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3 years ago

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A machine has a mechanical advantage of 4.5. What force is put out by the machine if the force applied to the machine is 800 N?

1 answer:

Ghella [55]3 years ago

3 0

If the machine's mechanical advantage is 4.5, that means that

Output force = (4.5) x (Input force) .

We know the input force, and we need to find the output force. Rather than wander around the room looking at the floor while our hair smolders, let's try putting the numbers we know into the equation I wrote up there. OK ?

Output force = (4.5) x (Input force)

Output force = (4.5) x (800 N)

Now dooda multiplication:

<em>Output force = 3,600 N</em> .

That's exactly what the question asked for. So we're done !

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Which statement is true about a planet’s orbital motion?

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Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

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3 years ago

During one cycle, a refrigerator removes 500 J from a cold reservoir and discharges 800 J to its hot reservoir. (a) What is its

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Answer:

a)COP = 1.67

b) W = 300 J

Explanation:

given,

refrigerator removes = 500 J from a cold reservoir

discharge = 800 J to its hot reservoir

a) COP = $\dfrac{Q}{w}$

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2 years ago

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A 12cm candle is placed 6cm from a converging lens with a focal length of 15cm. What is the height of the image of the candle? S

amm1812

Answer:

The height of the image of the candle is 20 cm.

Explanation:

Given that,

Size of the candle, h = 12 cm

Object distance from the candle, u = -6 cm

Focal length of converging lens, f = 15 cm

To find,

The height of the image of the candle.

Solution,

Firstly, we will find the image distance of the candle. Let it is equal to v. Using lens formula to find the image distance.

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

v is image distance

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-6)}\\\\v=-10\ cm$

If h' is the height of the image. Magnification is given by :

$m=\dfrac{h'}{h}=\dfrac{v}{u}$

$h'=\dfrac{vh}{u}\\\\h'=\dfrac{-10\times 12}{-6}\\\\h'=20\ cm$

So, the height of the image of the candle is 20 cm.

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3 years ago