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3 years ago

Any two application of gravity

Physics

1 answer:

Brrunno [24]3 years ago

5 0

Answer:

Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.

Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.

The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.

Hope that helps.

Explanation:

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If you ride your bicycle down a straight road for 500m then turn around and ride back, your distance is____ your displacement.

Helen [10]

Answer:

C-less than

Explanation:

Distance is total distance traveled (1000m here if you stop where you started).

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3 years ago

a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit

ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

$W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)$

We have

$\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}$

$\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}$

Then the total work is

$W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right) \approx \boxed{36.8\,\rm J}$

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2 years ago

A person is walking briskly in a straight line. The figure shows a graph of the person’s position x as a function of time t. Wha

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where $x(t)$ is the person's position at time $t$ . I've taken the liberty of estimating $x(6)\approx2.67\,\mathrm m$ . The closest choice among the answers is $0.8\,\dfrac{\mathrm m}{\mathrm s}$ .

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2 years ago

ou have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,

Dimas [21]

So far, since you moved into the apartment until the end of this much of the story, you haven't done ANY work on the dresser yet.

I'll admit that you pushed, groaned and grunted, sweated and strained plenty. You're physically and mentally exhausted, you're not interested in the dresser at the moment, and right now you just want to snappa cappa brew, crash on the couch, and watch cartoons on TV. But if you've done your Physics homework, you know you haven't technically done any work yet.

In Physics, "Work" is the product of Force times Distance.

Since the dresser hasn't budged yet, the Distahce is zero. So no matter how great the Force may be, it's multiplied by zero, so the Work is zero.

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3 years ago