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agasfer [191]
3 years ago
14

Two sources of in-phase coherent sound are located at the points (0 m, 2 m) and (-2 m, 0 m). An observer at the origin hears con

structive interference because she is equidistant from both sources. However, if she moves in the +x direction, she hears destructive interference for the first time when she reaches the point (0.6 m, 0 m). What is the frequency of the sound that the sources are emitting? a) 476 Hz b) 391 Hz c) 335 Hz d) 295 Hz e) 244 Hz
Physics
1 answer:
Korolek [52]3 years ago
3 0

Answer:

c) 335 Hz

Explanation:

Given that the two sources:

S₁ ≡ (0 m, 2 m)

S₂ ≡ (-2 m, 0 m)

The lady hears the destructive interference for the first time when she reaches the point (0.6 m, 0 m).

So, the path difference = Δx = |S₂P - S₁P|

S_2P=\sqrt {(-2-0.6)^2+(0-0)^2}=2.6\ m

S_1P=\sqrt {(0-0.6)^2+(2-0)^2}=2.0881\ m

So,

Δx = |2.6 - 2.0881| = 0.5119 m

For, first destructive interference, Δx = λ / 2

So,

Wavelength = 1.0238 m

Also,

c = ν×λ

Where,

c is the speed of sound wave having value as 346 m/s

ν is the frequency

λ is the wavelength

So,

<u>ν = c / λ = 346 m/s / 1.0238 m ≈  335 Hz</u>

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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68
liubo4ka [24]

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time      ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at²     ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

7 0
3 years ago
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Which term, taken from the celestial sphere, gives its name to a.m. and p.m.?
DaniilM [7]

Options are. Zenith, Great circle, Equinox, or Meridan
3 0
3 years ago
A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig
aleksandr82 [10.1K]

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

\frac{a + b}{20} = \frac{a}{7}

a = \frac{7}{13}b

Differentiating the above eqn w.r.t t:

\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}

Now, we know that:

Rate = \frac{db}{dt} = 5\ ft/s

Thus

\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s

5 0
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garik1379 [7]

Answer:

true

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8 0
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