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3 years ago

If the output work of a simple machine is ____ than the input work, the machine is said to have less than 100% efficiency.

Physics

1 answer:

bearhunter [10]3 years ago

6 0

Less than. A machine should have equal or more output than input to have a high efficiency. :)

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How much kinetic energy does an 80 kg man have while running at 3 m/s?

Soloha48 [4]

Hello!

$\large\boxed{KE = 360 J}$

Use the equation KE = 1/2mv² to solve for the kinetic energy of the man.

We are given the mass and velocity, so plug these values into the equation:

KE = 1/2(80)(3²)

KE = 1/2(720)

KE = 360 J

7 0

3 years ago

A ball was dropped from a height of 10 feet. Each time it hits the ground, it bounces 4/5 of its previous height. Find the total

Shtirlitz [24]

Answer:

d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

so we can say

$h_2 = \frac{4}{5}h$

so the distance covered is given as

$d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........$

here we know that

h = 10 feet

$d = h + 2(\frac{4}{5}h)(1 + \frac{4}{5} + (\frac{4}{5})^2 + ...........)$

$d = 10 + 2(\frac{4}{5}(10))(\frac{1}{1 - \frac{4}{5}})$

$d = 90 ft$

8 0

3 years ago

A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su

aniked [119]

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒ G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² = 4 ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

7 0

3 years ago

Chase is trying to push a 120-kilogram boulder along the street. He applies a force of 1,115 newtons, and he does this over a di

lord [1]

Answer:

The work done is equal to zero = 0

Explanation:

Let us remember that the definition of physical work is given by the product of force by the displacement of the body.

$Work=F*d\\\\where\\F= force [N]\\d=distance [m]\\replacing\\Work=1115*(0) = 0$