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Kitty [74]
3 years ago
7

Advantages, disadvantages and applications of dsb-sc​

Engineering
1 answer:
allochka39001 [22]3 years ago
3 0

Answer:

<h3>advantages: </h3>

<em>lower power consumption, modulation system is simple</em>

<h3>disadvantages<em>:</em></h3>

<em>complex detection</em>

<h3><em>applications:</em></h3>

analog TV systems: to transmit color information

<h3><em /></h3>

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Explanation:

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According to the decreasing order of toughness. list the following materials (note: the steels are assumed to have no cold work
fiasKO [112]

Answer:

1090 Steel >1040 Steel > Pure aluminium >Diamond.

Explanation:

Toughness:

  Toughness can be define as the are of load -deflection diagram up to fracture point.

Modulus of toughness can be defines as the area of stress-strain diagram up to fracture point.Modulus of toughness is the property of material.

So the decreasing order of toughness can be given as follows

1090 Steel >1040 Steel > Pure aluminium >Diamond.

8 0
3 years ago
Line.
Veronika [31]

Air supplied to a pneumatic system is supplied through the C. Actuator

Explanation

Pneumatic systems are like hydraulic systems, it is just that these systems uses compressed air rather than hydraulic fluid.  Pneumatic systems are used widely across the industries. these pneumatic systems needs a constant supply of compressed air to operate. This is provided by an air compressor. The compressor sucks in air at a very high rate from the environment and stores it in a pressurized tank. the Air is supplied thereafter with the help of a actuator valve that is a more sophisticated form of a valve.

From the above statement it is clear that Air supplied to a pneumatic system is supplied through the  Actuator

7 0
3 years ago
is released from under a vertical gate into a 2-mwide lined rectangular channel. The gate opening is 50 cm, and the flow rate in
SVEN [57.7K]

Answer:

hello your question is incomplete attached below is the complete question

answer: There is a hydraulic jump

Explanation:

First we have to calculate the depth of flow downstream of the gate

y1 = C_{c} y_{g} ----------- ( 1 )

Cc ( concentration coefficient ) = 0.61  ( assumed )

Yg ( depth of gate opening ) = 0.5

hence equation 1 becomes

y1 = 0.61 * 0.5 = 0.305 m

calculate the flow per unit width q

q = Q / b ----------- ( 2 )

Q = 10 m^3 /s

 b = 2 m

hence equation 2 becomes

q = 10 / 2 = 5 m^2/s

next calculate the depth before hydraulic jump y2 by using the hydraulic equation

answer : where  y1 < y2 hence a hydraulic jump occurs in the lined channel

attached below is the remaining part of the solution

4 0
3 years ago
99 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
qwelly [4]

Answer:

1. Can you tell me something about yourself?

2. What are you weaknesses?

3. If you would describe yourself in one word?

Explanation: Those questions above 1, 2, and 3 are not harmful to ask your client. Bit the last two 4 and 5 are very harmful, because you don't need to be all up in they business and you don't want to put a lot of pressure on your client.

Hope this helps☝️☝☝

7 0
3 years ago
Read 2 more answers
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0
3 years ago
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