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VLD [36.1K]
3 years ago
5

A heavy frog and a light frog jump straight up into the air. They push off in such away that they both have the same kinetic ene

rgy just as they leave the ground. Air resistance is negligible. Which of the following statements about these frogs are correct?
The heavier frog goes higher than the lighter frog
The heavier frog goes higher than the lighter frog
They collide and stick together.
the lighter frog is moving faster than the heavier frog.
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
5 0

Answer:

The lighter frog goes higher than the heavier frog.

The lighter frog is moving faster than the heavier frog

Explanation:

If both frogs have the same kinetic energy when they leave the ground, the following equality applies:

K(light) = K(heavy) = \frac{1}{2} *ml*vol^{2} = \frac{1}{2}*mh*voh^{2}

Now, if the only force acting on the frogs is gravity, when they reach to the maximum height, we can apply the following kinematic equation:

vf^{2} -vo^{2} = 2*a*hmax = vf^{2} -vo^{2} = 2*(-g)*hmax

When h= hmax, the object comes momentarily to an stop, so vf =0

Solving for hmax:

hmax =\frac{vo^{2} }{2*g}

As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.

As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.

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A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

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3 years ago
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total mass M= 280 kg

<span>The coefficient of kinetic friction between the sled and the snow  = 0.080</span>

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