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3 years ago

11

When the moon slowly moves away from the earth.How is it affecting the gravitational force between the earth and the moon?

1 answer:

sukhopar [10]3 years ago

3 0

Well, think about how the tides will be affected when the moon moves farther away. If the moon first started off very close the earth, we would have more tsunamis. (Scientists have found that the moon has possibly been closer to earth long ago.) While it moves away, soon there will no longer be many tides.

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the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the tab

JulsSmile [24]

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

<em>Mass of the first block, m₁ = 4.0 kg</em>
<em>Mass of the second block, m₂ = 2.0 kg</em>

The net force on the system of the two blocks is calculated as follows;

$m_2 g - T = m_1 a$

where;

<em>T </em><em>is the tension in the connecting string due weight of the first block</em>

$m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2$

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

Learn more about net force on two connected blocks here: brainly.com/question/13539944

5 0

3 years ago

How many neutrons does potassium have?

Sloan [31]

Answer:

the answer is 20 neutrons

Explanation:

6 0

3 years ago

Read 2 more answers

What kind of weather would an occuluded front likely bring

abruzzese [7]

An occluded front forms when a cold front catches up with a warm front. So it would bring Sun and warmth.

3 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

Two objects 5 kg and 7 kg are attached to the end of inextensible string which passes over frictionless pulley

Bumek [7]

Answer:

ans:

tenson(T) = 20 N

acceleration (a) = 2.86 m/s

Explanation:

T + mg = Mg

T = Mg - mg

T = g( M - m )

T = 10× ( 7-5 )

T = 20 N

again;

T = 20

Ma = 20

a = 20 / 7

= 2.86 m/s

3 0

3 years ago