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ratelena [41]
3 years ago
12

Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?

Physics
1 answer:
Licemer1 [7]3 years ago
8 0
Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

Foot-pounds are gross units. Better to work in SI units when you can!

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Answer:You can calculate it by finding the amount of work exerted.

Explanation:

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3 years ago
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A car turns a certain curve of radius 24.98 m with constant linear speed of
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Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

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7 0
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Answer:

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Therefore someone at a place might see a clear large shadow due to shinny sun reflection and another a small or dull Shadow at same time if the intensity of the sun or lighting source is going down.

Explanation:

The closer a body/object is to a lighting source the larger the shadow it produces, and the farther the body the smaller the shadow produced.

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ORIGIN OF MOON AND THE RELATIONSHIP WITH THE EARTH
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A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle
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Answer:

F_2 = 1.10 \mu N

Explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

F = \frac{kq_1q_2}{r^2}

now since it depends inverse on the square of the distance so we can say

\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}

now we know that

r_2 = 18.2 mm

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also we know that

F_1 = 2.45 \mu N

now from above equation we have

F_2 = \frac{r_1^2}{r_2^2} F_1

F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)

F_2 = 1.10 \mu N

5 0
3 years ago
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