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3 years ago

Which of the following are not conditions for producing work?

Physics

2 answers:

jeka943 years ago

8 0

Answer:

A and C

Explanation:

We know that

Work done,W=Fd

Where F=Applied force

d=Displacement of object

Momentum, p= mv

It is not necessary that momentum must be greater than force .If momentum is less than force then the work can be produced because work depend upon displacement and applied force.

$W=Fdcos\theta$

If the value of theta is not equal to zero the work can be produced if theta is equal to 30 ,45 or 60 degree.

Therefore, it is not necessary that atleast part of the applies force must act along the same line object moves.

If force is greater than weight then, the work can be produced .

If force is less than the weight then, the work cannot be produced .

Hence, the force must be very large.

TEA [102]3 years ago

6 0

Answer:D

Explanation:

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The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

$N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N$

$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

$v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s$

Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

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nignag [31]

Answer:

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Explanation:

given data

wavelength = 656.5 nm

rest wavelength = 656.8 nm

to find out

the speed of the star , is it approaching

solution

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so put all value and find

velocity of source = 3 × $10^{-9}$ × 3 × $10^{8}$ / 656.8 × $10^{-9}$

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and

it is approaching to earth because wavelength of star is decreasing than rest

and

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

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