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3 years ago

14

Two pipes of equal length are each open at one end. Each has a fundamental frequency of 470 Hz at 310 K. In one pipe the air tem

perature is increased to 315 K. If the two pipes are sounded together, what beat frequency results?

1 answer:

Tresset [83]3 years ago

7 0

Answer:

Given:

Fundamental frequency: 470Hz

T1:310k,T2:315k

Calculating velocity

Recall v=(331m/s)✓[T1/273k)

V=331✓(310/273)

V1=331*(1.0656)=352.72m/s

V2=331✓(315/273)=355.5m/s

Fundamental frequency=4L

F2=F1(V2/V1)

F2=470(355.5/352.72)=474.4Hz

Beat=[F2-F1]=474.4-470=4.4Hz

Explanation:

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3 years ago

43. A rocket sled accelerates at a rate of 49.0m/s2. Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component

lord [1]

(a) 3675 N

Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:

$F_x = m a_x$

where

$F_x$ is the horizontal component of the force

m is the mass of the passenger

$a_x$ is the horizontal component of the acceleration

Here we have

m = 75.0 kg

$a_x = 49.0 m/s^2$

Substituting,

$F_x=(75.0)(49.0)=3675 N$

(b) 3748 N, 11.3 degrees above horizontal

In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:

$R=mg=(75.0)(9.8)=735 N$

where we used

$g=9.8 m/s^2$ as acceleration of gravity.

So, this is the vertical component of the force exerted by the seat on the passenger:

$F_y = 735 N$

and therefore the magnitude of the net force is

$F=\sqrt{F_x^2+F_y^2}=\sqrt{3675^2+735^2}=3748 N$

And the direction is given by

$\theta = tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{735}{3675})=11.3^{\circ}$

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3 years ago

An apple is thrown across the cafeteria with a force of 10 N and at an acceleration of 6 m/s2. What is the mass of the apple? ..

Komok [63]

Answer:

Explanation:

F = m * a

The apple's acceleration is not influenced by the acceleration due to gravity for this question. In real life it most certainly is influenced by gravity.

F = m * a

F = 10 Newtons.

a = 6 m/s^2

m = 10/6 = 1.66 kg. Mighty large apple

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4 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago

Two parallel slits are illuminated by light composed of two wavelengths. Wavelength A is 564 nm and the other is wavelength B an

anygoal [31]

Answer:

423nm

Explanation:

To find the unknown wavelength you take into account the distance y to the maximum central fringe, for light fringes and dark fringes.

- for light fringes:

$dsin\theta=m\lambda\\\\sin\theta\approx\theta=\frac{y}{D}\\\\y=\frac{m\lambda_1D}{d}$

- for dark fringes:

$y=\frac{m\lambda_2/2 D}{d}$

The third-order bright fringe (m= 3) of wavelength A coincides with the fourth dark fringe (m=4) of the wavelength B. Hence you have that:

$\frac{(3)\lambda_1D}{d}=\frac{(4)\lambda_2D}{d}\\\\\lambda_2=\frac{3}{4}\lambda_1=\frac{3}{4}(564nm)=423nm$

hence, the wavelength B is 423nm

7 0

4 years ago

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