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Svet_ta [14]
2 years ago
15

An rv is traveling 60 km/h along a highway at night. a boy sitting near the driver of the rv turns a flashlight on and shines it

at another boy at the back end of the rv. what is the speed of the light relative to the boys? m/s what is the speed of light relative to a stationary observer on the side of the road? m/s
Physics
1 answer:
nika2105 [10]2 years ago
3 0

Speed of light relative to the boys = 300,000,000 m/s

Speed of light relative to a stationary observer on the side of the road =

<h3>300,000,000 m/s</h3><h3>What is special theory of relativity?</h3>
  • The theory of special relativity explains how speed affects space, time, and mass.
  • Small amounts of mass (m) can be interchangeable with large amounts of energy (E), as defined by the classic equation E = mc2, according to the theory, which offers a means for the speed of light to define the link between energy and matter.

<h3>What are the special relativity's guiding principles?</h3>
  • The laws of physics should be independent of the uniform motion of an inertial frame of reference, and the speed of light should be constant in any such frame, according to two fundamental tenets that constitute the basis of the special theory of relativity.

To learn more about special relativity visit:

brainly.com/question/12497537

#SPJ4

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7 0
1 year ago
Two ping pong balls have been painted with metallic paint and charged by contact with an Van de Graaff generator. The charge on
andre [41]

Answer:

0.035 N

Explanation:

Parameters given:

Charge q1 = -3.31x10^(-7) C

Charge q2 = -5.7x10^(-7) C.

Distance between them, R = 22 cm = 0.22 m

Electrostatic force between to particles is given as:

F = (k* q1 * q2) / R²

F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²

F = 0.035 N

6 0
3 years ago
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When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to
Serjik [45]

Answer:

E) d/sqrt2

Explanation:

The initial electric force between the two charge is given by:

F=k\frac{q_1 q_2}{d^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

d is the separation between the two charges

We can also rewrite it as

d=\sqrt{k\frac{q_1 q_2}{F}}

So if we want to make the force F twice as strong,

F' = 2F

the new distance between the charges would be

d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}

so the correct option is E.

8 0
2 years ago
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