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3 years ago

10

HELP!!!! PLEASE (and there will be more)

2 answers:

Alexus [3.1K]3 years ago

6 0

Answer:

i wanna say tranverse if not surface

Explanation:

dybincka [34]3 years ago

4 0

Answer : Surface Wave

Explanation:

Gravity waves along the surface of liquids, such as ocean waves. (btw it does look like ocean waves)

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An organism that cannot make its own food called

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heterotroph

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Which statement is true of a wave that’s propagating along the pavement and girders of a suspension bridge?

alexandr1967 [171]

The statement which is true of a wave that’s propagating along the pavement and girders of a suspension bridge is A. The wave is mechanical, with particles vibrating in a direction that is parallel to that of the wave, forming compressions and rarefactions.

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Which statement best describes what happens when more waves pass a certain point per second?

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Answer:D increase in frequency

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3 years ago

A note of frequency 200Hz has a velocity of 400m/s. what is the wavelength of the note

Xelga [282]

Answer:

$\huge\boxed{\sf \lambda = 2 m}$

Explanation:

<h3><u>Given data:</u></h3>

Frequency = f = 200 Hz

Velocity = v = 400 m/s

<h3><u>Required:</u></h3>

Wavelength = λ = ?

<h3><u>Formula:</u></h3>

v = fλ

<h3><u>Solution:</u></h3>

Put the givens in the formula

400 = (200)λ

Divide 200 to both sides

400/200 = λ

2 m = λ

λ = 2 m

$\rule[225]{225}{2}$

8 0

1 year ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago