Ask question

Ask question

All categories

2 years ago

5

Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d

ial is set to 5 msec/div. What is the peak-to-peak amplitude of the displayed signal? What is the frequency in khz?

1 answer:

denis-greek [22]2 years ago

8 0

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

$Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts$

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

Finding Frequency:

As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

$Time Period = 3 div * 5msec /div\\Time Perod = 15 msec$

Since,

$Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz$

You might be interested in

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

B)

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

2 years ago

A projectile is launched

Flauer [41]

The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

The given parameters;

initial horizontal velocity, vₓ = 16 m/s
initial vertical velocity, $v_y =0$
time interval 1 seconds

The components of the velocity can be horizontal or vertical velocity.

The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.

The vertical component of the velocity is calculated as;

$v_y = v_0_y -gt\\\\v_y = 0 - (1\times 9.8)\\\\v_y = -9.8 \ m/s$

The horizontal component of the velocity is constant since it is not affected by gravity.

The horizontal component of the velocity = 16 m/s

Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

Learn more here:brainly.com/question/20349275

4 0

2 years ago

At what wavelength would a star radiate the greatest amount of energy if the star has a surface temperature of 60,000 K?

kompoz [17]

Answer:

$\lambda=4.81\times 10^{-8}\ m$

Explanation:

We have,

The surface temperature of the star is 60,000 K

It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

$\lambda T=2.89\times 10^{-3}$

Here,

$\lambda$ = wavelength

$\lambda=\dfrac{2.89\times 10^{-3}}{60000}\\\\\lambda=4.81\times 10^{-8}\ m$

So, the wavelength of the star is $4.81\times 10^{-8}\ m$ .

7 0

2 years ago

The energy due to Earth pulling down on an object is called ___.

Slav-nsk [51]

Gravity is the energy due to Earth pulling down on an object.

6 0

3 years ago

a female vocalist with a soprano voice can sing as high as 1000 hz. given that the speed of sound is 345 m/s what is the wavelen

Neporo4naja [7]

0.345 m.

<h3>Explanation</h3>

The wavelength is the distance that the wave travels in each cycle. The wave travels 345 meters in each second. Let the wavelength of this wave be $\lambda$ . That's the distance the wave travels in one cycle.

The frequency of the sound wave is 1 000 Hz, meaning that there are 1 000 cycles in each second. The wave travels a distance of 1 000 wavelengths in one second. That would be a distance of $1,000\;\lambda$ .

From the speed of the wave, the wave travels 345 meters in one second. In other words,

$1,000\;\lambda = 345$ .

$\lambda = \dfrac{345}{1,000} = 0.345\;\text{m}$ .

To generalize:

$\lambda = \dfrac{v}{f}$ ,

where

$\lambda$ wavelength of the wave,
$v$ the speed of the wave, and
$f$ the frequency of the wave.

3 0

2 years ago