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stiks02 [169]
2 years ago
13

Help me PLEASE! 25 points!!!!! :)

Physics
1 answer:
NeTakaya2 years ago
7 0

Answer:

A.

Explanation:

All the arrows align to the point right

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Distance is total path travelled by an object during its entire journey. It is a scalar quantity i.e only magnitude.

Displacement is the shortest distance covered by an object. It is basically the change in position of object. It is a vector quantity i.e direction as well as magnitude.

When an object is travelling in a straight line and stops at the end point, then both distance and displacement are same.

When an object is travelling in a straight line and then changes its direction or we can say come backwards then the magnitude of distance and displacement are different.

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4 . A negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it . ( a ) what is the unk
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(a) The unknown charge kept at distance 0.250m directly below  -0.450 charge is  2.315 x 10⁻¹² C.

(b) The magnitude of the force is 29.16 x 10¹² N.

<h3>What is electrostatic force?</h3>

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/r²

where k = 9 x 10⁹ N.m²/C²

Given is a negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it .

In equation of the magnitude of force, only the magnitude of charges are taken.

Substituting the values into the expression ,we get the charge

0.150 = (9 x 10⁹ x 0.450x q)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the unknown charge is 2.315 x 10⁻¹² C.

b)Let the unknown charge be  0.450C to exert on -450C.

Substituting the values into the expression ,we get the electrostatic force

F = kq₁q₂/r²

F = (9 x 10⁹ x 0.450x 0.450)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the force is 29.16 x 10¹² N.

Learn more about electrostatic force.

brainly.com/question/9774180

#SPJ1

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1 year ago
If a 375 mL sample of water was cooled from 37.5 C to 0 C how much heat wad lost in joules
harkovskaia [24]
1 mL = 1 gram
Q=375*4.179*37.5

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