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blondinia [14]
2 years ago
6

The skateboarder in the drawing starts down the left side of the ramp with an initial speed of 6.4 m/s. Neglect nonconservative

forces, such as friction and air resistance, and find the height h of the highest point reached by the skateboarder on the right side of the ramp.
Physics
1 answer:
aleksandrvk [35]2 years ago
4 0

Answer:

2.087m

Explanation:

Using the formula for calculating the maximum height reached by an object expressed as:

H = u²/2g

u is the initial speed

g is the acceleration due to gravity

Given

u = 6.4m/s

g = 9.81m/s²

Substitute into the formula:

H =  6.4²/2(9.81)

H = 40.96/19.62

H = 2.087m

Hence the height h of the highest point reached by the skateboarder on the right side of the ramp is 2.087m

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NISA [10]

The work done by the shopping basket is 147 J.

<h3>When is work said to be done?</h3>

Work is said to be done whenever a force moves an object through a certain distance.

The amount of work done on the shopping basket can be calculated using the formula below.

Formula:

  • W = mgh

Where:

  • W = Amount of work done by the basket
  • m = mass of the shopping basket
  • h = height of the shopping basket
  • g = acceleration due to gravity.

Form the question,

Given:

  • m = 10 kg
  • h = 1.5 m
  • g = 9.8 m/s²

Substitute these values into equation 2

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  • W = 147 J.

Hence, The work done by the shopping basket is 147 J.

Learn more about work done here: brainly.com/question/18762601

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Where would a boat produce the highest concentration of carbon monoxide?
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Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h
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Disagree.
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A woman throws a javelin 35 mph at an angle 30 degrees from the ground. Neglecting wind resistance or the height the javelin thr
Anna71 [15]

Answer:

35 mph

Explanation:

The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.

When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.

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