The Type K thermocouple has a sensitivity of about 41 uV /°C, i.e. for each degree difference in the junction temperature, the o
1 answer:
Answer:
ΔTmin = 3.72 °C
Explanation:
With a 16-bit ADC, you get a resolution of 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:
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Answer:
a) T_A = , b) T_B = g [m₂ (
) + m₁ (
) ]
c) x = , d) m₂ = m₁ (
)
Explanation:
After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act
a) The tension of string A is requested
The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive
∑ τ = 0
T_A d - W₂ x -W₁ L/2 = 0
T_A =
b) the tension in string B
we write the expression of the translational equilibrium
∑ F = 0
T_A - W₂ - W₁ - T_B = 0
T_B = T_A -W₂ - W₁
T_ B = - g m₂ - g m₁
T_B = g [m₂ ( ) + m₁ (
) ]
c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system
T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0
at the point that begins to rotate T_B = 0
g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0
m₂ (d-x) = m₁ (0.5 L- d)
m₂ x = m₂ d - m₁ (0.5 L- d)
x =
d) The mass of the block for which it is always in equilibrium
this is the mass for which x = 0
0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}
m₂ = m₁
m₂ = m₁ ( )
