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allochka39001 [22]

3 years ago

6

7.How large is the moon compared to Earth? a. about the same diameter as Earth b. about one-half the diameter of Earth c. about

one-fourth the diameter of Earth d. about one-eighth the diameter of Earth

1 answer:

omeli [17]3 years ago

5 0

Answer:

C

Explanation:

-Google

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If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.

Ksenya-84 [330]

Answer:

twice

Explanation:

From magnification = height of image / height of object

Distance of image/ distance of object = magnification

If the distance and height of the object represents the initial light distance and the exposed surface respectively.

And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.

Hence the new image exposure would be twice as large.

If we use the formula our point of investigation is Height of image,

H2= D2/D1× H1

H2 = 2D2/D1 × H1

H2 = 2H1

6 0

3 years ago

A man can lift a mass of 200kg onThe surface of the earth. what is the amount of mass he can lift on the surface of the moon?

Arlecino [84]

Answer:

1,211.1 kg.

Explanation:

the force of gravity is less on the moon than on earth, so if the man can lift 200kg on earth, he could lift a greater amount on the moon because there is less resistance from gravity.

To know the amount of mass he can lift on the moon, we first need to know the amount of weight that is equivalent to those 200kg here on earth. This because the weight of the object is equal to the force that must be applied to lift it, and that force is applied by the man and it will be the same here and on the moon.

We calculate weight using the formula:

$w=mg$

where $w$ is the weight of the object (the force with which the earth attracts the object) $m$ is the mass and g the acceleration of gravity.

so

$w=200g$

for earth the acceleration due to gravity is: $g=9.81m/s^2$

thus:

$w=(200kg)(9.81m/s^2)\\w=1962N$

now we use this value to calculate the mass he can lift on the moon, since for the moon $g=1.62m/s^2$ .

we use the same equation, w =mg substituting w = 1962N and $g=1.62m/s^2$ :

$w=mg\\\\1962N=m(1.62m/s^2)\\\\m=\frac{1962N}{1.62m/s^2}\\\\ m=1,211.1kg$

he can lift 1,211.1 kg.

You can also find the result using the approximate value of the acceleration of gravity on the moon as g/6, where g is the acceleration on earth.

8 0

4 years ago

An electromagnetic wave consists of changing electric and magnetic fields.

NemiM [27]

I'm going to say that is it true because when electromagnetic waves occurs when a electric and magnetic field vibrates.

Hope this is correct and is what you are looking for!<span />

8 0

3 years ago

Read 2 more answers

Susan goes out to exercise. She runs for one hour at a constant speed and velocity. What is the overall net force acting on Susa

kiruha [24]

Option (A ) is correct.

Explanation:

susan is moving with constant velocity, so both the direction and magnitude of the velocity remains same. so the acceleration of susan =0. This is because an object gets accelerated when either the magnitude or direction of the speed changes.

now the force is given by

$F= ma$

F= force

m= mass

a= acceleration

Here a =0

so F= 0

so the net force on susan is zero.

3 0

3 years ago

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A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 0

3 years ago