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3 years ago

What is the force required accelerate a 5.0 kg object at 3.0m/s^ 2?

1 answer:

6 0

The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N

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Kelly sits on a rock. Her weight is an action force. Describe the reaction force.

Mumz [18]

An equal and opposite force acting upwards

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4 years ago

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Momentum is the mass of an object multiplied by its velocity.<br> a. True<br> b. False

Ostrovityanka [42]

A.) True. Momentum is the mass of an object multiplied by its velocity.

It is defined as a mass in motion. It refers to the quantity of motion an object has.

8 0

4 years ago

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A car is moving at a initial velocity of 20 meters per second accelerates at a rate of 1.5 meters per second squared for 4 secon

Brrunno [24]

Answered using calculus.
Antidifferentiated the acceleration to get velocity. Added variable c as we do not know if there was an extra number there yet.
Knowing that when time is 0, the velocity is 20, we can substitute those numbers into the equation and find that c = 20.
Now we have full velocity equation: v = 1.5t + 20
Now we substitute 4 into t to find out the velocity after 4 seconds. This gives us the final answer of 26m/s

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4 years ago

Which of the following is NOT an example of a medium.

andriy [413]

Answer:

D.) Metal

Explanation:

light can not pass through metal.

8 0

3 years ago

You throw a ball upward from ground level with initial upward speed v0. What is the max height of the trajectory?

Inga [223]

Answer:

The max height of the ball is y = -1/2 (v0²/g).

It takes the ball t = -2 · v0/g to hit the ground.

The speed of the ball when it hits the ground is v = -v0.

Explanation:

The height and velocity of the ball is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t

When the ball is at max height, the velocity is 0. So, let´s find the time at which the velocity of the ball is 0.

v = v0 + g · t

0 = v0 + g · t

t = -v0/g

Now, replacing t = -v0/g in the equation of height, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located on the ground)

y = v0 · t + 1/2 · g · t²

Replacing t:

y = v0 · (-v0/g) + 1/2 · g · (-v0/g)²

y = -(v0²/g) + 1/2 · (v0²/g)

y = -1/2 (v0²/g)

The max height of the ball is y = -1/2 (v0²/g). Remember that g is negative.

Since the acceleration of the ball is always the same, the time it takes the ball to impact the ground will be twice the time it takes to reach its max height, t = -2 v0/g.

However, let´s calculate that time knowing that at that time the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · t + 1/2 · g · t²

0 = t · ( v0 + 1/2 · g · t)

0 = v0 + 1/2 · g · t

-2 · v0/g = t

It takes the ball t = -2 · v0/g to hit the ground.

Let´s use the equation of velocity at final time (t = -2 · v0/g):

v = v0 + g · t

v = v0 + g · ( -2 · v0/g)

v = v0 - 2· v0

v = -v0

The speed of the ball when it hits the ground is v = -v0.

7 0

4 years ago