How much work is done by a crane that lowers 1000N of material a distance of 150

2 answers:

3 0

The crane does NO work.

Work = (force) x (distance)

= (1,000 N) x (150 meters ?) = 150,000 joules

But the work is in the direction of the force, and it isn't the crane
pushing the material down. It's gravity.

Gravity is doing the work.

If the material was being lowered by a cable wrapped around the shaft
of an electric generator, then you could use the work done by gravity to
generate some electrical energy, and then sell the energy.

Or, if the "material" happened to be water, you could let gravity lower it
through a turbine or a water wheel, and use the work done by gravity to
grind wheat.

Yes, the crane may be raising a sweat, working against gravity. The
purpose is only to prevent gravity from doing the work too fast.

Citrus2011 [14]4 years ago

3 0

Since workdone is Force times distance Given force-1000n Distance-150 Wd-1000 times 150 Wd-150000n.

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26. Keenan found the mass of a book to be 4.56*10^ -2 kg . What is the mass of the book in milligrams?

vagabundo [1.1K]

Taking into account the rule of three for the change of units, the mass of the book is 45600 miligrams.

First of all, the rule of three is a mathematical tool that helps you quickly solve proportionality problems.

Having three known values and one unknown, a proportional relationship is established between all of them in order to find the fourth term of the proportion.

If the relationship between the magnitudes is direct (when one magnitude increases, so does the other; or when one magnitude decreases, so does the other), the rule of three is applied as follows, where a, b and c are known values and x is the unknown to calculate:

a → b

c → x

So: $x=\frac{cxb}{a}$

Being 1 kg equivalent to 1000000 milligrams, In this case the rule of three is applied as follows: if 1 kg equals 1000000 milligrams, 4.56×10⁻² kg equals how many milligrams?

1 kg → 1000000 milligrams

4.56×10⁻² kg → x

So:

$x=\frac{4.56x10^{-2} kg x1000000 miligrams }{1 kg}$