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defon
2 years ago
9

What do you need to include in a free body diagram?

Physics
1 answer:
Licemer1 [7]2 years ago
3 0

Answer:

Interact with it

Explanation:

The body in a condensed form (often a dot or a box)

Straight arrows pointing in the direction in which forces act on the body are represented.

Moments are portrayed by curved arrows pointing in the direction in which they impact the body.

A coordinate system is a series of coordinates.

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How much work does it take to move a 50 μC charge<br> against a 12 V potential difference?
lukranit [14]
<span>work =V*Q =12*50*10^-6

The total work done will be equal to 

work = V.Q

which means 

w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>
8 0
2 years ago
Ask a member of the family to help you.Do the following activities and identify the skill/skills being excited . use a separate
Archy [21]

Answer:

Ok. Thanks.

I'll try it out.

3 0
2 years ago
The friction that prevents an object from moving when a force is applied is ________ friction.
uranmaximum [27]
Static friction is what you are looking for.
Kinetic friction is the force exerted on an already moving object, slowing it down.
3 0
3 years ago
Which metal and nonmetal families are the most different and why
nevsk [136]

Answer:

since they have got specific and special metallic and non metallic characteristics I guess

3 0
3 years ago
A force of 44 N will stretch a rubber band 88 cm ​(0.080.08 ​m). Assuming that​ Hooke's law​ applies, how far will aa 11​-N forc
Setler79 [48]

Answer:

<em>The rubber band will be stretched 0.02 m.</em>

<em>The work done in stretching is 0.11 J.</em>

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = <em>0.02 m</em>

<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.

potential energy stored = \frac{1}{2}ke^{2}

==> \frac{1}{2}* 550* 0.02^{2} = <em>0.11 J</em>

3 0
2 years ago
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