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Tanya [424]
3 years ago
12

A mass of 10 kg is at a point A on the table. It moved to a point B. If the line joining A and B is horizontal. What is the work

done on the object by the gravitational force ? Explain your answer
Physics
1 answer:
elena55 [62]3 years ago
6 0

Answer:

Work done by gravitational force = 0 joule

Explanation:

Work done is given by the relation

                        W  =  F   x    S   joule

Where,    F  -  the force applied in the form of push or pull

               S  -  displacement caused by the force

If a force is acting on a body and it doesn't cause any displacement, then work done will be zero.

Gravitational force acts on the body even if the body is at rest.

<em>Work done by the gravitational is applicable only if there is some vertical component of motion involved.</em>

In this case, some amount of work is done to move the body from point A to B.But, the body is displaced in the horizontal direction. No vertical motion is involved.

So, the work done due to gravitational force on a mass is zero joule.

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calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end
sdas [7]

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

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3 years ago
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What two types of evidence are used to classify orgianisms?
jeyben [28]
One is their traits and their characterists that they have in common 
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If a system has 475 kcal of work done to it, and releases 5.00 × 102 kJ of heat into its surroundings, what is the change in int
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First, we convert kcal to joules:
1 kcal = 4.184 kJ
475 kcal = 1987.4 kJ

Now, calculating the change in internal energy:

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ΔU = -500 + 1987.4
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