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3 years ago

How do I find the acceleration and if you could just put the equation in your answer. Thanks a lot.

Physics

1 answer:

Dafna1 [17]3 years ago

7 0

You use Newton's second law, which says that the net force acting on a body is equal to the product of the body's mass and the acceleration it undergoes. That is,

$\sum\vec F=m\vec a$

From the question, it sounds like the given force is the only force acting on the block, which means there is only going to be one direction of movement if the acceleration is constant. This allows us to ignore the vector notation and talk only in terms of magnitudes. The above equation reduces to

$F=ma$

So in this case,

$F=ma\implies a=\dfrac Fm$

and numerically, we get

$a=\dfrac{40\,\rm N}{12.5\,\rm kg}=3.2\,\dfrac{\rm m}{\mathrm s^2}$

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A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W

Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)$

$V^{2} = 24.5 m/s$

$V = \sqrt{24.5} \ m/s$

$V = 4.95 \ m/s$

V = ± 4.95 m/s

V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)$

$0 = U^{2} - 16.072 m/s$

$U^{2} = 16.072 m/s$

$U = \sqrt{16.072} \ m/s$

U = ± 4.01 m/s

U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

$F = \frac{mv - mu}{t}$

$F.t = m(v - u)$

⇒ Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,

∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )

F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )

F.t = 0.120 kg × 8.96 m/s

Impulse = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

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3 years ago

Moving a neutral wire in a(n) ____ field will induce a(n) _____.

frutty [35]

the answer is b) magnetic current

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4 years ago

Your skeleton. It is there for support, protection, and movement. But you could not move your bones without the help of your ___

lana66690 [7]

I believe the answer is D.) the muscular system

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3 years ago

Read 2 more answers

Do you think a baseball curves better at the top of a high mountain or down on a flat plain

Serhud [2]

A baseball will curve better on the flat plain if it is higher than sea level but low elevation.

Hope this helped!

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3 years ago

Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu

11111nata11111 [884]

Answer:

velocity = 62.89 m/s in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $10^5$ kg - m/s in the right direction.