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Makovka662 [10]

3 years ago

8

What is the independent variable and why?

2 answers:

Vedmedyk [2.9K]3 years ago

6 0

How long it takes to cook an egg. I think

valina [46]3 years ago

4 0

The independent variable is how long the egg takes to cook

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If the weight of displaced liquid and the weight of an object are the same, the object will:

liberstina [14]

Answer:

c

Explanation:

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____ charges are attract each other

timama [110]

Answer:

One positive and one negative

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Which best describes the relationship between the direction of energy and wave motion in a transverse wave?

sammy [17]

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A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What

iogann1982 [59]

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

$\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s$

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

$a=\omega^2 r$

At the instant t = 0,

$\omega=3.93 rad/s$

So the centripetal acceleration of the tip of the blades is

$a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2$

5 0

2 years ago

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

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

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

2 years ago