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3 years ago

This diagram shows two different forces acting on a skateboarder. The

Physics

2 answers:

algol133 years ago

4 0

Answer:

Option A. 0.36 m/s to the left

Explanation:

From the question given above, the following data were obtained:

Force resistance (Fᵣ) = 55.2 N

Applied force (Fₐ) = 31.8 N

Mass (m) = 65.2 kg.

Acceleration (a) =?

Next, we shall determine the net force acting on the skateboarder. This can be obtained as illustrated below:

Force resistance (Fᵣ) = 55.2 N

Applied force (Fₐ) = 31.8 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₐ

Fₙ = 55.2 – 31.8

Fₙ = 23.4 N to the left

Finally, we shall determine the acceleration of the skateboarder. This can be obtained as follow:

Mass (m) = 65.2 kg.

Net force (Fₙ) = 23.4 N to the left

Acceleration (a) =?

F = ma

23.4 = 65.2 × a

Divide both side by 65.2

a = 23.4 / 65.2

a = 0.36 m/s

Since the net force acting on the skateboarder is to the left. Therefore, the acceleration of the stakeholder will also be towards the left (i.e 0.36 m/s to the left)

guapka [62]3 years ago

4 0

Answer:

A. 0.36 m/s to the left

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4 years ago

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3 years ago

A ball is launched from ground level and hits the ground again after an elapsed time of 4 seconds and after traveling a horizont

jeka94

Answer:

1)a. It is constant the whole time the ball is in free-fall.

2)b. = 14 m/s

3) e. = 19.6 m/s

Explanation:

1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.

2) speed = distance/time

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4 years ago

A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as

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Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

Acceleration (a) is defined as the time rate of change of velocity

$a = \frac{v_{2} - v_{1} } {t}$

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Final velocity = v₂ = 0 m/s

Initial velocity = v ₁ = 0 m/s

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a = 0 ms⁻²

b.)

Given data

As the space shuttle start from rest, So initial velocity is zero

Initial velocity = v₁ = 0 ms⁻¹

Final velocity = v₂ = 4600 ms⁻¹

Time = t = 8 min = 480 s

By the definition of Acceleration (a)

$a = \frac{v_{2} - v_{1} } {t}$

$a = \frac{4600 - 0 } {480}$

a = 9.58 ms⁻²

c.)

Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

Initial velocity = v₁ = 0 ms⁻¹

Final velocity = v₂ = 4600 ms⁻¹

Time = t = 10 min = 600 s

By the definition of Acceleration (a)

$a = \frac{v_{2} - v_{1} } {t}$

$a = \frac{4600 - 0 } {600}$

a = 7.67 ms⁻²

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