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mafiozo [28]
2 years ago
15

Particles q1, 92, and q3 are in a straight line.

Physics
1 answer:
Kisachek [45]2 years ago
7 0

The answer to the question is 1.125  Newton

Formula for electrostatic force is F = ( K q1 q2 )/ r²

where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.

Total net force on q3 will be the summation of electrostatic force between q1 and q3 and electrostatic force between q2 and q3, as all the three charges are of same sign and lie in the same line.

Electrostatic force between q1 and q3

r will be 0.500 + 0.500 = 1 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 1²

F₁₃ = 2.25  × 10⁻¹ N

Electrostatic force between q2 and q3

r will be 0.500 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 0.5²

F = (225 × 10⁻³) / (25 × 10⁻²)

F₂₃ = 9 × 10⁻¹ N

Total force on q3 will be F₁₃ + F₂₃

Total force on q3 =  ( 2.25  × 10⁻¹ ) + (9 × 10⁻¹ ) N

Total force on q3 =  ( 2.25  × 10⁻¹ ) + (9 × 10⁻¹ ) N

Total force on q3 =  ( 11.25  × 10⁻¹ ) N

Total force on q3 =  1.125  N

Thus after solving we got the net force on q3 as 1.125  Newton

Learn more about Electrostatic Force here:

brainly.com/question/23121845

#SPJ10

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A 2 kg, frictionless block is attached to a horizontal, ideal spring with spring constant 300 N/m. At t = 0 the spring is neithe
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Answer:

Explanation:

Given that,

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M = 2kg

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Velocity v = 12m/s

At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0

xo = 0

It velocity is 12m/s at t=0

Then, it initial velocity is

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Then, amplitude is given as

A = √[xo + (Vo²/ω²)]

Where

xo is the initial amplitude =0

Vo is the initial velocity =12m/s

ω is the angular frequency and it can be determine using

ω = √(k/m)

Where

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m is the mass of object = 2kg

Then,

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ω = 12.25 rad/s²

Then,

A = √[xo + (Vo²/ω²)]

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An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti
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Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

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  • Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
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       F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)

b)

  • The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

       F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N  (2)

c)

  • According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

       a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2  (3)

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  t = √(162.7/4.9)

The horizontal distance traveled in that time is ...

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