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

mestny [16]

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

Time = Distance / speed

Time = 7 / 3×10⁸

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

8 0

2 years ago

How many atoms of each element are in the chemical formula NH4NO3?

Igoryamba

<span>2 Nitrogen, 4 Hydrogen, 3 Oxygen

9 atoms per molecule.

NH4, ammonium (not to be confused with ammonia NH3) is a 1+ ion and NO3 is 1-.
</span>

4 0

3 years ago

Read 2 more answers

A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc

Hoochie [10]

Answer:

a= - 6.667 m/s²

Explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v= 0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

$a=- \dfrac{20^2}{2\times 30} \ m/s^2$

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be - 6.667 m/s² .

6 0

3 years ago

The position-time equation for a certain train is

astraxan [27]

Answer:

$a=4.8m/s^2$

Explanation:

Hello,

In this case, since the acceleration in terms of position is defined as its second derivative:

$a=\frac{d^2x(t)}{dt^2}=\frac{d^2}{dt^2}(2.9+8.8t+2.4t^2)$

The purpose here is derive x(t) twice as follows:

$a=\frac{d^2x(t)}{dt^2}=\frac{d}{dt}(8.8+2*2.4*t)\\ \\a=4.8m/s^2$

Thus, the acceleration turns out 4.8 meters per squared seconds.

Best regards.

8 0

4 years ago

A real image is four times as far from a lens as is theobject.

saveliy_v [14]

Answer:

1.25 focal lengths

Explanation:

The lens equation states that:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the object distance

q is the image distance

In this problem, the image is 4 times as far from the lens as is the object: this means that

$q=4p$

If we substitute this into the lens equation and we rearrange it, we get

$\frac{1}{f}=\frac{1}{p}+\frac{1}{4p}=\frac{4+1}{4p}=\frac{5}{4p}\\p=\frac{5}{4}f=1.25 f$

so, the object distance measured in focal lengths is

1.25 focal lenghts

3 0

4 years ago