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3 years ago

Please this is my senior work I’m trying to pass so help please What is Hooke’s Law

Physics

1 answer:

3241004551 [841]3 years ago

7 0

Answer:

It is a law of mechanics and physics discovered by Robert Hooke. This theory of elasticity says the extension of a spring is proportional to the load applied to it. Many materials obey this law as long as the load does not exceed the material's elastic limit.

Explanation:

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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

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

7 0

3 years ago

The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t

EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

$Ay = By = \frac{w * 6}{2} = 3w$

$P_c_r = 3w * F.S = 3w * 2.0 = 6w$

$I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7$

To find the maximum intensity, w, let's take the Pcr formula, we have:

$P_c_r = \frac{\pi^2 E I}{(KL)^2}$

Let's take k = 1

$E = 200*10^9$

Substituting figures, we have:

$6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}$

Solving for w, we have:

$w = \frac{67266.84}{6}$

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

$\sigma _c_r = \frac{w}{A}$

$\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y$ . This means it is safe

The maximum intensity w = 11.211KN/m

3 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLEST

DochEvi [55]

Answer:

1.62 m/s²

Explanation:

4 0

2 years ago

Would tap water likely freeze and boil at those exact temperatures? (temperature freezes at 0 degree celius and boils at 100 deg

Kitty [74]

Explanation:

Tap water will not boil and freeze at those exact temperature which are the boiling and freezing points.

O°C and 100°C are the boiling points of pure water.

Pure water is a pure substance with a definite composition made up of 2 atoms of hydrogen and 1 atom of oxygen.

Pure water boils at 100°C and freezes at O°C ,

Tap water is not pure water.
It has some impurities in it which causes causes the elevation of the boiling point of liquids and widening of the boiling range.
Tap water is treated water that contains some other dissolved chemicals.

learn more:

Molecular formula brainly.com/question/4574324

#learnwithBrainly

6 0

3 years ago

Define principal focus of a lence

Irina18 [472]

Answer:

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7 0

3 years ago