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3 years ago

GIVING BRAINLIST!!!!

Physics

2 answers:

Lena [83]3 years ago

6 0

ANSWER The charged object will lose its electrons and the uncharged object will absorb it

34kurt3 years ago

3 0

A negative object so it takes in the charged object and nothing will happen until something else would touch it I think

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what aspect of receiver analog-to-digital conversion determines the maximum receive bandwidth of a direct digital conversion sdr

Zepler [3.9K]

Answer:

The sample rate determines the maximum receive bandwidth of a direct digital conversion SDR. An analog signal must be sampled at twice the rate of the highest frequency component of the signal by an analog-to-digital converter so that the signal can be accurately reproduced.

Explanation:

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2 years ago

The circuit in Fig. P4.23 utilizes three identical diodes having IS = 10−14 A. Find the value of the current I required to obtai

Sliva [168]

Answer:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have: $I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

Explanation:

For this case we have the figure attached illustrating the problem

We know that the equation for the current in a diode id given by:

$I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}$

For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode $v_1 + v_2 + v_3= 2$ and each voltage is the same v for each diode, so then:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have:

$I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

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3 years ago

A plane traveled west for 4.0 hours and covered a distance of 4400 kilometers whats the velocity

Elina [12.6K]

The velocity is 1100 kilometers per hour

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B it’s Intail velocity

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RSB [31]

Answer:

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Explanation:

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3 years ago