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Ulleksa [173]
3 years ago
14

How many nanoseconds does it take light to travel 3.50 ft in vacuum?

Physics
1 answer:
Fiesta28 [93]3 years ago
6 0
Answer:3.56 nanosecond

In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:

time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft=  3.56 10^{-9} second or 3.56 nanosecond
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What force acts between earth and the moon
ololo11 [35]

Explanation:

Gravity

Gravity is one major force that creates tides. In 1687, Sir Isaac Newton explained that ocean tides result from the gravitational attraction of the sun and moon on the oceans of the earth (Sumich, J.L., 1996).

4 0
3 years ago
Heat is distributed through the atmosphere by winds<br> True<br><br> False
amm1812

Answer:

True!

Explanation:

Hope this helps!

5 0
3 years ago
PLSS HELP NOWW!!!!!!(01.01 LC) Which is an example of the force of attraction between two objects that have mass? (2 points) a M
Lyrx [107]

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6 0
3 years ago
It is night. Someone who is 4 feet tall is walking away from a street light at a rate of 8 feet per second. The street light is
KiRa [710]

Answer: 4 ft/s

Explanation:

Given

height of man=4 ft

speed of person v=8 ft/s

height if street light=12 ft

Let x be the distance between person and street light and y be the length of his shadow

From diagram

as the two triangle ADE and ABC are similar therefore we can say that

\frac{4}{12}=\frac{y}{x+y}

\frac{1}{3}=\frac{y}{x+y}

x+y=3y

x=2y

differentiate above Equation w.r.t time we get

\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} y}{\mathrm{d} t}

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6 0
3 years ago
A cello string vibrates in its fundamental mode with a frequency of 335 1/s. The vibrating segment is 28.5 cm long and has a mas
Inga [223]

Answer:

The tension in string is found to be 188.06 N

Explanation:

For the vibrating string the fundamental frequency is given as:

f1 = v/2L

where,

f1 = fundamental frequency = 335 Hz

v = speed of wave

L = length of string = 28.5 cm = 0.285 m

Therefore,

v = f1 2L

v = (335 Hz)(2)(0.285)

v = 190.95 m/s

Now, for the tension:

v = √T/μ

v² = T/μ

T = v² μ

where,

T = Tension

v = speed = 190.95 m/s

μ = linear mass density of string = mass/L = 0.00147 kg/0.285 m = 5.15 x 10^-3 kg/m

Therefore,

T = (190.95 m/s)²(5.15 x 10^-3 kg/m)

<u>T = 188.06 N</u>

4 0
3 years ago
Read 2 more answers
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