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svet-max [94.6K]
3 years ago
8

large truck exerts 7000 n of force on a piston with an area of 0.4m squared and the is ton can only support 16,000 pa of pressur

e. how much pressure is exerted by the truck on the piston
Physics
1 answer:
sasho [114]3 years ago
6 0
7000 N / 0,4m^2  = 17 500 Pa.
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Orchestra instruments are commonly tuned to match an A-note played by the principal oboe. The Baltimore Symphony Orchestra tunes
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Answer:

Δλ = 3*10⁻³ m.

Explanation:

  • At any wave, there exists a fixed relationship between the speed of  the wave, the wavelength, and the frequency, as follows:

       v = \lambda* f  (1)

       where v is the speed, λ is the wavelength and f is the frequency.

  • Rearranging terms, we can get λ from the other two parameters, as follows:

       \lambda = \frac{v}{f}  (2)

  • Since v is constant for sound at 343 m/s, we can find the different wavelengths at different frequencies, as follows:

        \lambda_{1} =\frac{v}{f_{1}} = \frac{343m/s}{440(1/s)} = 0.779 m  (3)

        \lambda_{2} =\frac{v}{f_{2}} = \frac{343m/s}{442(1/s)} = 0.776 m  (4)

  • The difference between both wavelengths, is just the difference between (3) and (4):

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       ⇒ Δλ = 3*10⁻³ m.

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Chord progressions that move to resting points that release tension are called
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A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

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Then:

$d u=-\sin (\omega t) d t

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Replacing that in our integral we get:

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Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

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Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

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