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Lubov Fominskaja [6]

3 years ago

9

Assuming the mass of the moon is 7.4 ×10^22 kg and it's radius is 1 .7 × 10^6m. Calculate the gravitational energy of a 1kg mass

at the moon's surface

1 answer:

nirvana33 [79]3 years ago

3 0

Replace 1kg for the mass of the moon

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A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo

cestrela7 [59]

Answer:

Volume of balloon = 1000 cm^3

Explanation:

The head of a normal person can be assumed as a sphere with radius 10 cm.

Volume of sphere $=\frac{4}{3} \pi r^3$ , where r is the radius.

We have approximate radius = 10 cm.

Approximate volume of head $=\frac{4}{3} \pi r^3=\frac{4}{3} *\pi* 10^3=4188cm^3$

In the given options the closest value to the approximate volume is 1000 cm^3.

So, volume of head = Volume of balloon = 1000 cm^3

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3 years ago

1. What part (or parts) of this system store potential energy?

choli [55]

Answer:

<em>➢</em><em>when you crank you make kinetic energy and then the kinetic energy makes potential energy.</em>

Explanation:

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5 0

3 years ago

2. In the pendulum lab, the period (swing time) was O a. the independent variable O b.graphed on the vertical-axis O c.graphed o

Nadya [2.5K]

B is appropriate answer

7 0

3 years ago

Bob has been Springfield's top meteorologist for ten years. He received data from satellite imaging and computer models that a h

djyliett [7]

The correct letter of the answer would be letter b, Bob is using his personal knowledge of the local area to make his forecast. It does not look like he's ignoring the computer for he's been the top meteorologist for ten years, meaning to say, he would use his knowledge about the local area and tell them his knowledge about the weather.

7 0

3 years ago

Read 2 more answers

Starting from rest, a particle that is confined to move along a straight line is accelerated at a rate of 5.0 m/s2. Which one of

Elanso [62]

Answer:

c) The slope is not constant and increases with increasing time.

Explanation:

The equation for the position of this particle (starting from rest is)

$s = at^2/2 = 5t^2/2 = 2.5t^2$

We can take derivative of this with respect to time t to get the equation of slope:

$s' = (2.5t^2)' = 2*2.5t = 5t$

As time t increase, the slope would increases with time as well.

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4 years ago