4 years ago

What is the speed of light?

Physics

2 answers:

ozzi4 years ago

7 0

The speed of Light is 300,000,000 Meters/Seconds

Alona [7]4 years ago

3 0

C is correct....
300,000,000m/s

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If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, in

motikmotik

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

$F_{initial} = \frac{kq_1q_2}{r^2}$

Here,

k = Coulomb's constant

$q_{1,2}$ = Charge at each object

r = Distance between them

As the distance is doubled so,

$F_{final} = \frac{kq_1q_2}{( 2r )^2}$

$F_{final} = \frac{ kq_1q_2}{ 4r^2}$

$F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}$

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3 years ago

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Statements 1, 3, and 5 are true.

(A, C, and E)

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Ksenya-84 [330]

The answer is nucleus

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Look at figure D . Explain what will happen if a larger mass it put on the force meter

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it will experience great force

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3 years ago

The focal length of a lens is inversely proportional to the quantity (n-1), where n is the index of refraction of the lens of th

Ainat [17]

Answer:

46.22 cm

Explanation:

The focal refraction, fr is given by

$fr = \frac {c}{(1.572 -1)} = \frac {c}{0 .572}$

The focal red light is given by

$fv = \frac {c}{(1.605 - 1)} = \frac {c}{0.605}$

$\frac {fv}{fr} = \frac {0.572}{0 .605} = 0.945455$

$\frac {1}{fr} = \frac{1}{image} + \frac {1}{object}$ and making fr the subject we obtain

$fr = \frac {image * object}{(image + object)} = \frac {24.00 * 55} {(24.0 + 55)} = 16.70886 cm$

fv = 0.945455* 16.70886 cm = 15.79747 cm

$image = \frac {object * f} {(object - f)} = \frac {15.79747 * 24.0}{(24.0 - 15.79747)} = 46.22222 cm$

Therefore, violet image is approximately 46.22 cm

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4 years ago