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kap26 [50]
3 years ago
15

For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. What is the

very first frequency of audible sound for these speakers to produce destructive interference at the listener?The speed of sound is 344 m/sec. (1 point)
166 Hz
331 Hz
662 Hz
1323 Hz
Physics
1 answer:
alekssr [168]3 years ago
7 0

You'll get destructive interference if both waves are the same frequency but the peaks of one wave overlap the troughs of the other wave.

That can only happen if one wave has to travel (1/2 wavelength) farther than the other one to reach your ears.  So we want to find the lowest frequency for which 52 cm is 1/2 of a wavelength ... the wavelength is 104 cm.

Frequency = (speed) / (wavelength)

Frequency = (344 m/s) / (104 cm)

Frequency = (344 m/s) / (1.04 meter)

Frequency = (344 / 1.04) per second

Frequency = 330.8 Hz .

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Someone please help<br> Just need answers
marishachu [46]

Answer:

As follows,

Explanation:

KE=1/2mv^2

In 1st question,

KE=1/2mv^2=1/2*0.05*12=0.3 J [50g=0.05 kg]

In 2nd question,

KE=1/2mv^2

6.8=1/2*0.046*v^2

v=sqrt(6.8/0.023)

v=17.19

In 3rd question,

KE=1/2mv^2

63/392=m

m=0.16kg=160g

For 4th,

a.

1st case,

KE=1/2mv^2=1/2*28*2.4^2=80.64

2nd case,

KE=1/2mv^2=1/2*28*3.7^2=191.66

Change in KE=191.66-80.64=11.02

b.Speed/velocity gained

6 0
3 years ago
If a barometer reads 772 mm hg, what is the atmospheric pressure expressed in pounds per square inch?
VLD [36.1K]
15.23.....................
8 0
3 years ago
Read 2 more answers
Will give correct answer brainliest
Tasya [4]

Answer:

the answer is a

Explanation:

3 0
3 years ago
Do magnets have to touch each other in order to experience a magnetic<br> force? Explain
tiny-mole [99]
Magnets don’t need to touch each other in order to be magnetic because all objects have an attraction however, the closer the magnets get, the more magnetic it is.
4 0
3 years ago
The 1000-lb elevator is hoisted by the pulley system and motor M. If the motor exerts a constant force of 500 lb on the cable, d
Nezavi [6.7K]

Answer:

the power that must be supplied to the motor is  101.45 hp

Explanation:

Given that,

The weight of elevator = 1000lb

The motor exerts a constant force of 500 lb on the cable

The load has been hoisted s = 15 ft starting from rest

The motor has an efficiency of e = 0.65

According to the equation of motion:

F = ma

3(500) - 1000 = \frac{1000}{32.2} * aa = 16.1 ft/s^2

v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (15-0)\\\\v = 21.98m/s

To calculate the output power:

P(out) = F. v

P(out) = 3 (500) * 21.98

P(out) = 32970 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = P(out) / P(in)

0.65 = 32970 / P(in )

P(in) = 50,723 lb.ft/s

P(in) =  50,723 / 500 hp

    = 101.45 hp

the power that must be supplied to the motor is  101.45 hp

5 0
4 years ago
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