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denis23 [38]
2 years ago
13

You jump off a truck and accelerate toward the surface of the Earth. Does the Earth accelerate toward you?

Physics
1 answer:
Triss [41]2 years ago
7 0

When we jump from the truck and accelerate towards the earth surface, the earth also accelerates towards us but it's acceleration is very negligible.

To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.

<h3>What's the gravitational force between the earth and a person?</h3>
  • Gravitational attraction force is GMm/r² between the earth and a person.
  • M= mass of the earth

m= mass of the person

r= separation between them.

<h3>What's the acceleration of the earth towards the person when he jumps from a truck?</h3>
  • According to Newton's second law, Force = M×acceleration
  • Acceleration= Force / M
  • Here, Force = GMm/r²,

so acceleration of earth= Gm/r²

  • As this acceleration is very small, so we can't notice it.

Thus, we can conclude that the earth also accelerates towards us.

Learn more about the gravitational force here:

brainly.com/question/72250

#SPJ4

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The best answer would be C. 

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3 0
2 years ago
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
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laila [671]

Answer:

1.45544 J

Explanation:

See attachment

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21. If the Sun's rays were at 45° to a vertical pillar, how would
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Answer:

Let the height of the pole AB = x m. ∴ Length of shadow OB ol the pole AB = x m. Let the angle of elevation be ө, i.e. Hence, the angle of elevation of the Sun's altitude is 45°.

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Answer:

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