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3 years ago

9

At what distance from the centre of the earth its acceleration due to gravity becomes one third only? [Mass and radius of earth

= 6x10 24 kg & 6.4x10 6 m resp.]

2 answers:

Anarel [89]3 years ago

5 0

235 is the answer sir

Kipish [7]3 years ago

3 0

Answer:

235

Explanation:

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A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled

ad-work [718]

Answer:

shown below

Explanation:

2 x 10⁷ as a number is 20,000,000

20,000,000 - 10 = 19,999,990

It went 19,999,990 m/h

in km/h:

19,999,990 / 1000 = 19,999.99 km/h

in km/s

19,999,990 / 3,600,000 = ~5.56 km/s

in m/s

19,999,990 / 3600 = ~5555.56 m/s

7 0

2 years ago

Which of the following is an example of matter

miskamm [114]

For the first one, leaves is matter.
for the second one, ability to react
hope this helps

8 0

3 years ago

Read 2 more answers

Two charges that are separated by one meter exert 1-n forces on each other. if the magnitude of each charge is doubled, the forc

Goshia [24]

The electrostatic force between the two charges is
$F=k_E \frac{q_1 q_2}{r^2}$
where q1 and q2 are the magnitudes of the two charges, and r the distance between them.

We can see from the formula that F is proportional to the product between the two charges:
$F \sim q_1 q_2$
so, if the magnitude of each charge is doubled, the new force will get a factor 4:
$F' \sim (2 q_1 )(2 q_2 )=4 q_1 q_2 =4 F$
So, the new force will be 4 times the original force:
$F' = 4 \cdot 1N= 4N$

5 0

3 years ago

Question 10 of 10Which statements are true?Check all that apply.O A. If heat is added to a substance, the temperature alwaysincr

inna [77]

The true statements are B and C.

This comes from the following facts:

Even when heat is added to a substance in a change of phase the temperature does not increases since this heat is latent heat.

3 0

1 year ago

If a ????=87.5 kgm=87.5 kg person were traveling at ????=0.900????v=0.900c , where ????c is the speed of light, what would be th

Diano4ka-milaya [45]

Answer:

$\frac{K.E_r}{K.E}=2.875$

Explanation:

Given:

mass, m = 87.5kg

Velocity, V = 0.900c

now,

the relativistic kinetic energy id given as:

$K.E_r=(\gamma-1)mc^2$ ...........(1)

where,

$\gamma$ = relativistic factor, given as; $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Now, the classical kinetic energy is given as:

$K.E = \frac{1}{2}mv^2$ ..........(2)

Dividing the equation (1) by (2) we get

$\frac{K.E_r}{K.E}=\frac{(\gamma-1)mc^2}{\frac{1}{2}mv^2}$

or

$\frac{K.E_r}{K.E}=\frac{(\gamma-1)c^2}{\frac{1}{2}v^2}$

substituting the values in the equation we get,

$\frac{K.E_r}{K.E}=\frac{(\frac{1}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}-1)c^2}{\frac{1}{2}\times(0.90c)^2}$

or

$\frac{K.E_r}{K.E}=2.875$

5 0

3 years ago