How many nanoseconds does it take light to travel 2.50 feet in a vacuum

1 answer:

5 0

A handy-dandy "rule of thumb" that I and many other engineers
have been using for years is:

   1 nanosecond ===> 1 foot.

Let's figure it out and see whether that's even close:

Speed of light in vacuum = 299,792,458 meters per second.

   (299,792,458 m/s) x ( 1 sec / 10⁹ nanosec) x (3.28084 ft/meter)

   = (299,792,458 x 1 x 3.28084 x 10⁻⁹) ft/nanosec

   = 0.983571 ft/nanosec .

   That's 1.64% less than 1.000000 .
   The engineer's handy-dandy rule of thumb is valid.

2.5 feet would take

   (2.5 x 0.983571 sec) =   2.45893 nanoseconds

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3 years ago

Read 2 more answers

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$