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Free_Kalibri [48]

2 years ago

14

The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone'

s velocity just before
it strikes the ground?

1 answer:

Svetlanka [38]2 years ago

4 0

Answer:

79

Explanation:

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Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

Degger [83]

B. kinetic energy increases and potential energy decreases

5 0

4 years ago

How much more matter is in a golf ball than a ping pong?

melisa1 [442]

Explanation:

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3 years ago

A 0.15 kg mass is suspended from a vertical spring and descends a distance of 4.6 cm, after which it hangs at rest. An additiona

ch4aika [34]

Answer:

The total extension = 19.9 cm

Work done in stretching the spring through both displacement = <em>0.633 J.</em>

Explanation:

From Hook's law,

F = ke .................................... Equation 1.

Where F = force or weight, k = force constant of the spring, e = extension.

making k the subject of the equation,

k = F/e .......................... Equation 2

Given: F = W = mg = 0.15(9.8) = 1.47 N, e = 4.6 cm = 0.046 m.

k = 1.47/0.046

k = 31.96 N/m.

When an additional mass of 0.5 kg is suspended,

Total mass suspended(M₁) = 0.15+0.5 = 0.65 kg

Total Weight (W₁) = 0.65(9.8) = 6.37 N.

k = 31.96 N/m

Substituting into equation 1

6.37 = 31.96e

e = 6.37/31.96

e = 0.199 m

e = 19.9 cm

Thus the total extension = 19.9 cm

a.

Work done in stretching the spring through both displacement

W = 1/2ke²..................... Equation 3

Where W = work done, k = spring constant, e = extension.

Given: k = 31.96 N/m, e = 0.199 m

Substituting into equation 3

W = 1/2(31.96)(0.199)²

<em>W = 0.633 J.</em>

5 0

3 years ago

The absolute magnitude of a star is the amount of light received by Earth.

frutty [35]

B.false......absolute magnitude

<span>Measure of the amount of light a star actually gives off.....thats the definition....hop this is helpful</span>

<span>
</span>

6 0

4 years ago

Read 2 more answers

The closely packed cones in the fovea of the eye have a diameter of about 2 Î¼m. For the eye to discern two images on the fovea

Marina CMI [18]

Answer:

The distance between the object is $l=0.0056\ cm$

Explanation:

The free body diagram of this setup is on the first uploaded image

From the question

The diameter of closely packed cones in the fovea of the eye is = $2 \mu m$

The distance of separation by one cone(not excited ) is $d = 4\mu m = 4*10^{-4}cm$

The distance between the two point-like object is l

The diameter of the eye is D = 2 cm

The distance of the two point-like object from the near point of the eye is A = 28 cm

From the diagram we see that the light from the two point-like object form a triangle of similar base l and d and height D and A

So for a triangle with similar base we have that

$\frac{l}{A} =\frac{d}{D}$

$\frac{l}{28} = \frac{4*10^{-4}}{2}$

making l the subject we have

$l = \frac{28 *4*10^{-4}}{2}$

$l=0.0056\ cm$

5 0

3 years ago

Read 2 more answers