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-Dominant- [34]
3 years ago
11

Part complete Sound with frequency 1240 Hz leaves a room through a doorway with a width of 1.11 m . At what minimum angle relati

ve to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.
Physics
1 answer:
Sedbober [7]3 years ago
6 0

Answer:

14.43° or 0.25184 rad

Explanation:

v = Speed of sound in air = 343 m/s

f = Frequency = 1240 Hz

d = Width in doorway = 1.11 m

Wavelength is given by

\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343}{1240}\\\Rightarrow \lambda=0.2766\ m

In the case of Fraunhofer diffraction we have the relation

dsin\theta=\lambda\\\Rightarrow \theta=sin^{-1}\frac{\lambda}{d}\\\Rightarrow \theta=sin^{-1}\frac{0.2766}{1.11}\\\Rightarrow \theta=14.43^{\circ}\ or\ 0.25184\ rad

The minimum angle relative to the center line perpendicular to the doorway will someone outside the room hear no sound is 14.43° or 0.25184 rad

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Which statement is true? Hydroelectric power is generated from moving water. Oil, a renewable resource, is processed to make jet
Mrac [35]

Answer: hydroelectric power is generated from moving water

Explanation: hydroelectric power is generated from moving water

6 0
4 years ago
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Calculate the amount of air in a room 6m long, 5m wide and 3mm high.​
sattari [20]

Answer:

0.09kg of air

Explanation:

The dimensions of the room are given

change the height to meters by dividing it by thousand.

For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).

Volume refers to the amount of space inside a box or a object.

The amount of air is equal to the volume.

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2 years ago
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A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s
sergey [27]

Answer:

Y = V / f      where Y equals wavelength

4 Y1 = V / f1       for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2   for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1      dividing equations

f2 = 2 f1  

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

8 0
3 years ago
You have two points in a soil. Point A is at 75 cm; point B is at 25 cm above the reference. The capillary potential energy at p
user100 [1]

Answer:

The potential energy at point A is 17.1675 J

Explanation:

The capillary potential is the work expended to bring up a unit mass of liquid to a point in a capillary region from a level liquid surface. It is the capillary potential that facilitates the movement of moisture within soil capillaries

In meteorology it is used to describe the level of saturated soil above the water table

Potential energy is the energy inherent in a body by virtue of its position, therefore the potentials of both point A and B are

Point A, elevation = 75 cm capillary potential = -100 cm

Point B, elevation = 25 cm capillary potential = -200 cm

The total potential energy at point A is

Elevation above reference - capillary potential =75-(-100) = 175 cm

which gives per unit mass

PE = m × g × h = 1 kg × 9.81 m/s ² × 1.75 m = 17.1675 kg·m²/s² = 17.1675 J

8 0
3 years ago
The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?
vitfil [10]

The expression of the electric flux is

\Phi = \frac{Q}{\epsilon_0}

Here,

Q = Total charge enclosed in the closed surface

\epsilon_0 = Permittivity due to free space

Rearranging to find the charge,

Q = \epsilon_0 \Phi

Replacing with our values we have finally

Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)

Q = 1.6284*10^{-8} C (\frac{10^9nC}{1C})

Q = 0.1684nC

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

7 0
3 years ago
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