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ziro4ka [17]
3 years ago
12

A child with a mass of 20 kg sits at a distance of 2 m from the pivot point of a seesaw. where should a 16-kg child sit to balan

ce the seesaw g

Physics
1 answer:
Maksim231197 [3]3 years ago
4 0
In the above problem, we need to find mass of the second child, so that the Center of Mass remains at the origin( pivot).

CM= m1r1+m2r2/m1+m2
0= 20*-2+16*r2/20+16
r2= 40/16
r2= +2.5 m

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A student practicing for a track meet ran 250 m in 30 seconds. What was her average speed?
DochEvi [55]

Answer:

8.33 meters/sec.

Explanation:

distance = 250 meters

time = 30 sec.

velocity = distance / time

            = 250 meters

                   30 sec.

            = 8.33 meters/sec.

6 0
3 years ago
Why does Quito, Equator has very little changes to the daylight hours<br> throughout the year?
marin [14]

Because of the position on the equator, the change in rotation of the Earth on its axis throughout the year doesn't affect it much. Unlike the poles, Quito is almost constantly in direct view of the sun. So, because of lack of change in rotation, the daylight hours are hardly varied as Quito is almost constantly in more or less the same spot in relation to the sun.

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3 years ago
If the merry-go-round makes one revolution in 10 seconds, what is the child’s linear speed?
Anton [14]
The child's linear speed is
              
    <em> (pi / 5) x (the child's distance from the center of the ride, in feet)</em>

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5 0
3 years ago
You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget
aleksklad [387]

Answer:

35.7 m

Explanation:

Let

\mid A\mid=18.5 m

\mid B\mid=41 m

We have to find the distance between Joe's and Karl'e tent.

A_x=Acos\theta

A_y=Asin\theta

Substitute the values then we get

A_x=18.5cos23^{\circ}=17 m

A_y=18.5sin 23^{\circ}=7.2 m

B_x=41cos37.5^{\circ}=32.5 m

B_y=41sin37.5^{\circ}=-24.96 m

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

B=A+C

C=B-A

C_x=B_x-A_x=32.5-17=15.5 m

C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m

\mid C\mid=\sqrt{C^2_x+C^2_y}

\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m

Hence, the distance between Joe's and Karl's tent=35.7 m

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3 years ago
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