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tensa zangetsu [6.8K]
3 years ago
14

The frequency of a sound wave was measured to be 351 Hz. If you know that the speed of sound in air was 343 m/s, then determine

the wavelength of the sound wave.
Physics
1 answer:
netineya [11]3 years ago
4 0

Answer:

wavelength of the sound wave will be 0.977 m

Explanation:

We have given frequency of the sound wave f = 351 Hz

Speed of the sound wave v = 343 m/sec

We have to find the wavelength of the sound wave

We know that velocity of the sound wave is given by v=\lambda f

So wavelength of the sound wave \lambda =\frac{v}{f}=\frac{343}{351}=0.977m

So wavelength of the sound wave will be 0.977 m

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U = 3, 9 , v = 4, 2 (a) find the projection of u onto v. (b) find the vector component of u orthogonal to v.
ExtremeBDS [4]

Answer:

(a) At U = 3, 9 , v = 4, 2, the projection of u onto v is w1=<2,8>

(b)At U = 3, 9 , v = 4, 2,  the vector component of u orthogonal to v is w2 = <4,-1>

Explanation:

A

The projection of u onto v is given by:

w1= projvu = (u⋅v||v||2)v

Given that  u= <6,7> and v=<1,4>, we can find the projection of u onto v as shown below:

w1= projvu = (u⋅v||v||2v=(<6,7>⋅<1,4><1,4>⋅<1,)

=(6⋅1+7⋅41⋅1+4⋅4)<1,4>

=3417<1,4>

=<2,8>

Part B

The vector component of u orthogonal to v is given by:

Using the given vectors and the projection found in part (a), we can find the vector component of u orthogonal to v as shown below:

w2=u−projvu

=<6,7≻<2,8>

=<(6−2),(7−8)>

=<4,−1>

To learn more about vector component, click brainly.com/question/17016695

#SPJ4

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2 years ago
The physical structure of the earth’s rock is changed by _____.
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It could be stress or strain
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Q = cmAT
castortr0y [4]

Answer: Q=3000 cal

Explanation:

We are given the following formula:

Q=m. c. \Delta T   (1)

Where:

Q=3000 cal is the amount of heat

m=300g  is the mass  of water

c=1 cal/g \°C  is the specific heat of water

\Delta T  is the variation in temperature, which in this case is  \Delta T=30\°C-20\°C=10\°C  

Rewriting equation (1) with the known values at the right side, we will prove the result is 3000 cal:

Q=(300g)(1 cal/g \°C)(10\°C)   (2)

Q=3000 cal   This is the result

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A. Kelvin. It is 32 degrees fahrenheit

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3 years ago
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