if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

3 years ago

An amplifier has a 50 watt output and a 5 watt input. what is the gain in decibels for this amplifier, rounded to the nearest de

Marrrta [24]

Gain in decibels is given by;

Gain db = 10*log (Po/Pi), where Po = Power output, Pin = Power input

Substituting;

Gain in db = 10 * log (50/5) = 10 db

6 0

2 years ago

Consider an ideal monatomic gas of N particles with mass m in thermal equilibrium at a temperature T. The gas is contained in a

harina [27]

Answer:

$K.E.=\dfrac{3}{2}KT$

Explanation:

Given that

Number of particle =N

Equilibrium temperature= T

Side of cube = L

Gravitational acceleration =g

The kinetic energy of an atom given as

$K.E.=\dfrac{3}{2}KT$

Where

Equilibrium temperature= T

Boltzmann constant =K

K =1.380649×10−23 J/K

3 0

2 years ago

A television camera lens has a 17-cm focal length and a lens diameter of 6.0 cm. what is its number?

IRINA_888 [86]

Answer:

= 2.83

Explanation:

F number (N) is given by the formula;

F- number = f/D

where f = focal length of lens and D = diameter of the aperture

Therefore;

F number = 17 cm/6 cm

<u> = 2.83</u>

3 0

2 years ago