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3 years ago

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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]3 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,

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

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Studentka2010 [4]

Answer:

$V(t)= 240V* H(t-5)$

Explanation:

The heaviside function is defined as:

$H(t) =1 \quad t\geq 0\\H(t) =0 \quad t$

so we see that the Heaviside function "switches on" when $t=0$ , and remains switched on when $t>0$

If we want our heaviside function to switch on when $t=5$ , we need the argument to the heaviside function to be 0 when $t=5$

Thus we define a function f:

$f(t) = H(t-5)$

The $-5$ term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

$V(t)= 240V* H(t-5)$

I have made a sketch for you, and added it as attachment.

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3 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

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Answer:

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Ballon volume of 3200ml of xenon gas is at a gauge pressure of 122kPa and a temperature of 27c. What is the volume when the ball

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Given:

The initial volume of the gas, V₁=3200 ml=3.2×10⁻³ m³

The initial pressure of the gas, P₁=122 kPa

The initial temperature of the gas, T₁=27 °C=300 K

The final temperature, T₂=65 °C=338 K

The final pressure, P₂=112 kPa

To find:

The final volume of xenon gas.

Explanation:

From the combined gas law,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Where V₂ is the volume after it is heated.

On rearranging the above equation,

$V_2=\frac{T_2P_1V_1}{T_1P_2}$

On substituting the known values,

$\begin{gathered} V_2=\frac{338\times112\times10^3\times3.2\times10^{-3}}{300\times112\times10^3} \\ =3.61\text{ m}^3 \end{gathered}$

Final answer:

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