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Rufina [12.5K]
3 years ago
6

The gravitational force of attraction between two objects would increase by

Physics
1 answer:
OleMash [197]3 years ago
6 0
The gravitational force of attraction between two objects would be increased by "decreasing the distance between two objects"

Hope this helps!
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HELP ASAP! <br>why is rolling friction much smaller than sliding friction? Also give an example.​
IceJOKER [234]

Answer:

Rolling friction is much smaller than sliding friction because Rolling friction is considerably less than sliding friction as there is no work done against the body that is rolling by the force of friction. For a body to start rolling a small amount of friction is required at the point where it rests on the other surface, else it would slide instead of roll.

Rolling Friction example: Anything with weels (cars,skateboards) or a ball rooling.

Sliding Friction example: Bicycle brakes,skinning your knee walking,writing.

6 0
3 years ago
When hot and cold air meet, the hot air rises to the top. What causes the hot air to rise???
vekshin1
Hot air rises<span> because when you </span>heat air<span> (or any other gas for that matter), it expands. When the </span>air<span> expands, it becomes less dense than the </span>air<span>around it. The less dense </span>hot air<span> then floats in the more dense cold </span>air<span> much like wood floats on water because wood is less dense than water.</span>
3 0
3 years ago
C. A helium atom has a diameter of approximately 9.8 • 10-11 meters. What is the diameter of a helium atom in nanometers? Given
Studentka2010 [4]

1\ nm = 10^{-9} m\\\\1\ m =\frac{1\ nm}{10^{-9}}

Since the diameter of helium atom is approximately 9.8\times 10^{-11}\ m, therefore the diameter of helium atom in nano meter,

9.8\times 10^{-11}\ m=9.8\times 10^{-11} \times\frac{1\ nm}{10^{-9}}=0.098\ nm

4 0
4 years ago
What are two parts of an atom
seraphim [82]

Answer:

Nucleus And electron cloud

Explanation:

Hope this helps

7 0
3 years ago
A string of 26 identical Christmas tree lights are connected in series to a 120 V source. The string dissipates 73 W. What is th
spin [16.1K]

To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

R_{eq}= \frac{V}{I}

Here,

V = Voltage

I = Current

While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,

P=VI \rightarrow I = \frac{P}{V}

I =0.608 A

Applying Ohm's law

R_{eq} = \frac{120V}{0.608A}

R_{eq} = 197.4\Omega

Therefore the equivalent resistance of the light string is 197.4\Omega

6 0
3 years ago
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