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3 years ago

In what situation you would prefer to use a successive approximation ADC over flash ADC?

Engineering

1 answer:

Levart [38]3 years ago

8 0

Answer and Explanation:

Both both successive approximation normally known SAR type and flash type ADC are used for the conversion of analog signal to digital signal successive approximation type ADC are preferred over flash type when we need fix conversion time and better accuracy and if we have don/t matter with speed of the conversion because flash type has better speed of conversion than SAR type

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3 years ago

What is the effect of the workpiece specific cutting energy on the cutting forces, and why?

ella [17]

Explanation:

Specific cutting energy:

It the ratio of power required to cut the material to metal removal rate of material.If we take the force required to cut the material is F and velocity of cutting tool is V then cutting power will be the product of force and the cutting tool velocity.

Power P = F x V

Lets take the metal removal rate =MRR

Then the specific energy will be

$sp=\dfrac{F\times V}{MRR}$

If we consider that metal removal rate and cutting tool velocity is constant then when we increases the cutting force then specific energy will also increase.

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4 years ago

A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s

monitta

Answer:

a. $\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$ which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

$v_1 = \dfrac{4}{3} \times \pi \times r^3$

$\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3$

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

$\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}$

We have;

$\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$

$\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )$

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

$\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}$

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. $W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}$

$p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }$

p₂ = 100000/0.941 = 106.265 kPa

$W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J$

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

4 0

3 years ago

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. A

irina [24]

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.

So; we can say :

$D_{pursuit} =D_{police}$

By using the second equation of motion to find the distance S;

$S= ut + \dfrac{1}{2}at^2$

$D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)$

$D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)$

$D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)$

$D_{police} = ut _P + \dfrac{1}{2}at_p^2$

where ;

u = 0

$D_{police} = \dfrac{1}{2}at_p^2$

$D_{police} = \dfrac{1}{2}*(1.96)*(t+12)^2$

$D_{police} = 0.98*(t+12)^2$

$D_{police} = 0.98*(t^2 + 144 + 24t)$

$D_{police} = 0.98t^2 + 141.12 + 23.52t$

Recall that:

$D_{pursuit} =D_{police}$

$(187.8+15.65 \ t-1.525 t^2)= 0.98t^2 + 141.12 + 23.52t$

$(187.8 - 141.12) + (15.65 \ t - 23.52t) -( 1.525 t^2 - 0.98t^2) = 0$

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

t = -6.16 OR t = 3.02

Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs

From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;

Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s

Total time required for the police car to over take the automobile = 15.02 sec

8 0

3 years ago

When calculating gear ratios, which formula is different from the others?

BartSMP [9]

Answer:

The number of threads in a worm is the number of teeth in a worm. The speed transmission ratio of a worm and worm gear set is obtained by dividing the number of teeth of the worm gear by the number of threads of the worm.

6 0

3 years ago