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iren2701 [21]
3 years ago
5

A heavy object falls with the acceleration as a light object during free fall. why?

Physics
1 answer:
alexgriva [62]3 years ago
6 0

<u>A heavy object falls with the acceleration as a light object during free fall because:</u>

Heavy things have a large gravitational force and also have less acceleration. So both the effects exactly cancel and make the falling objects to have the same acceleration irrespective of mass.

Free fall is a unique motion which has only gravitational force that acts on an object. Objects that undergo free fall experience only have the influence of gravity and not any other force.

So when we apply newton's second law of gravity which is:

           \text {acceleration}=\frac{\text {Force}}{\text {mass}}

where,

F is the force

m is the mass

a is the acceleration

For example: When a 1000 kg elephant and a 1 kg rat fall from the same height,

The acceleration can be calculated as follows:

For elephant: F = 10000 N and m = 1000 kg. So,

           \text {acceleration}=\frac{10000}{1000}=10\ \mathrm{m} / \mathrm{s}^{2}

For rat :  F = 10 N and m = 1 kg

Thus, \text {acceleration}=\frac{10}{1}=1\ \mathrm{m} / \mathrm{s}^{2}

Hence, it shows that both the animals have the same acceleration irrespective of their mass.

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Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.
kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees

ELECTRIC FORCE (F)

F = \frac{KQq}{d^{2} }

Where K = 9 x 10^{9} Nm^{2}/C^{2}

The distance between q_{1} and q_{3} can be calculated by using Pythagoras theorem.

d = \sqrt{33^{2} + 33^{2}  }

d = 46.7 cm = 0.467 m

For force F_{1}, substitute all the parameters into the formula above

F_{1} = (9 x 10^{9} x 3 x 1)/0.467^{2}

F_{1} = 2.7 x 10^{10}/0.218

F_{1} = 1.24 x 10^{11} N

For force F_{4}, substitute all the parameters into the formula above

F_{4} = (9 x 10^{9} x 3 x 4)/0.33^{2}

F_{4} = 1.08 x 10^{11}/0.1089

F_{4} = 9.92 x 10^{11} N

For force F_{2}, substitute all the parameters into the formula above

F_{2} = (9 x 10^{9} x 3 x 2)/0.33^{2}

F_{2} = 5.4 x 10^{10}/0.1089

F_{2} = 4.96 x 10^{11} N

Summation of forces on Y component will be

F_{y} = F_{4} - F_{1} Sin 45

F_{y} = 9.92 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{y} = 9.04 x 10^{11} N

Summation of forces on X component will be

F_{x} = F_{2} - F_{1} Cos 45

F_{x} = 4.96 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{x} = 4.08 x 10^{11} N

Net Force = \sqrt{F_{x} ^{2} + F_{y} ^{2}  } }

Net force = \sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2}  }

Net force = 9.9 x 10^{11} N

The direction will be

Tan ∅ = F_{y}/F_{x}

Tan ∅ = 9.04 x 10^{11} / 4.08 x 10^{11}

Tan ∅ = 2.216

∅ = Tan^{-1}(2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

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The heat flows into my body and I start with a warming feeling going down my throat as it starts there I get this warm feeling and I break out in a coldsweat from consuming a warm drink a cold drink will help cool me down :

Explanation:

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A guyot is ________. A. any portion of the ocean floor that is topographically higher than surrounding sea floor B. an extinct o
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Answer:

c.an extinct oceanic hot-spot volcano that has subsided below sea level

Explanation:

- Marine geology defines a guyot as an isolated underwater volcanic mountain with a flat top more than 200m below the surface of the sea.

-The flat top is due to years of wave erosion.

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If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
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Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

  • \rm y = 0.02\sin(30x-200 t).

The vertical displacement of a wave is given in generalized form as

\rm y = A\sin(kx -\omega t).

<em>where</em>,

  • A = amplitude of the displacement of the wave.
  • k = wave number of the wave = \dfrac{2\pi }{\lambda}.
  • \lambda = wavelength of the wave.
  • x = horizontal displacement of the wave.
  • \omega = angular frequency of the wave = \rm 2\pi f.
  • f = frequency of the wave.
  • t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

\rm A=0.02.\\k=30.\\\omega =200.\\

therefore,

\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.

It is the required frequency of the wave.

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