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Agata [3.3K]
3 years ago
9

How many nanoseconds does it take light to travel 1.00 ft in vacuum? (This result is a useful quantity to remember.)

Physics
1 answer:
Svetach [21]3 years ago
4 0

To solve this problem, we will define speed as the amount of distance traveled per unit of time. We will clear the value of time and in parallel we will convert the Units to an international system to facilitate the calculation since we know the speed of light in a vacuum. The speed defined in terms of distance and time would be

v = \frac{x}{t} \rightarrow t = \frac{d}{v}

The speed of light in a vacuum is  3 * 10 ^ 8m / s and 1ft is equivalent to 0.3048m, so the estimated time would be

t = \frac{d}{v}

t = \frac{0.3048m}{3*10^8m/s}

t = 1.02*10^{-9}s

We know that 1 second is equivalent to 10 ^ 9ns, therefore

t =  1.02*10^{-9}s (\frac{10^9ns}{1s})

t = 1.02ns

Therefore the time is 1.02ns

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2. An electric motor transfers 2003 in 10.0 s. If it then transfers the same amount of energy in 8.0 s, has its
Marizza181 [45]

Answer:

decreased

Explanation:

because it transfered from 10.0s

to 8.0s

5 0
2 years ago
A hammer taps on the end of a 4.10-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two p
xz_007 [3.2K]

Answer:

Speed of sound inside metal is ≅ 8200 \frac{m}{s}

Explanation:

Given :

Length of metal bar x = 4.10 m

From general velocity equation,

 v = \frac{x}{t}

Where v = speed of sound in air = 343 \frac{m}{s}

For finding time from above equation,

  t = \frac{x}{v}

 t = \frac{4}{343}

t = 0.01166 sec

Since pulses are separated by  t_{o} =  11.1 \times 10^{-3} = 0.0111 sec

So we take time difference,

\Delta t = t_{} -t_{o}  = 0.0005

So speed of sound in metal is,

 v = \frac{x}{\Delta t }

 v = \frac{4.10}{0.0005}

 v = 8200 \frac{m}{s}

4 0
3 years ago
You push a heavy crate down a ramp at a constant velocity. Only four forces act on the crate. Which force does the greatest magn
Brilliant_brown [7]

The friction force does the greatest magnitude of work on the crate

Consider all four forces. The normal force does no work at all, since there is no motion in the direction of that force, perpendicular to the ramp. The force of gravity is smaller than the force of friction, since you still need to push the crate to get constant velocity. The force of you pushing is also smaller than the force of friction, since you are moving down a ramp, and are therefore assisted by gravity. Therefore the force doing greatest magnitude of work is the force of friction. Note that, even though the frictional work is negative, it still has the greatest magnitude

Learn more about friction force here:

brainly.com/question/4618599

#SPJ4

6 0
2 years ago
An object is falling toward Earth's surface, and its acceleration due to gravity is measured. What is NOT needed to calculate Ea
mario62 [17]

I think it's weight of the object.

7 0
3 years ago
Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,
adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

               x_cm = 1 /M   ∑ x_{i} m_{i}

               y_cm = 1 /M   ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

    x4=?

   y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

    M = 2.00 + 2.20 + 3.40

    M = 7.6 kg

    0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

     x4 = -6.8 / 6

     x4 = -1,133 m

Axis y

    0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

    y4 = -11/6

    y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0
4 years ago
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