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Alla [95]
2 years ago
5

If this charged soot particle is now isolated (that is, removed from the electric field described in the previous part), what wi

ll be the magnitude E of the electric field due to the particle at distance 1.00 meter from the particle?
Physics
1 answer:
kicyunya [14]2 years ago
7 0

The magnitude of the electric field due to the particle at the given distance is 9\times 10^9 q.

<h3>Electric field strength</h3>

The electric field strength of a charged particle is the force per unit charge in the given field.

The electric field strength of a charge is given as;

E = \frac{F}{q} \\\\E = \frac{kq}{r^2}

where;

  • k is Coulomb's constant
  • q is the charge
  • r is the distance

E = \frac{9\times 10^9 \times q}{1^2} \\\\E = 9\times 10^9 q

Thus, the magnitude of the electric field due to the particle at the given distance is 9\times 10^9 q.

Learn more about electric field here: brainly.com/question/14372859

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If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice
Savatey [412]

Answer:

remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).

The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.

The apparent brightness of a star is measured in watts per square meter.

If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).

Some of the examples of stars are;

- Canopus.

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- Antares.

- Vega.

8 0
3 years ago
A car starting from rest has a constant acceleration of 4 meters per second per second. How fast will it be going in 5 seconds?
irinina [24]

Answer:

18 m

Explanation:

Given : vo = 0 m/s ; t = 3 s; a = 4 m/s^2 ; d = ? m ; average velocity = ? m/s ; fonal velocity = ? m/s

solving for the final velocity, v

v = a * t

v = 4 m/s^2 * 3 s

v = 12 m / s

Solving for the average velocity. avg v

avg v = (vo + v) / 2

avg v = (0 m / s + 12 m/s) / 2

avg v = 6 m / s

Solving for the distance traveled after 3 s

d = avg v * t

d = 6 m / s * 3 s

d = 18 meters

In the first 3s the car travels 18 meters.

4 0
2 years ago
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