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NeX [460]
2 years ago
11

A boy standing on a 19.6 meter tall bridge sees a motorboat approaching the bridge at a constant speed. When the boat is 27 mete

rs from the bridge, the boy drops a stone to the water below. If the stone strikes the water 3.0 meters in front of the boat, at what speed is the boat traveling?
Physics
1 answer:
azamat2 years ago
8 0

Answer:

A. 12 m/s

Explanation:

Let’s remember that the definition of velocity is the variation of position of an object respect with to time. We know that the boy dropped the stone when the boat was 27 meters from the bridge and the stone hit the water 3 meters in front of the boat. So, the Boat must have traveled x=27 m-3m=24 m. The next step is calculating the amount of time that took the boat to make that travel; coincidentally, it is the same time that takes the stone to reach the water.

The equation that describes the motion of the stone is:

y = y_0 + v_0 * t+1/2 * a * t^2

The boy drops the stone from rest, so we can say that v_0=0. We can fixate the reference line on top of the bridge, so y_0=0 as well. The equation will be then:

-19,6 m = -1/2 * 9,8 m/s^2  * t^2

t^2= -(19,6 m)/(-4,9 m/s^2) = 4,012 s^2

t=√(4,012 s^2) = 2,003 s

Knowing the time that takes the stone to reach the water, that is the same that time that the boat uses to travel the 24 meters. The velocity of the boat is:

v = ∆x/∆t = (27 m-3 m)/(2,003 s-0s) = 11,9816 m/s ≈ 12 m/s

Have a nice day! :D

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Q = ba⁴ * ε₀

Explanation:

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2 years ago
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Answer:

0.456033049

Explanation:

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Making \mu the subject we obtain

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B) form a straight line with the Moon in the middle.

Explanation:

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2 years ago
Vector A⃗ has magnitude 8.00 m and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
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Answer:

293.7 degrees

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A = - 8 sin (37) i + 8 cos (37) j

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B = a i+ b j , where and a and b are constants to be found

A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

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