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tatyana61 [14]
3 years ago
8

If a piece of space debris is too large to be a meteoroid and too small to be a planet, it could be

Physics
2 answers:
rewona [7]3 years ago
8 0

The answer is B. an asteroid

Vaselesa [24]3 years ago
3 0

an asteroid was the correct answer


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Joe has a mass of 40 kg, and Sally has a mass of 25 kg. They sit in class with a separation of 6 meters. Find the gravitational
goldfiish [28.3K]

Answer:

hmmmmmmmm

Explanation:

mmmmmmmmmmmmmmmmmmmmmmm  pay attention in class kid

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PLZ HELP FAST! Which two statements help explain why digital storage of data is so reliable?
Shkiper50 [21]

Answer:

C and D

Explanation:

But really, You should be able to answer this with the tech knowledge of a tomato.  You're given four answers, and are to choose which two are explain the reliability of digital storage.

The first two describe bad nasty things, the second two describe beneficial things.

So logically....

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3 years ago
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2. In the activities below indicate if friction is useful or not useful? Explain your answer. i. Writing yes it is useful becaus
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Answer:

Please find the answer in the explanation

Explanation:

Friction is a force that opposes motion. One or two of the advantages of friction are break and ability of an object to walk.

Writing yes it is useful because when your writing because friction helps you see what your writing

ii. Rubbing. Yes, it is useful.

friction make it possible for two object to rub each other

iii. Skiing. No. It is not useful because With presence of friction, skiing will not be possible.

iv. Rotating a wheel No. It is not useful because Friction will oppose the rotation of the wheel.

8 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

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3 years ago
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The energy diagram shows the changes in energy during a chemical reaction.
yanalaym [24]

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D

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