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Lapatulllka [165]
4 years ago
9

Finding the spring constant as shown, spring 3, which has an unknown spring constant k3, replaces spring 2. the mass of the weig

ht and pulley are unchanged: m=5.8 kg and mp=1.7 kg . however, because of the different spring constant the distance the pulley hangs below the ceiling is now, h=112.5 cm . the unloaded length of spring 3 is ℓ′3=27.5 cm ; after hanging the mass and pulley, the spring's length is ℓ3=34.5 cm . determine the spring constant of spring 3, k3.
Physics
1 answer:
nevsk [136]4 years ago
6 0
Replaces spring 2. the mass of the weight and pulley are unchanged: m=5.8 kg and mp=1.7 kg
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You push very hard on a heavy desk, trying to move it. You do work on the desk:
frozen [14]

Answer:

(C) Only if it starts moving

Explanation:

We know that work done is given by

W=F.d=Fdcos\Theta

So there are two case in which work done is zero

First case is that when force and displacement are perpendicular to each other

And other case is that when there is no displacement

So for work to be done there must have displacement, if there is no displacement then there is no work done

So option (c) will be the correct option

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3 years ago
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3 years ago
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You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength
Ray Of Light [21]

Answer:

wavelength \lambda = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = \frac{D\lambda}{d}    ....................1

\lambda is Wavelength of light and  D is Distance between screen and slit and d is slit width

so put here value and we get

\lambda = \frac{2.96*0.325*10^{-6}}{2.20}    

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4 0
3 years ago
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
Sati [7]

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

5 0
4 years ago
Which of these statements is correct about Newton's first law of motion?
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Answer:

Answer D: It describes the relationship between motion and force.

Explanation:

The answer is D because a law is something that describes something in nature, but does not try to explain how or why is occurs (that is a theory). Options B and C sound more like theories, while option A sounds like a definition. Option D is correct because a law describes without explaining.

4 0
3 years ago
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