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Alexus [3.1K]
3 years ago
5

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

ame location as Rock X. Both rocks fall for several seconds before landing on the ground directly below the cliff. Frictional forces are considered to be negligible. After Rock Y is released from rest several seconds after Rock X is released from rest, what happens to the separation distance S between the rocks as they fall but before they reach the ground, and why? Take the positive direction to be downward.
Physics
1 answer:
babunello [35]3 years ago
7 0

Answer:

Separation increases at all times that rock X falls because it falls with a greater speed

Explanation:

For both rocks, let initial velocity ∪=0

To find the displacement at any given time interval of Δt then

S= ∪Δt +0.5gΔt²

Since rock X is first released followed by Y, then X has a greater speed than Y therefore the distance covered by X is longer. This is because despite 0.5gΔt² being same for both rocks at any time Δt but rock X having already attained some velocity, its ∪Δt  is more hence the separation S increases. Conclusively, S increases at all times that rock X falls since rock X falls with a greater velocity than rock Y

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Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro
Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

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f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

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3 years ago
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