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A water-skier is pulled behind a boat by a rope that is at an angle of 13° and has a tension of 490 N. The water-skier has a mass of 49 kg.

What are the magnitudes of the x- and y- components of the tension?

What is the normal force acting on the skier?

Answers

Answer 1

Answer:

normal force= 480.2 x component=477.4  y component= 110.4

Explanation:

I'm sorry I just did this today in class but i think normal force has to be equal to mg or the mass * gravity of the person and that would be 480.2 newtons. I multiplied 9.8 (gravity) by the mass of 49g

I think for the x and y components you can make a triangle with the angled string and then use sohcahtoa to solve for the tension. I used cos(13) = x/490 to solve for x component and then I used the pythagoran theroem to get the remaining side which would be : a^2 + 477.4^2 = 490^2

hope this can help and that it is correct. good luck

Answer 2

The magnitudes of the x- and y- components of the tension is 477.44N and 110.23N

The normal force acting on the skier is 480.2N

The magnitude of the x- and y- components of the tension is expressed as:

x = Tcos 13°y = T sin  13°

Given the following

Tension T = 490N

x = 490 cos13° = 477.44N

y = 490sin13° =  110.23N

Hence the magnitudes of the x- and y- components of the tension is 477.44N and 110.23N

The normal force acting on the skier is equal to the weight.

N = W = mg

N = W = 49 * 9.8

N = W = 480.2N

Hence the normal force acting on the skier is 480.2N

Learn more on force here: https://brainly.com/question/12970081


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A rock is kicked horizontally off a dock into the bay 3
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Answers

Answer:

The true statement is;

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Explanation:

The given information are;

The direction in which the rock is kicked = Horizontally

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The time taken for the rock to hit the water is given by the following equation for projectile motion;

y = y₀ + v₀× sin(θ₀)×t - 1/2×g×t²

Where;

θ = The angle the rock makes with the horizontal when it was kicked = 0

y = The height through which the rock drops

y₀ = The initial height of the rock

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g = The acceleration due to gravity ≈ 9.81 m/s²

For the different speeds, v₁, v₂, and v₃, we have;

y = y₀ + v₁× sin(0)×t - 1/2×g×t² = y₀ - 1/2×g×t²

y = y₀ + v₂× sin(0)×t - 1/2×g×t² = y₀ - 1/2×g×t²

y = y₀ + v₃× sin(0)×t - 1/2×g×t² = y₀ - 1/2×g×t²

Therefore, the time it takes for the ball to hit the water is given as follows;

[tex]t = \sqrt{\dfrac{(y_0 - y) \times 2}{g} }[/tex]

Which is the same for each case

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[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]

[tex]v_y[/tex] = The vertical velocity = v₀ × sin(θ₀) - g×t = v₀ × sin(0) - g×t = -g×t

[tex]v_y[/tex] = -g×t

Therefore, [tex]v_y[/tex], is the same for all three times

While vₓ = v₁, v₂, and v₃, for each of the three times.

Therefore, [tex]v = \sqrt{v_x^2 + v_y^2}[/tex]  the speed with which the rock hits the water is different for all three times.

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Answers

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Answers

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Hello I have the same question but I couldnt figure it out but i looked it up and this is what found it seemed to help me alot.

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right on EDGE2021

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hope this helps

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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hope it helps

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Answers

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Answers

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Have a great day:)

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