A wave with a period of 0.008 second has a frequency of?

Physics

2 answers:

boyakko [2]3 years ago

8 0

As per the question the time period of the wave is 0.008 s.

we have to calculate the frequency of the wave.

Before going to calculate the frequency of the wave first we have to understand the frequency.

Frequency of a wave is defined as the number of waves passing through a point in unit time.

Time period of a wave means time taken by a wave for one complete vibration or oscillation.

The relation between frequency and wave is given as frequency is the reciprocal of time period.

Mathematically we can say $frequency[ f]=\frac{1}{T}$

$= \frac{1}{0.008s}$

=125 Hz [ans]

tatyana61 [14]3 years ago

6 0

The formula for frequency is f = 1/T where f is frequency and T is period in seconds.
You have you period which is 0.008s and that is all you will need to solve or frequency in a wave:
f = 1/2
f = 1/0.008s
f = 125Hz

You might be interested in

What is the strength of the electric field ep 0.90 mm from a proton?

dem82 [27]

The electric field strength is inversely related to the square of the distance.so the strength of the electric field is given by

$E=\frac{1}{4\pi \epsilon _{0} } \frac{q}{r^{2} } = \frac{k q}{r^{2} }$

Here, $\frac{1}{4\pi \epsilon _{0} } = k$ is constant depend upon medium and its value is $9.0 \times10^{9} \ N m^2/C^2$ and q is charge and r is the distance.

Given $r = 0.90 mm = 9.0 \times 10^{-4} m$ and we know the charge of proton, $q = 1.6\times 10^{-19} \ C$ .

Therefore,

$E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\ E= 1.77 \times 10^{-3} N/C$