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3 years ago

Amount of work done by a rotating object

2 answers:

5 0

The work done by a rotating object can be calculated by the formula Work = Torque * angle.

This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.

An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.

The formula is Work = torque * angle.

Torque = 1000 N*m

Angle = [50 revolutions] * [2π radians/revolution] = 100π radians

=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.

riadik2000 [5.3K]3 years ago

5 0

There is no question here try to add a list of options and repost it

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One horsepower (1 hp) is the unit of power based on the work that a horse can do in one second. This is defined, in English unit

nikdorinn [45]

Power is defined as rate of work done which means it is work done in 1 second of time

Now it is given that we have a horse that will give power 745.7 W

So it will do work of 745.7 J in 1 second of time

now if we wish to find the work done by horse in 0.55 s

so we can say

$W = Power \times time$

$W = 745.7 \times 0.55$

$W = 410.14 J$

So it will do total work of 410.14 J in 0.55 s of time

7 0

3 years ago

Why don’t we feel the gravitational force of a large object such as a skyscraper semi-truck?

Kobotan [32]

Answer:

Se the explanation below

Explanation:

We do not feel these forces of these bodies, because they are very small compared to the force of Earth's attraction. Although its mass is greater than that of a human being, its mass is not compared to the Earth's mass. In order to understand this problem we will use numerical data and the universal gravitation formula, to give validity to the explanation.

Force exerted by the Earth on a human being

$F=G*\frac{m_{1}*m_{2}}{r^2}$

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 5.97*10^24[kg]

r = distance from the center of the earth to the surface or earth radius = 6371 *10^3 [m]

Now replacing we have

$F = 6.673*10^{-11} *\frac{80*5.97*10^{24}}{(6371*10^{3})^{2} } \\F = 785[N]$

Force exerted by a building on a human being

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 300000 [ton] = 300 *10^6[kg]

r = distance from the building to the person = 2[m]

$F = 6.673*10^{-11}*\frac{80*300*10^6}{2^{2} } \\F= 0.4 [N]$

As we can see the force exerted by the Earth is 2000 times greater than that exerted by a building with the proposed data.

8 0

3 years ago

A force that pushes or pulls is known as

xeze [42]

Answer:

A force that pushes or pulls is known as Newton's third law of Motion.

Explanation:

Newton's Third Law of Motion. Newton's Third Law of Motion states that for each action, there's an equal and opposite reaction. What this suggests is that pushing on an object causes that object to keep off against you, the precise same amount, but within the other way.

7 0

3 years ago

Read 2 more answers

A total of 25.6 kJ of heat energy is added to a 5.46 L sample of helium at 0.991 atm. The gas is allowed to expand against a fix

bagirrra123 [75]

Answer:

(a) W = 1329.5 J = 1.33 KJ

(b) ΔU = 24.27 KJ

Explanation:

(a)

Work done by the gas can be found by the following formula:

$W = P\Delta V$