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prisoha [69]
3 years ago
14

The value read at an analog input pin using analogRead() is returned as a binary number between 0 and the maximum value that can

be stored in [X] bits. 1. This binary number is directly linearly proportional to the input voltage at the analog pin, with the smallest and largest numbers returned corresponding to the minimum and maximum ADC input values, respectively. At a Vcc of 3.3 V, analogRead(A0) returns a value of 1023. Approximately what voltage is present at the pin A0 on the MSP430F5529?
Engineering
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer: The approximate voltage would be the result from computing analogRead(A0)*3.3 V / 1023

Explanation:

Depending on the binary read of the function analogRead(A0) we would get a binary value between 0 to 1023, being 0 associated to 0V and 1023 to 3.3V, then we can use a three rule to get the X voltage corresponding to the binary readings as follows:

3.3 V ---------> 1023

X  V------------>analogRead(A0)

Then X=\frac{3.3 \times analogRead(A0)}{1023} Volts

Thus depending on the valule analogRead(A0) has in bits we get an approximate value of the voltage at pin A0, with a precission of 3,2mV approximately (3.3v/1023).

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RoseWind [281]

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

T_0 = T+\frac{V^2}{2c_p}

Where

T = Static temperature

V = Velocity of Fluid

c_p = Specific Heat

Re-arrange to find the static temperature we have that

T = T_0 - \frac{V^2}{2c_p}

T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})

T = 672.88K

Now the pressure of helium by using the Adiabatic pressure temperature is

P = P_0 (\frac{T}{T_0})^{k/(k-1)}

Where,

P_0= Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}

P = 0.399Mpa

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

3 0
3 years ago
A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u
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Answer:

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\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}

where,

\gamma= Specific Heat ratio of air

M = Mach number upstream

We know that \gamma _{air}=1.4

Applying values we get

\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)

Similarly the temperature downstream is obtained by the relation

\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}

Applying values we get

\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C

The Mach number downstream is obtained by the relation

M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616

3 0
4 years ago
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aleksandrvk [35]

Answer:

All the detailed steps are mentioned in pictures.

Explanation:

See attached pictures.

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Isormophous phase diagram
shusha [124]

Answer:

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Hope it help you friend

6 0
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If a sky diver decides to jump off a jet in Arkansas
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Answer:

The answer to this question can be defined as follows:

Explanation:

The sky driver began his sky journey from Arkansas, drove across the Tennessee River then landed in North Carolina. He returned to both the north in the very same direction. He began with NC, traveled through Tennessee, eventually lands in Arkansas. But North Carolina has been in the third state on which skydiver was traveling over, and It's also more than 700 miles from Arkansas to the NC.

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