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larisa [96]
4 years ago
10

A guitar string with a linear mass density of 2.0 g/m is stretched between two vertical rods that are 65 cm apart. The string ho

lds a standing wave, fixed at both ends, with three antinodes when it oscillates at a frequency of 430 HzWhat is the frequency of the fifth harmonic of this string?
Physics
1 answer:
Rudiy274 years ago
7 0

Answer:

717 Hz

Explanation:

<u>solution:</u>

The wave with three antinodes has m = 3. Thus, f_3 = 3f_1 and so the fundamental frequency of the string is  

f_1 =f_3/3

    =430 Hz/3

    =143 Hz

Thus, the frequency of the fifth harmonic is

f_5 = 5*f_1

      = 5*143 Hz

     = 717 Hz

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A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit
Phoenix [80]

As per Bernuolli's Theorem total energy per unit mass is given as

\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E

now from above equation

P = 4390 N/m^2

\rho = 0.999 * 10^3

v = 7.22 * 10^{-4} m/s

H = 23.6 m

now by above equation

\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E

E = 235.7 J/kg

Part B)

Now energy per unit weight

U = \frac{E}{g}

U = \frac{235.7}{9.8}

U = 24 m

7 0
4 years ago
1. What types of natural phenomena could serve as time standards?<br>​
Alex Ar [27]
In practice, something that follows a very predictable pattern can be used as a time standard. This include things like radioactive decay, planetary orbit, and the speed of light, among others.
8 0
3 years ago
Read 2 more answers
1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t
densk [106]

Answer:

F = 36 N

Explanation:

Given that,

Charge, q₁ = +8 μC

Charge, q₂ = -5 μC

The distance between the charges, r = 10 cm = 0.1 m

We need to find the magnitude of the electrostatic force. The formula for the electrostatic force is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\F=36\ N

So, the magnitude of the electrostatic force is 36 N.

3 0
3 years ago
The change in internal energy during one complete cycle of a heat engine A. equals the net heat flow into the engine. B. equals
Stels [109]

Answer:

B. equals zero

Explanation:

Given data

one complete cycle = heat flow

solution

we have given that when heat engine complete 1 cycle change in energy = net heat flow

that is always equal to zero

from first law of thermodynamics that

ΔU = Q + W

we know ΔU is the change internal energy in system and Q is net heat transfer in system and W is  net work done in system

therefore change of internal energy during one cycle

ΔU = Ufinal -  Uinitial

ΔU  = Uinitial  -  Uinitial  = 0

7 0
3 years ago
How much heat is required to raise 100 grams of water (c= 4.18) by 5 degrees Celsius?
Andrei [34K]

Answer:

Heat capacity, Q = 2090 Joules.

Explanation:

Given the following data;

Mass = 100 grams

Specific heat capacity = 4.18 J/g°C.

Temperature = 5°C

To find the quantity of heat required;

Heat capacity is given by the formula;

Q = mct

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

t represents the temperature of an object.

Substituting into the formula, we have;

Q = 100*4.18*5

Heat capacity, Q = 2090 Joules.

7 0
3 years ago
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