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andrew-mc [135]
3 years ago
6

A stone is dropped from a tower of height 3600cm.Find the velocity by which it strike the ground.Also,find the time taken to rea

ch the ground.​
Physics
2 answers:
andriy [413]3 years ago
5 0

Answer:

Explanation:

Here h=36m, a=g=10m/s, u=0

Use s=ut+(1/2)at^2

36=0*t+(10t^2)/2

time=2.68 sec

As from the formula of final velovity, v=(2gh)^1/2

v=(2*10*36)^1/2

v=(720)^1/2

v=26.83………This is the final velocity

enyata [817]3 years ago
5 0

Answer:

Explanation:

s = vit + 1/2 gt*2

36 = 0 + 1/2 x 9.8 x t*2

t*2 = 7.34

t = 2.7sec

now,

2gs = vf*2 - vi*2

2 x 9.8 x 36 = vf*2 - 0

vf*2 = 705.6

vf = 26.5m/s

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A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
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a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

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Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

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v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

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∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

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h=h_{f} = 198km = 198000 m

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a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

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                                                          = -ΔE                                                            

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