explanation:
When a satellite is orbiting the earth, the centripetal force is balanced by the gravitational force as :
...........(1)
Where
M is the mass of the earth
m is the mass of the planet
From equation (1), the speed of the satellite depends only on the mass of the earth and the orbital radius.
So, If a payload of material is added until it doubles the satellite's mass, the earth's pull of gravity on this satellite will double but the satellite's orbit will not be affected. It is true.
vf = 10 m/s. A ball with mass of 4kg and a impulse given of 28N.s with a intial velocity of 3m/s would have a final velocity of 10 m/s.
The key to solve this problem is using the equation I = F.Δt = m.Δv, Δv = vf - vi.
The impulse given to the ball with mass 4Kg is 28 N.s. If the ball were already moving at 3 m/s, to calculate its final velocity:
I = m(vf - vi) -------> I = m.vf - m.vi ------> vf = (I + m.vi)/m ------> vf = I/m + vi
Where I 28 N.s, m = 4 Kg, and vi = 3 m/s
vf = (28N.s/4kg) + 3m/s = 7m/s + 3m/s
vf = 10 m/s.
.
Answer:
f = 25 lbs
m = 5 lbs
a =?
f = ma
25 = 5 a( divide both sides by 5)
a = 5(lbs)
There are 60 seconds in a minute.
This means that 5 minutes would be 60 times 5.
60×5 = 300
There are 300 seconds in 5 minutes.
Answer:
The maximum height reached by the water is 117.55 m.
Explanation:
Given;
initial velocity of the water, u = 48 m/s
at maximum height the final velocity will be zero, v = 0
the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².
The maximum height reached by the water is calculated as follows;
v² = u² + 2gh
where;
h is the maximum height reached by the water
0 = u² + 2gh
0 = (48)² + ( 2 x -9.8 x h)
0 = 2304 - 19.6h
19.6h = 2304
h = 2304 / 19.6
h = 117.55 m
Therefore, the maximum height reached by the water is 117.55 m.
