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ArbitrLikvidat [17]
3 years ago
12

The Type K thermocouple has a sensitivity of about 41 uV /°C, i.e. for each degree difference in the junction temperature, the o

utput changes by 41 microVolts • If you have a 16-bit ADC, what is the smallest temperature change you can detect if the ADC range is 10 V?
Physics
1 answer:
Novosadov [1.4K]3 years ago
3 0

Answer:

ΔTmin = 3.72 °C

Explanation:

With a 16-bit ADC, you get a resolution of 2^{16}=65536 steps. This means that the ADC will divide the maximum 10V input into 65536 steps:

ΔVmin = 10V / 65536 = 152.59μV

Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:

\Delta Tmin = \frac{\Delta Vmin}{41 \mu V/C}= 3.72 C

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The three components of velocity in a velocity field are given by u = Ax + By + Cz, v = Dx + Ey + Fz, and w = Gx + Hy + Jz. Dete
Alexxandr [17]

Answer:

The relationship is only between the coefficients A, E and J which is:

A + E + J = 0. The remaining coefficients can be anything without any constraints.

Explanation:

Given:

The three components of velocity is a velocity field are given as:

u = Ax + By + Cz\\\\v = Dx + Ey + Fz\\\\w = Gx + Hy + Jz

The fluid is incompressible.

We know that, for an incompressible fluid flow, the sum of the partial derivatives of each component relative to its direction is always 0. Therefore,

\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0

Now, let us find the partial derivative of each component.

\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(Ax+By+Cz)\\\\\frac{\partial u}{\partial x}=A+0+0=A\\\\\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(Dx+Ey+Fz)\\\\\frac{\partial v}{\partial y}=0+E+0=E\\\\\frac{\partial w}{\partial z}=\frac{\partial }{\partial z}(Gx+Hy+Jz)\\\\\frac{\partial w}{\partial z}=0+0+J=J

Hence, the relationship between the coefficients is:

A+E+J=0

There is no such constraints on other coefficients. So, we can choose any value for the remaining coefficients B, C, D, F, G and H.

6 0
3 years ago
A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

#SPJ1

6 0
2 years ago
Practice 3: Label the correct phase that would result from the Moon and Earth in these positions.
Anna71 [15]

Answer:

both position I think in nor

5 0
3 years ago
Read 2 more answers
Which statement correctly describes the relationship between thermal energy and particle movement?
zhuklara [117]

As thermal energy increases, there is more particle movement. As thermal energy increases, there is more particle movement. As thermal energy increases, there is less particle movement.

Sure hope this helps you

5 0
2 years ago
Please Please Please help on these 3 questions :) 20 POINTS!! - NO LINKS PLEASE
GenaCL600 [577]

Answer:

Catapult on the ground: Normal, gravity

Catapult (I'm assuming launching marshmallow): Reaction of Force Applied

Marshmallow: Force Applied

Explanation:

This is the forces that act on a stationary object and a launched object. The catapult may also experience a force friction if your teacher is taking a more practical sense.

3 0
3 years ago
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