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3 years ago

8

A waiter is carrying a tray above his head and walking at a constant velocity. If he applies a force of 5.0 newtons on the tray

and covers a distance of 10.0 meters, how much is the work being done?

2 answers:

salantis [7]3 years ago

7 0

the other answer is wrong, the answer is A. 0 joules for plato users, i got 100 on the test

k0ka [10]3 years ago

4 0

W = f × x = 5 × 10 = 50

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An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t

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3 years ago

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest

shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer: 3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f = 5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

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3 years ago

A rock with a mass of 0.3 kg falls from the top of a cliff. If it takes the rock 2.5 s to reach the ground, what was the impulse

kaheart [24]

Impulse=force x time
force=mass x acceleration due to gravity
force=
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impulse =3000 x 2.5= ( sorry i don't have a calculator right now so you must calculate this yourself)
I converted from kg to g because it is the standard.
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Elanso [62]

A is the correct answer

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3 years ago

Read 2 more answers