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Umnica [9.8K]
4 years ago
5

A water skier is pulled behind a boat. When the skis push down on the water, what is the reaction force?

Physics
2 answers:
IrinaVladis [17]4 years ago
8 0
The answer is C) The water pushed up on the skis

The water reacts to the downward force of the skis by pushing back up against the skis. 
statuscvo [17]4 years ago
8 0

Answer:

the water pushes up on the skis

Explanation:

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Based on the law of conservation of energy, which statement is false?
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I think it’s Energy is lost when machines don’t work right.

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3 years ago
A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl
aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

D₂= 167.21 cm

Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0
3 years ago
_______ measures an object's change in position per unit time.
Deffense [45]

I believe your answer should be C. Speed.

6 0
3 years ago
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Answer:

idk

Explanation:

7 0
3 years ago
Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direc
Lerok [7]

Answer:

424.26 m/s

Explanation:

Given that Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direction along which Q is flying. Find the magnitude of velocity of the air craft P relative to air craft Q

The relative speed will be calculated by using pythagorean theorem

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Relative speed = sqrt( 180000 )

Relative speed = 424.26 m/s

Therefore, the magnitude of velocity of the air craft P relative to air craft Q is 424.26 m/s

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