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irina [24]
4 years ago
5

The Reynolds number for flow over a fixed horizontal flat plate is constant throughout its length. a) True b) False

Engineering
1 answer:
ANTONII [103]4 years ago
5 0

Answer: False

Explanation: Reynolds number is the number for determining he flow pattern of the behavior of fluids.It judges the flow of the fluid ans states whether it is of turbulent or laminar natured. This number has the property of kinematics as well as statics. It can also be defined as the ratio between the inertial force to viscous force.It's length changes along with change length of the horizontal plate .Thus, the statement given is false.

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MITM can present in two forms: eavesdropping and manipulation. Discuss the process involved when an attacker is eavesdropping an
Nikitich [7]

Answer / Explanation:

Eavesdropping attack is also sometimes refereed to as sniffing attack. It is simply the process by which an attacker or hacker tries to penetrate very suddenly into an unaware individuals network or server with the intention to steal information transmitted over the network or server through that computer.  

To prevent such attack, there are several mean which include installing network monitoring software to check who else is connected to the network but the most common method of preventing such attack is to encrypt the Hypertext Transfer Protocol (http) and the way to do this is by securing it with a sort of security key.

On installing the security key, the network becomes encrypted and secured such that whatever network transmitted over the network becomes encrypted and unable to read. The protocol then converts to (https).

5 0
3 years ago
The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and
garik1379 [7]

Solution :

The spring is expanded by 2 times of the block when it moves down an inclined by x times.

Here, $x_1$ = 39 mm

        x_2 = 225 mm

a). From the work energy principal,

   Work forces = kinetic energy

$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$

$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$

$9.75= 7.5(V^2_2-0.08^2)$

$1.3= V^2_2-0.08^2$

$V_2=1.14\ m/s$

b). calculating the distance travelled by the block before it comes to rest.

Substitute the value of V_2 in (1),

$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$

$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$

$98.43x - 100(4x^2+0.156x)+0.048=0$

$98.43x - 400x^2-15.6x+0.048=0$

$82.83x - 400x^2+0.048=0$

$  400x^2- 82.83x-0.048=0$

x = 0.20 m

4 0
3 years ago
Why is it important for engineers to consider both short and long term implications of their work?
polet [3.4K]
The right answer should be A).
7 0
3 years ago
THEME: What is the impact of technology on architecture?
abruzzese [7]

Answer:

With increased technological knowledge and consequent decreased factors of ignorance, the structures have less inert masses and therefore less need for such decoration. This is the reason why the modern buildings are plainer and depend upon precision of outline and perfection of finish for their architectural effect.

8 0
4 years ago
A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t
iren2701 [21]

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, P_{1}= 1 bar

v_{2} = 4 bar

pV¹°¹ = constant

From superheated propane table, at  P_{1}= 1 bar andT_{1}  = 27⁰C

v_{1} = 0.557 m³/kg

v_{2} = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1  \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557  \\\\V2 = 0.158 m^3/kg

At  = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\  \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\   \\W = -19.75KJ

(b) Heat transfer = ?

Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ

8 0
4 years ago
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