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zepelin [54]
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?
Physics
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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Lostsunrise [7]

Answer:

Corpuscular theory of light

Explanation:

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3 years ago
a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




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3 years ago
Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30
barxatty [35]
For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

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a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
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2(36,000 mph²)(x) = 60² - 30²
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3241004551 [841]
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