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3 years ago

What is required when setting up a smart phone as a WIFI hotspot?

1 answer:

5 0

How to create a personal hot spot on an iPhone?

Go to Settings | Cellular | Personal Hotspot.

Tap the slider next to Allow Others to Join. ...

Your Wi-Fi Password will be shown right underneath the Allow Others to Join option. ...

Now, on another device, such as a laptop, go to the Wi-Fi section and search for nearby networks.

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Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following

torisob [31]

Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

a. The distance from the leading edge at which the transition will occur.

b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer

c. Which fluid has a higher heat transfer

Calculation:

The transition from the lamina to turbulent begins when the critical Reynolds

number reaches $5\times 10^5$

$(a). \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16 \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\$

$(b).\; \text{For the lamina boundary layer momentum boundary layer thickness is given by}:\\\frac{\delta}{x} =\frac{5}{\sqrt{R_e}}\;\;\;\;\quad\text{for}\; R_e$ $(c). \frac{\delta}{\delta_t}={P_r}^{\frac{r}{3}}\\\text{For air} \;P_r \;\text{equivalent 1 hence both momentum and heat dissipate with the same rate for oil}\; \\P_r >>1 \text{heat diffuse very slowly}\\\text{So heat transfer rate will be high for air.}\\\text{Convective heat transfer coefficient will be high for engine oil.}$

7 0

3 years ago

Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.

-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

r = 0.5*θ

θ = 0.5*t^2 ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

dr/dt = 0.5*dθ/dt

d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

dθ/dt = t

d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

tan Ψ = r / (dr/dθ) = 1 / 0.5

Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

a_r = d^2r / dt^2 - r * dθ/dt

a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

a_θ = r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction: N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction: F - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

N_c = 3.03 N

F - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

F = 1.81 N

4 0

3 years ago

Describe the placement of the views in a multi view drawing

Marianna [84]

Answer:

like a mountain place thanks #careonlearning

8 0

3 years ago

Can i have answer of this question please?

cestrela7 [59]

uh its a tough one mate

3 0

3 years ago

Compare and contrast ""centralized"" and ""decentralized"" routing algorithms. (What are the advantages and disadvantages of eac

pantera1 [17]

Answer:

Comparison between centralized and decentralized routing algorithms:

The major similarity between both centralized and decentralized routing algorithms is that they are both communication serving systems. They both utilize node system(e.g, computer) and are both a communication liking system( e.g, Cable).

Contrasting between centralized and decentralized routing algorithms:

There are few differences between these two type of communication link or path way system but to name a couple of them,

For centralized, there is a single client server distribution node which simply means that one or more client server system are connected to a central processing server.

This also means that if the central processing server or pathway fails, it leads to the failure of the entire system. That is, there is no sending, responding or general processing of any form of requests. Example of a system that uses the centralized routing algorithm is the google search engine.

WHILE:

for the decentralized routing algorithms, there are multiple client server distribution pathway and each server makes its own decision. Here,there is no single entity that receives and responds to the request therefore,failure of any form of central path way processing node does not lead to the failure of the whole system unlike for the centralized system.

Advantages of centralized routing algorithm:

It can be easily protected or secured due to the nature of the system. If the central node is been secured, it generally translate to the different client node being secured.

It is easy to disconnect a connected client node from the central pathway node or central server as the case may be.

Disadvantages of centralized routing algorithm:

The client nodes are totally dependent on the central node or server so if there is a failure in the central server, the client node is then totally shut down.

Advantages of decentralized routing algorithm:

There is a random distribution of data on all the processing node or server which automatically creates a form of balance within the system. This leads to minimal or no down time processing client request.

Disadvantages of decentralized routing algorithm:

Due to the nature or fact that there are multiple processing system for different client node, it is difficult to detect which client node or request processing server is faulty. This can lead to delay in the fixing of fault in the system should it arise.

Why do we prefer a ""decentralized"" algorithm for routing messages through the internet?

The major reason why decentralized algorithm routing is preferred is because of the level of security attached to it. Each processing servers are secured independently and there is privilege of utilizing independent networking system.

5 0

3 years ago