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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:

You might be interested in

A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted

alexira [117]

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - $D^{2}$ /Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf = free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

= 240v/h

q = Dj ( U - $U^{2}$ /Uf)

2100 = 240 ( U - $U^{2}$ /55)

Solve for U

U = 44m/h

5 0

3 years ago

Having an adequate ___________ provides a driver with the time and space necessary to react to potential and immediate hazards i

ra1l [238]

Answer:

Space Cushion

Explanation:

Having an adequate Space Cushion provides a driver with the time and space necessary to react to potential and immediate hazards in the driving scene.

8 0

2 years ago

Steel bar A of 5 mm diameter is subject to a stress of 500 MPa. Aluminum bar B of 10 mm diameter is subject to a stress of 150 M

weqwewe [10]

Answer:

aluminum bar carrying a higher load than steel bar

Explanation:

Given data;

steel abr

diameter = 5 mm

stress = 500 MPa

aluminium bar

diameter = 10 mm

stress = 150 MPa

we know

stress = laod/area

for steel bar

$500 = \frac{P}{\frac{\pi}{4} 5^2}$

solving for P

P = 9817.47 N

for Aluminium bar

$150 = \frac{P}{\frac{\pi}{4} 10^2}$

solving for P

P = 11790 N

aluminum bar carrying a higher load than steel bar

6 0

3 years ago

Supón que tienes que calcular el centro de gravedad de una pieza

Anton [14]

Answer:huh?

Explanation:

4 0

4 years ago

two pints of ethyl ether evaporate over a period of 1,5 hours. what is the air flow necessary to remain at 10% or less of ethyl

lesya692 [45]

The air flow necessary to remain at the lower explosive level is 4515. 04cfm

<h3>How to solve for the rate of air flow</h3>

First we have to find the rate of emission. This is solved as

2pints/1.5 x 1min

= 2/1.5x60

We have the following details

SG = 0.71

LEL = 1.9%

B = 10% = 0.1 a constant

The molecular weight is given as 74.12

Then we would have Q as

403*100*0.2222 / 74.12 * 0.71 * 0.1

= Q = 4515. 04

Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm

Read more on the rate of air flow on brainly.com/question/13289839

#SPJ1

7 0

2 years ago