Which is an example of qualitative data?

To practice Problem-Solving Strategy 16.1 Standing waves. An air-filled pipe is found to have successive harmonics at 800 HzHz ,

bazaltina [42]

Answer:

Length of the pipe = 53.125 cm

Explanation:

given data

harmonic frequency f1 = 800 Hz

harmonic frequency f2 = 1120 Hz

harmonic frequency f3 = 1440 Hz

solution

first we get here fundamental frequency that is express as

2F = f2 - f1 ...............1

put here value

2F = 1120 - 800

F = 160 Hz

and

Wavelength is express as

Wavelength = Speed ÷ Fundamental frequency ................2

here speed of waves in air = 340 m/s

so put here value

Wavelength =340 ÷ 160

Wavelength = 2.125 m

so

Length of the pipe will be

Length of the pipe = 0.25 × wavelength ......................3

put here value

Length of the pipe = 0.25 × 2.125

Length of the pipe = 0.53125 m

Length of the pipe = 53.125 cm

7 0

2 years ago

Parasaurolophus was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.5-m-long hollow tube in the cr

Oksi-84 [34.3K]

Answer:

f1 = 58.3Hz, f2 = 175Hz, f3 = 291.6Hz

Explanation:

lets assume speed of sound is 350 m/s.

frequencies of a standing wave modes of an open-close tube of length L

fm = m(v/4L)

where m is 1,3,5,7......

and fm = mf1

where f1 = fundamental frequency

so therefore: f1 = 350 x 4 / 1.5

f1 = 58.3Hz

f2 = 3 x 58.3

f2 = 175Hz

f3 = 5 x 58.3

f3 = 291.6Hz

5 0

3 years ago

A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed

GrogVix [38]

Answer:

a. $\tau = 2.1161 s$

b. $V(18.8 \ s) = 0.0256 \ V$

Explanation:

The equation for the voltage V of discharging capacitor in an RC circuit at time t is:

$V(t) = V_0 e^{(- \frac{t}{\tau}) }$

where $V_0$ is the initial voltage, and $\tau$ is the time constant.

For our problem, we know

$V_0 = 185 \ V$

and

$V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

So

$185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

$e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }$

$ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })$

$- \frac{10 \ s}{\tau} = ln (\frac{1.64 \ V}{ 185 \ V })$

$\tau = \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }$

This gives us

$\tau = 2.1161 s$

and this is the time constant.

At t = 18.8 s we got:

$V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }$

$V(18.8 \ s) = 0.0256 \ V$

4 0

2 years ago

Someone answer these questions please???

chubhunter [2.5K]

Yeah sure someone else in answering rn

8 0

2 years ago

Read 2 more answers

What is the approximate energy required to raise the temperature of 1.00 L of hydrogen by 90 °C? The pressure is held constant a

Oliga [24]

Answer:

Q = 116.8 J

Explanation:

Here given that the temperature of 1 L hydrogen is increased by 90 degree C at constant pressure condition.

So here we will have

$Q = n C_p \Delta T$

here we know that

n = number of moles

$n = \frac{1}{22.4}$

$n = 0.0446$

for ideal diatomic gas molar specific heat capacity at constant pressure is given as

$C_p = \frac{7}{2}R$

now we have

$Q = (0.0446)(\frac{7}{2}R)(90)$

$Q = 116.8 J$

3 0

2 years ago