Can someone please answer this ? it doesn’t make sense to me

How do you find average speed on a graph

coldgirl [10]

Answer:

speed equals distance over time 50 divided by 5.

5 0

2 years ago

How much of the universe is believed to consist of ordinary atoms that can be detected?

Margaret [11]

The universe believed to be has no limit in space and the elements, molecules and different kind of gases are found in this area. So the percentage of that is believed that and ordinary atom can be detected is C. 3% i hope you understand my answer.

4 0

2 years ago

A television camera lens has a 17-cm focal length and a lens diameter of 6.0 cm. what is its number?

IRINA_888 [86]

Answer:

= 2.83

Explanation:

F number (N) is given by the formula;

F- number = f/D

where f = focal length of lens and D = diameter of the aperture

Therefore;

F number = 17 cm/6 cm

<u> = 2.83</u>

3 0

2 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$