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Misha Larkins [42]
3 years ago
13

Sometimes, when the wind blows across a long wire, a low-frequency "moaning" sound is produced. The sound arises because a stand

ing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density = 0.0120 kg / m ) sustains a tension of 341 N because the wire is stretched between two poles that are 16.86 m apart. The lowest frequency that an average, healthy human ear can detect is 20.0 Hz. What is the lowest harmonic number n that could be responsible for the "moaning" sound?
Answer is not n= 4.47
Physics
1 answer:
Crazy boy [7]3 years ago
6 0

Answer:

The smallest integer is n = 4

Explanation:

Using the equation V= Sqrt(F/Linear density)

V= Sqrt(341/0.0120)

V= Sqrt(28416.7)

V= 168.57m/s

Path distance =[ (n +1)/2]lambda

But V= f(Lambda)

n lambda/2 =L

n = f2L/V

n = (20 × 2 × 16.86) / 168.57

n = 4.0007

The smallest integer is n= 4

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Answer:cho  v₀ =0s  

α=Δv/Δt

Explanation:

\frac{0-495}{2,16-1,78}

=-1302,631579

chuyển động chậm dầnđều

3 0
3 years ago
Light is shining perpendicularly on the surface of the earth with an intensity of 680 W/m^2. Assuming that all the photons in th
andrew11 [14]

Answer:

3.066×10^21  photons/(s.m^2)

Explanation:

The power per area is:

Power/A = (# of photons /t /A)×(energy / photon)

E/photons = h×c/(λ)

photons /t /A = (Power/A)×λ /(h×c)  

photons /t /A = (P/A)×λ/(hc)

photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)

                      = 3.066×10^21

Therefore, the number of photons per second per square meter 3.066×10^21  photons/(s.m^2).

4 0
3 years ago
Kuiper Belt objects are composed of which substances? Select all that apply.
levacccp [35]
IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron

8 0
4 years ago
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b
Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

Q=t^3-2t^2+4t+4

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4

At t = 1 s,

Current,

I=3(1)^2-4(1)+4\\\\I=3\ A

So, the current at t = 1 s is 3 A.

For lowest current,

\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s

Hence, this is the required solution.

7 0
4 years ago
A cylinder of compressed gas has a pressure of 488.2 kPa. The next day the cylinder of gas has a
Luda [366]

Answer:

20 °C

Explanation:

Ideal gas law:

PV = nRT

Rearranging:

P / T = nR / V

Since n, R, and V are constant:

P₁ / T₁ = P₂ / T₂

488.2 kPa / T = 468 kPa / 281.15 K

T = 293.29 K

T = 20.1 °C

Rounded, the temperature was 20 °C.

6 0
4 years ago
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