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VLD [36.1K]
3 years ago
10

A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.

Physics
1 answer:
sladkih [1.3K]3 years ago
5 0

Explanation:

Given that,

Charge 1, q_1=6.75\ nC=6.75 \times 10^{-9}\ C

Charge 2, q_2=4.46\ nC=4.46\times 10^{-9}\ C

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}

F=6.84\times 10^{-8}\ N

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

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A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring
sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

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