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kotegsom [21]
4 years ago
10

An object thrown straight up a distance ymaxymax. After tt seconds, it falls back and is caught again, just as it reaches the he

ight from which it was thrown. Claudia says its average velocity was zero, and Hossein says its average speed was 2ymaxt2ymaxt . What would you say to help them out?
Physics
1 answer:
Mazyrski [523]4 years ago
6 0

Answer:

Avg.velocity=(Δy/ Δt)  =(net displacement/ total time for journey)

Δy = 0

Δt = t

so avg. velocity = 0/t =0

Avg. speed =(total distance traveled/ total time for journey)

total distance = up +down = Ymax+Ymax= 2 Ymax

total time = t

avg. speed = 2 Ymax/t

Explanation:

Since there is no net displacement from the original position,velocity is zero. Claudia is right!

while it covered some distance in time t so its speed is not as qouted by Hossien

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Answer:

E = 1.45 x 10⁹ J = 1.45 GJ

Explanation:

According to the law of conservation of energy:

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where,

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Therefore,

E = (400\ kg)(9.81\ m/s^2)(500000\ m)-\frac{1}{2} (400\ kg)(1600\ m/s)^2\\E = 1.962\ x\ 10^9\ J - 0.512\ x\ 10^9\ J

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8 0
3 years ago
A toy of mass 0.155 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 305 N/m
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Answer:

310.38\times 10^{-4}j

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The total energy at any position of the motion is give by E=\frac{1}{2}mv^2+KX^2  here \frac{1}{2}mv^2 is energy due to motion and KX^2  is energy due to spring elongation

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