How to find gravitational potential energy?

Which one of the following is the correct option for fill in the

serg [7]

Answer:

mechanical advantage!

Explanation:

The Mechanical advantage of a machine is the factor by which the machine changes the input force.

When a a machine multiplies an input force, that's called a mechanical advantage.

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Defenition of Mechanical Advantage

brainly.com/question/16617083?referrer=searchResults

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Hope this helps! <3

5 0

3 years ago

A stunt woman attempts to swing from the roof of a m 24 high building to the bottom of an identical building using a m 24 rope,

MakcuM [25]

Answer:

Height from ground is 8 m where string will break

Explanation:

Let the string makes some angle with the vertical after some instant of time

So here we have

$T - mg cos\theta = \frac{mv^2}{R}$

$T = mg cos\theta + \frac{mv^2}{R}$

now by energy conservation we have

$\frac{1}{2}mv^2 = mg(R cos\theta)$

$mv^2 = 2mgR cos\theta$

$T = mg cos\theta + 2mgcos\theta$

$T = 3mg cos\theta$

For string break down we have

$T = 2mg = 3mgcos\theta$

$cos\theta = \frac{2}{3}$

Now height from the ground is given as

$h = R - Rcos\theta$

$h = 24 - 24(\frac{2}{3})$

$h = 8 cm$

7 0

3 years ago

When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a

Oxana [17]

Answer:

5 mg, $5\cdot 10^{-3}g$

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

$m=5\cdot 10^{-3}g$

If we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

$1 g : 1000 mg = 0.005 g : x$

and we find

$x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg$

4 0

3 years ago

the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path

Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:

$V=\sqrt{G\frac{M}{r}}$