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ella [17]
3 years ago
6

A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potent

ial energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)
A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.

At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.
Physics
1 answer:
Sergeu [11.5K]3 years ago
3 0

Answer: c. 40m

Explanation:

See picture

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Since there are no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, we would assume that momentum is conserved. This means that

total momentum of car before collision = total momentum of car after collision.

Recall,

momentum = mass x velocity

From the information given,

mass of car before collision = 2000

velocity of car before collision = 3

Thus,

total momentum of car before collision = 2000 x 3 = 6000

Also,

mass of sack = 500

mass of car and sack after collision = 500 + 2000 = 2500

velocity after collision = v

momentum after collision = 2500 x v = 2500v

Since momentum is conserved, then

6000 = 2500v

v = 6000/2500

v = 2.4

the speed of the flatcar is 2.4 m/s

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Describe at least four of the requirements (facts) that any model of solar system formation must include (according to the nebul
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A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released
statuscvo [17]

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

  • <em>mass of the block, m = 2.4 kg</em>
  • <em>height of the block, h =  5 m</em>
  • <em>compression of the spring, x = 25 cm = 0.25 m</em>

The spring constant is calculated as follows;

F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2  - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2  = mgh -   \frac{1}{2} kx^2\\\\mv^2   = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

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Answer:

Stars are forming in the spiral arms so there are high mass, hot, blue stars in the arms.

Explanation:

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