All categories

3 years ago

Which statement will be true if you increase the frequency of a periodic wave

Physics

2 answers:

Natasha_Volkova [10]3 years ago

8 0

Answer:

The number of waves per second will increase

Explanation:

The frequency of a wave is how many times the wave is repeated in a given time.

The larger the frequency, the more times per unit of time (in this case seconds) the wave will be repeated.

In addition, the frequency is inversely proportional to the wavelength, that is, at a higher frequency we get a lower wavelength. What this tells us is that the number of waves per second will increase.

In conclusion, when raising the frequency of the wave, the wave is repeated more times per second, which translates to a greater number of waves per second.

WITCHER [35]3 years ago

3 0

The number of waves per second will increase

You might be interested in

Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by circulating wa

iren2701 [21]

Answer:

A. 6.36 lbm/s

b. $T_2=341\textdegree F$

Explanation:

a. Given the following information;

#Compressor inlet:

Air pressure, $P_1=14.7psia,T_1=60\textdegree F$

#Compressor outlet:

$Air \ Pressure, P_2=150psia\\\\Air \ volume \ flow \ rate, \dot V_1=5000ft^3/min$

#Cooling rate,

$q_{out}=10Btu/lbm, \dot W=700hp$

# From table A-1E

Gas constant of air $R=0.3704\ psia.ft^3/lbm.R$

Specific enthalpy at $P_1=520R-h_1=124.27Btu/lbm$

Using the mass balance:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{0.3704\times520}{14.7}\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=\frac{5000/60}{13.1025}\\\\\dot m=6.36\ lbm/s$

Hence, the mass flow rate of the air is 6.36lbm/s

b The specific enthalpy at the exit is defined as the energy balance on the system:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\\dot W_{in}+ \dot mh_1=\dot Q_{out}+\dot m_2\\\\h_2=\frac{\dot W_{in}-\dot Q_{out}}{\dot m}+h_1\\\\h_2=\frac{700\times 0.7068-10\times 6.36}{6.36}+124.27\\\\h_2=192.06\ Btu/lbm\\\\#at \ h_2=182.06Btu/lbm, T_2=801R=341\textdegree F$

Hence, the temperature at the compressor exit $T_2=341\textdegree F$

5 0

4 years ago

(help please no links )

trapecia [35]

Answer:

Hello, how's your day going?

if humanity came together and made a base on the moon, it would be revolutionary. The point of a base on the moon would have multiple purposes. for example, some think that the moon contains valuable metals such as iron and titanium. a base would serve as a place for workers harvesting metals to rest. Obviously or not most of the iron harvesting would be done automatically by robots and such.

If such a base were constructed on the moon, it would be the begining of people living on other worlds and would be a great start for a base on Mars.

Hope it helped

Spiky Bob

5 0

3 years ago

A capacitor is being charged from a battery and through a resistor of 10 kΩ. It is observed that the voltage on the capacitor ri

swat32

Answer:

$C = 2.48 \times 10^{-4} Farad$

Explanation:

As per the equation of voltage on capacitor we know that

$V = V_{max}(1 - e^{-\frac{t}{\tau}})$

now we know that voltage reached to its 80% of maximum value in 4 second time

so we will have

$0.80 V_{max} = V_{max}(1 - e^{-\frac{4}{\tau}})$

$0.20 = e^{-\frac{4}{\tau}}$

$-\frac{4}{\tau} = ln(0.20)$

$-\frac{4}{\tau} = -1.61$

$\tau = 2.48$

as we know that

$\tau = RC$

$(10 k ohm)(C) = 2.48$

$C = 2.48 \times 10^{-4} Farad$

4 0

3 years ago

Caroline, a piano tuner, suspects that a piano's B4 key is out of tune. Normally, she would play the key along with her B4 tunin

Sergeeva-Olga [200]

Answer:

455.9 Hz, Flat

Explanation:

The beat frequency is basically the difference between the frequency of the B4 note and the frequency of the A4 tuning fork. This means the B4 note is 15.9 Hz off of 440Hz, and 67.4 Hz off of 525.3 Hz. As B4 is between the two notes, it would make sense to find its frequency by adding 15.9 to 440 and subtracting 67.4 from 523.3, both if which give us a frequency 455.9 Hz for the B4 key. This is because the note doesn't change for the different turning forks so both differences should result in the same frequency. Because the note should be 493.9 frequency but instead has a frequency of 455.9 Hz, it is flat because the frequency is lower than it is supposed to be.

Hope this helped!

7 0

4 years ago

How might an ecologist Measure population density of pine trees in an actual forest?

ruslelena [56]

<span>Ans : The best way to estimate the population density of pine trees in an actual forest is to count the number of trees in a small area and than multiply to find the number in a large area. To get the most accurate estimate, your sample area should be typical of the larger area. Suppose you count 10 pine tree in 100 sq meters of the forest. If the entire forest were 100 times the size, you would multiply your count by 100 to estimate the total population density, or 1000 (10X100) pine trees.</span>

4 0

4 years ago