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3 years ago

Objects in free fall accelerate due to ______

Physics

1 answer:

Charra [1.4K]3 years ago

3 0

Answer:

The correct answer is A.

Objects in free fall accelerate due to <u>gravity</u>.

Explanation:

Momentum can help an object to keep its state of motion at a constant velocity when no external force is applied. It can never accelerate the object.

According to the laws of motion, we know that acceleration is produced in a body only when a Force is applied in the direction of motion of body.

During a free fall, only the force of gravity is acting on an object and that too in the direction of its motion. Hence, the acceleration produced in a free falling object is due to gravity.

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a ball is dropped from rest at a height of 45.0 m above the ground. ignore the effects of air resistance. What is the speed of c

NemiM [27]

So, the final velocity of the ball when it is 10.0 m above the ground approximately <u>26.2 m/s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses the principle of final velocity in free fall. Free fall occurs only when an object is dropped (without initial velocity), so the falling object is only affected by the presence of gravity. In general, the final velocity in free fall can be expressed by this equation :

$\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}$

With the following condition :

v = final velocity (m/s)
h = height or any other displacement at vertical line (m)
g = acceleration of the gravity (m/s²)

<h3>Problem Solving</h3>

We know that :

$\sf{h_1}$ = initial height = 45.0 m
$\sf{h_2}$ = final height = 10.0 m
g = acceleration of the gravity = 9.8 m/s²

Note :

At this point 10 m above the ground, the object can still complete its movement up to exactly 0 m above the ground.

What was asked :

v = final velocity = ... m/s

Step by Step

$\sf{v = \sqrt{2 \times g \times \Delta h}}$

$\sf{v = \sqrt{2 \times g \times (h_1 - h_2)}}$

$\sf{v = \sqrt{2 \times 9.8 \times (45 - 10)}}$

$\sf{v = \sqrt{19.6 \times 35}}$

$\sf{v = \sqrt{686}}$

$\boxed{\sf{v \approx 26.2 \: m/s}}$

<h3>Conclusion</h3>

So, the final velocity of the ball when it is 10.0 m above the ground approximately 26.2 m/s.

The relationship between acceleration and the change in velocity and time in free fall brainly.com/question/26486625

5 0

2 years ago

002 10.0 points note: report your answer in joules here. a proton and an antiproton, both at rest with respect to one another, m

Ulleksa [173]

The mass of the proton is:
$m_p = 1.67 \cdot 10^{-27} kg$
and the mass of the antiproton is exactly the same, so the total mass of the two particles is $2m_p$ .

In the annihilation, all the mass of the two particles is converted into energy, and the amount of this energy is given by Einstein's equivalence between mass and energy:
$E=Mc^2$
where M is the mass converted into energy and c is the speed of light. In this example, $M=2m_p$ , therefore the energy released is
$E=2m_p c^2 = 2 (1.67 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2=3 \cdot 10^{-10}J$

3 0

3 years ago

What is the value of g

Mariana [72]

It depends on the environment around you ... the number and size of
the masses around you, and your distance from their center of mass.

If you happen to be on the surface of the Earth, then

   g = 9.8 m/s²
   or
   g = 9.8 newton/kilogram .

It's different on other planets. Be sure to look it up
before you leave, so you'll know what to expect.

5 0

4 years ago

A machine part has the shape of a solid uniform sphere of mass 205 g and diameter 4.10 cm. It is spinning about a frictionless a

kondor19780726 [428]

Answer:

Answer is in the following attachment.

Explanation: