Ask question

Ask question

All categories

3 years ago

6

A standing wave with a fundamental frequency 1.25x10 2 Hz is set up on a string of length 0.750 meters. What is the frequency of

the third harmonic?

1 answer:

Nezavi [6.7K]3 years ago

8 0

Answer:

<em>500Hertz</em>

Explanation:

Given the fundamental frequency of the string given as 1.25x10²Hz

Harmonics are integral multiples of the fundamental frequency.

In strings, the third harmonic F3 is expressed in terms of the fundamental frequency expressed as;

F3 = 4F0

F0 is the fundamental frequency;

Given

F0 = 1.25x10²Hz

F3 = 4(1.25x10²)

F3 = 4(125)

F3 = 500Hz

<em>Hence the frequency of the third harmonic is 500Hertz</em>

<em></em>

You might be interested in

A car travels with a velocity 16m/s and accelerates 5m/s^2. Calculate the final velocity when it has travelled 36.9m

marissa [1.9K]

v²=u²+2as

v²=16²+2×5×36.9

v²=625

√v²=√625

v=25

--------

--------

4 0

3 years ago

What is the velocity of an object that has a momentum of 4,000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.

julia-pushkina [17]

<h2>Answer: 34.78 m/s</h2>

Explanation:

The momentum $p$ is given by the following equation:

$p=m.V$ (1)

Where:

$m$ is the mass of the object

$V$ is the velocity of the object

Finding the velocity from (1):

$V=\frac{p}{m}$ (2)

$V=\frac{4000kg.m/s}{115kg}$

<u>Finally:</u>

$V=34.78m/s$ >>>This is the velocity of the object

4 0

3 years ago

Read 2 more answers

How can the difference in the brightness of spectral lines be explained?

Ksenya-84 [330]

<span>more lines = a lot of electrons returning back to ground state from same level</span>

5 0

3 years ago

4) Write down the transformation of energy in torch light.<br>

Elenna [48]

Answer:

<u>The transformation of energy in a torch light is as follows:</u>

1) When the torch is turned ON, the chemical energy in the batteries is converted into electrical energy.

2) The electrical energy is converted into heat and light energy. (We feel the torch to be hot after some time and we can see the light energy)

Hope this helped!

<h2>~AnonymousHelper1807</h2>

7 0

3 years ago

A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0

3 years ago