All categories

3 years ago

How would the force between two charged particles change if one of the charges were to triple in strength (3x stronger)?

Physics

1 answer:

Andreyy893 years ago

8 0

Answer:

The correct option is (d). "3x stronger".

Explanation:

The force between two charged particle is given by :

$F=k\dfrac{q_1q_2}{r^2}$

If one charge is tripled, $q_1'=3q$

New force will be :

$F'=\dfrac{kq_1'q_2}{r^2}\\\\=3\times F$

It means if one of the charges were to triple in strength, then the force will become 3 times of the initial force.

You might be interested in

The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?

Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2$

$For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2$

The second distance, r₂, can be determined from sound intensity formula given as;

$I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m$

Therefore, the second distance of the sound from the source is 431.78 m.

7 0

3 years ago

Which is a compound machine? a broom a screwdriver a slide a pair of scissors

8090 [49]

A pair of scissors .....................................

4 0

3 years ago

A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyon

jek_recluse [69]

Answer:

x₂ = 1.33 m

Explanation:

For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall

Στ = 0

W₁ x₁ - W₂ x₂ = 0

where W₁ is the weight of the woman, W₂ the weight of the table.

Let's find the distances.

Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.

x₁ = 2.5 -1.5 = 1 m

The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative

x₂ = $\frac{M_1g }{m_2 g} \ x_1$

let's calculate

x₂ = $\frac{100}{75} \ 1$

x₂ = 1.33 m

7 0

3 years ago

The distance versus time graph for Object A and Object B are shown.

barxatty [35]

Object B but not objective A

8 0

3 years ago

Read 2 more answers

The ability to cause change is defined as:

ehidna [41]

The question is poorly written.
ALL of the choices cause a change in something.

A). Force causes a change in velocity.

B). and C). Power and energy cause changes requiring work.

D). Impulse causes a change in momentum.

5 0

4 years ago