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3 years ago

15

Document the XSS stored exploit script: Use the View Source feature of the web page and create a screenshot of the few lines cod

e that could have prevented an XSS stored exploit had the programmer used them. (Show the relevant lines of code, not the entire script)

1 answer:

Natali [406]3 years ago

5 0

Answer:

Hold on let me ask my brother

Explanation:

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One of our wifi network standards is IEEE 802.11ac. It can run at 6.77 Gbit/s data rate. Calculate the symbol rate for 801.11ac

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Answer: Symbol rate, Fs = 0.846

Explanation:

The attachment below shows the calculations

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You talk with the owner and he likes the idea of using two large glulam beams as shown to carry the joist loads. Design the glul

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What soil horizon contains mostly clay silt and sand?

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2 years ago

An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions. If the

xenn [34]

Answer:C 0.12 V

Explanation:

Given

Concentration of $Fe^{2+} M_1=0.40 M$

Concentration of $Ni^{2+} M_2=0.002 M$

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$\Delta V=V_2-V_1-\frac{0.0592}{2}\log (\frac{M_1}{M_2})$

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3 years ago

. Determine the state of stress at point A on the cross-section at section a-a of the cantilever beam. Show the results in a dif

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2 years ago

Consider the circuit below where R1 = R4 = 5 Ohms, R2 = R3 = 10 Ohms, Vs1 = 9V, and Vs2 = 6V. Use superposition to solve for the

VladimirAG [237]

Answer:

The value of v2 in each case is:

A) V2=3v for only Vs1

B) V2=2v for only Vs2

C) V2=5v for both Vs1 and Vs2

Explanation:

In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.

Also, what the problem asks is the value V2 in each case, where:

$V_2=I_2R_2=V_{ab}$

If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.

In the first case we can use an equivalent resistance between R2 and R3:

$V_{ab}'=I_1'R_{2||3}=I_1'\cdot(\frac{1}{R_2}+\frac{1}{R_3})^{-1}$

And

$V_{S1}-I_1'R_1-I_1'R_4-I_1'R_{2||3}=0 \rightarrow I_1'=0.6A$

$V_{ab}'=I_1'R_{2||3}=3V=V_{2}'$

In the second case we can use an equivalent resistance between R2 and (R1+R4):

$V_{ab}''=I_3'R_{2||1-4}=I_3'\cdot(\frac{1}{R_2}+\frac{1}{R_1+R_4})^{-1}$

And

$V_{S2}-I_3'R_3-I_3'R_{2||1-4}=0 \rightarrow I_3'=0.4A$

$V_{ab}''=I_3'R_{2||1-4}=2V$

If we consider both batteries:

$V_2=I_2R_2=V_{ab}=V_{ab}'+V_{ab}''=5V$

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3 years ago