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2 years ago

What will be the mass of a body at the center of the earth as compared to other places on

2 answers:

8 0

The mass of the object does not change by moving it to another place. ... At the center of the earth the net gravitational force is zero, so the weight will be zero, but its masses will remain same. Hence the mass at the centre of earth will be equl to 50 kg.

mestny [16]2 years ago

6 0

Answer:

Explanation:

It's because the mass of the object doesn't change by moving it to next place.

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Identify the normal force on the shopping cart after 75 newtons of groceries are added to the cart

Anna71 [15]

If the shopping cart is parallel to the ground, the weight is acting vertically downward, then force is equal to weight. The normal force will be acting perpendicular to the ground,and acting upward opposite of the weight.

3 0

3 years ago

Read 2 more answers

An atomic nucleus is composed of A) protons. B) protons and neutrons. C) protons and electrons. D) protons, neutrons, and electr

8090 [49]

Answer:

Explanation:

The nucleus is made up of protons and neutrons, and electrons float around the neucleus in the electron cloud.

3 0

2 years ago

Read 2 more answers

Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of

slamgirl [31]

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

$v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2$

Substituting the velocities in the equation

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0

3 years ago

Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i

densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0

3 years ago

To add to the complexity, those precious atoms face the peril of disintegrating into the void. Luckily, by now, they are feeling

miskamm [114]

Answer:

B The most basic atoms were formed due to the force of gravity.

Explanation:

Since the statement says that the atoms could have faced the peril of disintegrating into the void, this means that, they could have been destroyed by movement away into the void.

But, it also says that by now, they are feeling the influence of gravity to bring them safely together. This statement shows that gravity brings them (the atoms) together and thus doesn't allow them disintegrate into the void.

<u>So, a reader can thus infer that the most basic atoms were formed due to the force of gravity since it doesn't allow the atoms disintegrate into the void.</u>

So, B is the answer.

5 0

2 years ago