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alexira [117]
3 years ago
15

The fall of a body on the earth's surface cannot be a complete free fall why ?​

Physics
1 answer:
leonid [27]3 years ago
3 0

Explanation:

because the boy has larger surface area due to which he offers the larger air resistance which decreases the acceleration so, he will fall towards the earth's surface approximately with constant velocity.

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which statement below is FALSE? A Gravity acts on any object with mass. B The closer objects are, the stronger the gravitational
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Explanation:

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Guys are our body made out of 70 percet water?
babunello [35]

Answer:

I'm pretty Sure. Yea look.

Explanation:

Up to 60% of the human adult body is water. According to H.H. Mitchell, Journal of Biological Chemistry 158, the brain and heart are composed of 73% water, and the lungs are about 83% water. The skin contains 64% water, muscles and kidneys are 79%, and even the bones are watery: 31%.

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Galileo performed an experiment to measure the speed of light by timing how long it took light to travel from a lamp he was hold
krok68 [10]

Answer:

The time it takes light to cover 1.5 km was too short to be measured by Galileo's instruments.

Explanation:

The speed of light is c=3*10^8m/s, which means the time it takes to cover a distance of 1.5 km (or 1,500m) will be

t= \dfrac{1500m}{3*10^8m/s}

t= 0.000005s

which is \dfrac{1}{200000} of a second! This time delay could in no way be measured by Galileo considering the fact that he was using his heartbeat to measure time!

6 0
3 years ago
Help w this science question please..?
alisha [4.7K]
C)Visible Light-the hint:Visible light.
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3 years ago
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

\vec{E} = 4.5\times 10^{4} N/C

7 0
3 years ago
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