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jeyben [28]
3 years ago
6

Two speakers are spaced 15 m apart and are both producing an identical sound wave. You are standing at a spot as pictured. What

would be the frequency produced by the speakers to create perfectly constructive interference? Assume n = 1 and v = 343 m/s
213.04 Hz
256.70 Hz
186.68 Hz
233.14 Hz
Physics
2 answers:
Alexus [3.1K]3 years ago
7 0

Answer:

The correct answer is option 213.04 Hz

Explanation:

Hello!

Let's solve this!

In this link we will find the image of the problem.

https://smart-answers.com/physics/question14138735

Regarding that image, we will first calculate the distance from my position to S1 and then to S2. Then the difference between these results.

We will use pitagoras.

S1 = \sqrt{10^{2}+22^{2}  }

S1 = 24.17

S2 = \sqrt{5^{2}+22^{2}  }

S2 = 22.56

The difference will be:

24.17-22.56 = 1.61 m

Constructive interference:

Δr=n*λ

λ=1.61 m (for n = 1)

Then we will calculate the frequency:

f = v / λ

f = (343m / s) /1.61m

f = 213.04 Hz

So the correct answer is option 213.04 Hz

e-lub [12.9K]3 years ago
4 0

Answer:

213.04

Explanation:

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MA_775_DIABLO [31]

Answer:

<u>Resolving</u><u> </u><u>horizontally</u><u>.</u> :

\sum d_{x} =  - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\  { \underline{d _{x} =  -  5.702 \: m}} \\  \\  \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0  \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y}  =  - 11.476 \: m}}

therefore, for resultant:

d =  \sqrt{ {d _{y} }^{2} + d _{x}  {}^{2}  }

substitute:

d =  \sqrt{ {( - 5.702)}^{2} +  {( - 11.476)}^{2}  }  \\  \\ d =  \sqrt{164.211}  \\  \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}

6 0
2 years ago
Item 19 Question 1 A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet a
sergey [27]

Answer:

y = 104.4 ft

Explanation:

As we know that we board in the car of ferris wheel at the bottom position

So we will have

final height of the car at angular displacement given as

y = y_o + R + R sin(270 - 255)

y = y_o + R + R sin 15

here we know that

y_o = 10 ft

R = 75 ft

so we have

y = 10 + 75 + 75 sin15

y = 104.4 ft

4 0
2 years ago
A very powerful vacuum cleaner which has a hose of circular cross section can lift a brick of mass 12 kg when the hose is placed
valkas [14]

Answer:

A). 1.9 cm

Explanation:

m = Mass of brick = 12 kg

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of hose

A = Area = \pi r^2

F = Force = mg

Let us assume that the pressure required to lift the brick would be atmospheric pressure

P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{F}{\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{F}{\pi P}}\\\Rightarrow r=\sqrt{\dfrac{12\times 9.81}{\pi\times 101325}}\\\Rightarrow r=0.01923\ m=1.9\ cm

The radius of the hose should be 1.9 cm

6 0
2 years ago
two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco
Reika [66]

Answer:

t=750s

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt

x_f is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

x_0_1=6km\\x_0_2=0

We have x_f_1=x_f_2. Thus, solving for t:

x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s

8 0
2 years ago
A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this
baherus [9]
20 characters longer later and woah
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