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VikaD [51]
3 years ago
12

A ball is droped from a height of 16m how much time will pass before the ball hits the ground​

Physics
1 answer:
sergey [27]3 years ago
4 0

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

h= \frac{1}{2}g{t}^{2}

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

g = 9.8m{s}^{-2}

By substitution we obtain,

16= \frac{1}{2}\times 9.8{t}^{2}

\implies32=9.8{t}^{2}

Diving through by 9.8

\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}

\implies{t}^{2} =3.265

square root both sides, we obtain

\implies t= \sqrt{3.265}

t=1.8s

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andrew-mc [135]

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Power = 1.4 kW  =  1,400 watts

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-- <u>For the CFL bulb:</u>

Power = 11 watts

Time = 10.5 hours = 37,800 seconds

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-- The toaster uses energy at 127 times the rate of the CFL bulb.

-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.

-- The toaster is used for 0.0086 times as long as the CFL bulb.

-- The CFL bulb is used for 116.7 times as long as the toaster.    

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A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous
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Answer:

42244138.951 m

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r = Radius of orbit from center of earth

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m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

v=\frac{2\pi \times r}{24\times 3600}

The Centripetal force of the satellite is balanced by the universal gravitational force

m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0
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