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Vladimir [108]
4 years ago
10

A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be able to make it tilt back and f

orth without the heavier child simply sinking to the ground. Given that her heavier child of weight W is sitting a distance L to the left of the pivot, at what distance L1 must she place her second child of weight w on the right side of the pivot to balance the seesaw?

Physics
1 answer:
horsena [70]4 years ago
7 0

Answer:

L₁ = W×L / w

Explanation:

The scenario is shown in the image below.

<u>At the pivot point, the torque acting on this point must be zero so that there will be easy back and forth without the heavier child.</u>

<u>Torque created by lighter child + Torque created by the heavier child = 0</u>

Thus,

According to the axis system, the heavier child is left to the pivot (origin), so,

W×(- L ) + w× L₁ = 0

So,

<u>L₁ = W×L / w</u>

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Answer:

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Given parameters:

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Insert the parameters and solve for q;

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Holden is trying to determine the velocity of his race car. He went 20 meters east, turned around, and went 40 meters west. He t
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Change in v= 9.8 m/s2xt. The diagram shows a ball falling toward Earth in<br> a vacuum.
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Answer:

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Young Jeffrey is bored, and decides to throw some things out of the window for fun. But since he is also very curious, he decide
Inga [223]

Answer:

a) Stone            v_{y} = - 7.25 ft / s ,  vₓ = 0.362 ft / s

b) tennis ball    v_{y} =  -3.16 ft / s ,   vₓ = 0.634 ft / s

c) golf ball        v_{y} = - 1,536 ft / s, vx = 0.634 ft / s

2) golf ball

Explanation:

1) The average speed is defined with the displacement interval in the given time interval

           v =( x_{f}-x₀) / Δt

let's use this expression for each object

a) Stone

  It tells us that it is released from y₀ = 10 ft and reaches the floor at

t = 0.788 s, but in the problem they tell us that the calculation is for t = 1.38 s

           v_{y} = (0-10) / 1.38

           v_{y} = - 7.25 ft / s

 in this interval a distance of x_{f} = 0.500 ft was moved away from the building (x₀ = 0 ft)

          vₓ = (0.500- 0) / 1.38

          vₓ = 0.362 ft / s

In my opinion it makes no sense to keep measuring the time after the stone has stopped.

b) tennis ball

It leaves the building at a height of y₀= 10ft and at the end of the period it is at a height of y_{f} = 5.63 ft, all this in a time of t = 0.788 + 0.591 = 1.38 s

       

the average vertical speed is

            v_{y} = (5.63 - 10) / 1.38

            v_{y} = -3.16 ft / s

for horizontal velocity the ball leaves the building xo = 0 reaches the floor

x₁ = 0.500 foot and when bouncing it travels x₂ = 0.375 foot, therefore the distance traveled

         x_{f} = x₁ + x₂

         x_{f} = 0.500 + 0.375

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         vₓ = (0.875 - 0) / 1.38

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the horizontal displacement x₀ = 0 ft to the point xf = 0.875 ft

          vₓ = (0.875 -0) / 1.38

          vₓ = 0.634 ft / s

2) in this part we are asked for the instantaneous speed at the end of the time interval

a) the stone is stopped so its speed is zero

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b) the tennis ball

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c) The golf ball

they do not indicate that it is still rising. Therefore its vertical speed is greater than zero

         v_{y} > 0

horizontal speed is constant

         vₓx = 0.634 ft / s

the total velocity of the object can be found with the Pythagorean theorem

         v = √ (vₓ² + v_{y}²)

When reviewing the results, the golf ball is the one with the highest instantaneous speed at the end of the period

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