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Lubov Fominskaja [6]

3 years ago

11

If the speed of sound in air is 343 m/s and the wavelength of this note is 2.62 m, what is the frequency of this C3 note? (Round

your answer to three significant figures.)

1 answer:

snow_lady [41]3 years ago

8 0

Answer:

<em>The frequency of of the note = 131 Hz.</em>

Explanation:

<em>Frequency:</em><em> Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)</em>

v = λf ............................ Equation 1

Making f the subject of the equation,

f = v/λ .......................... Equation 2

Where v = Speed, λ = wavelength, f = frequency

<em>Given: v = 343 m/s, λ = 2.62 m.</em>

<em>Substituting these values into equation 2</em>

<em>f = 343/2.62</em>

<em>f = 131 Hz</em>

<em>Thus the frequency of of the note = 131 Hz.</em>

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Imagine that a hypothetical life form is discovered on our moon and transported to Earth. On a hot day, this life form begins to

almond37 [142]

Answer:

The heat of vaporization 580 cal/g times 602g = cal in human and do the same for life form.

Explanation:

4 0

4 years ago

A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base

umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0

4 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago

Two point charges 3q and −8q (with q > 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so

riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q 8q / L²

Q = 8q (x² / L²)

so charge required = - 8q (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

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3 years ago

Range of motion is the distance an object can travel when separated from another object. Please select the best answer from the

Lady_Fox [76]

Based on the given statement above, the correct answer would be FALSE. It is not true that range of motion is the distance an object can travel when separated from another object because range of motion or ROM is the distance--linear or angular--<span>that a movable object may normally travel while properly ATTACHED (not separated) to another. Hope this answer helps.</span>

7 0

3 years ago