As the distance from a charged particle, "q", increases, the electric potential decreases.
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Electric potential between particles</h3>
The electric potential between particles is the work done in moving a unit charge from infinity to a certain point against the electrical resistance of the field.
V = Kq/r
where;
- K is Coulomb's constant
- q is the magnitude of the charge
- r is the distance between the charges
Thus, from the formula above, as the distance from a charged particle, "q", increases, the electric potential decreases.
Learn more about electric potential here: brainly.com/question/14306881
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Answer:
m = 1 kg
Explanation:
Given that,
The force constant of the spring, k = 39.5 N/m
The frequency of oscillation, f = 1 Hz
The frequency of oscillation is given by the formula as formula as follows :
So, the mass that is attached to the spring is 1 kg.
The net force does not depend on the mass.
We have 12N to the right, and 19N to the left.
The net force is (19.0-12.0)N=7.0N to the left.
Answer:
10.88 km
Explanation:
We shall represent displacement in terms of i , j unit vectors in the direction of east and north .
4.5 km due west
D₁ = - 4.5 i
6.7 km at an angle of 27° south of west
D₂ = - 6.7 cos27 i - 6.7 sin27j
= - 6.7 x .89 i - 6.7 x .45 j
= - 5.96i - 3 j
Total displacement
= D₁ + D₂
= - 4.5 i - 5.96i - 3 j
= -10.46 i - 3j
Magnitude = √ ( 10.46² + 3²)
= √ ( 109.41 + 9)
= √ 118.41
= 10.88 km .
Isolate the variable by dividing each side by factors that don't contain the variable.
m
=
32.021
¯
3
s
