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Solnce55 [7]
2 years ago
13

How long does it take light from a camera to travel to the face of someone standing 7 meters away?

Physics
1 answer:
mestny [16]2 years ago
8 0

The time taken for the light to travel from the camera to someone standing 7 m away is 2.33×10¯⁸ s

Speed is simply defined as the distance travelled per unit time. Mathematically, it is expressed as:

<h3>Speed = distance / time </h3>

With the above formula, we can obtain the time taken for the light to travel from the camera to someone standing 7 m away. This can be obtained as follow:

Distance = 7 m

Speed of light = 3×10⁸ m/s

<h3>Time =?</h3>

Time = Distance / speed

Time = 7 / 3×10⁸

<h3>Time = 2.33×10¯⁸ s</h3>

Therefore, the time taken for the light to travel from the camera to someone standing 7 m away is 2.33×10¯⁸ s

Learn more: brainly.com/question/14988345

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Explanation:

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What velocity must a car with a mass of 1280 kg have in order to have the same momentum as a 2230?
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3 years ago
An 80-cm uniform 10-kg bar is resting on two scales, one at either end. A smaller 4-kg mass (m) is placed at a distance of d = 2
Varvara68 [4.7K]

Answer

given,

length of bar = 80 cm

mass of the bar = 10 kg

smaller mass = 4 kg

distance = 20 cm

s_1 + s_2 = 10 + 4

s_1 + s_2 = 14\ kg

taking moment about B

s_1 \times 0.8 - 10 \times 0.4 - 4 \times 0.6 = 0

s_1 \times 0.8 = 6.4

s_1 = 8\ N

s_2 = 14 - s_1

s_2 = 14 - 8

s_2 = 6 N

difference between two scale = 8 - 6

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7 0
3 years ago
A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall
kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

F = q(\vec v \times \vec B)

so here we have

F = qvB sin90

since force is perpendicular to the velocity so here it must be centripetal force

here we have

\frac{mv^2}{R} = qvB

so we have

R = \frac{mv}{qB}

R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}

R = 4.06 cm

so here it will move in circle with radius 4.06 cm

8 0
3 years ago
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