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Georgia [21]
3 years ago
7

Two boys want to balance a seesaw perfectly. One boy weighs 120 pounds and is sitting four feet from the fulcrum. The other boy

weighs 80 pounds. Where should the lighter boy sit to balance the seesaw?
Physics
1 answer:
OLEGan [10]3 years ago
6 0

Basically a distance multiplied by a weight that is equal to the distance that is going to be multiplied by the weight. (for the equation we will use X for the distance).

equation: 4 x1 20 = ? x 80

Now step one: 4(120) = X(80)  

Or another way is 480 = 80X  

480/80 = X  

48/8= X  

X = 6

I hope this could help! Sorry if it didn't make much sense otherwise!


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Answer: the 30gram will hit the ground first

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3 years ago
An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be
Inessa05 [86]

Explanation:

The given data is as follows.

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where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

                    = V \times (1 - \frac{d}{R})

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

             = \frac{V}{r} \times 0.1532

So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

5 0
4 years ago
A wagon is accelerating down a hill. Which statement is true?
Ira Lisetskai [31]
Potential energy decreases and kinetic energy increases.

Potential energy is related to the height, since the wagon is going downhill, height decreases and potential energy decreases.

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