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gizmo_the_mogwai [7]
3 years ago
7

Explicar el funcionamiento de un multímetro analógico.

Engineering
1 answer:
Whitepunk [10]3 years ago
4 0

Answer:

Un multímetro analógico funciona como un medidor de bobina móvil de imán permanente (PMMC) para tomar mediciones eléctricas

Explanation:

El multímetro analógico es un medidor o galvanómetro D'Arsonval que funciona según el principio de los medidores de bobina móvil de imán permanente (PMMC)

Un multímetro analógico está formado por un puntero de aguja unido a una bobina móvil colocada entre el polo norte y sur de un imán permanente dispuesto de tal manera que, cuando una corriente eléctrica fluye a través de la bobina, genera una fuerza de campo magnético que interactúa con el imán fuerza de campo de los imanes permanentes que hace que la bobina se mueva junto con el puntero de la aguja sobre un dial graduado

Para controlar el movimiento del puntero de la aguja, de modo que el par requerido para producir una cantidad de movimiento por corriente detectada por el multímetro, se colocan dos resortes a través de la bobina para proporcionar resistencia al movimiento en ambas direcciones y para permitir la calibración del multímetro analógico.

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6 0
3 years ago
if both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will
egoroff_w [7]

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will: a) increase during a climb.

<h3>What is a ram air input?</h3>

A ram air input can be defined as an air intake system which is designed and developed to use the dynamic air pressure that is created due to vehicular motion, or ram pressure, in order to increase the static air pressure within the intake manifold of an internal combustion engine of an automobile.

This ultimately implies that, a ram air input allows a greater mass-flow of air through the engine of an automobile, thereby, increasing the engine's power.

In conclusion, indicated airspeed will increase during a climb when both the ram air input and drain hole of the pitot system become blocked.

Read more on pilots here: brainly.com/question/10381526

#SPJ1

Complete Question:

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

a) increase during a climb

b) decrease during a climb

c) remain constant regardless of altitude change

6 0
2 years ago
Damage reports should contain the information needed for what?
SIZIF [17.4K]

Answer:

<h2>it's damage kindly be careful</h2>
3 0
3 years ago
What does Faraday's law of induction states?​
Karolina [17]

Explanation:

This relationship, known as Faraday's law of induction (to distinguish it from his laws of electrolysis), states that the magnitude of the emf induced in a circuit is proportional to the rate of change of the magnetic flux that cuts across the circuit.

6 0
3 years ago
A building window pane that is 1.44 m high and 0.96 m wide is separated from the ambient air by a storm window of the same heigh
sdas [7]

Answer:

the rate of heat loss by convection across the air space = 82.53 W

Explanation:

The film temperature

T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C

to kelvin = (5 + 273)K = 278 K

From  the " thermophysical properties of gases at atmospheric pressure" table; At T_f = 278 K ; by interpolation; we have the following

\frac{278-250}{300-250}= \frac{v-11.44(10^{-6})}{15.89(10^{-6})-11.44(10^{-6})}  → v 13.93 (10⁻⁶) m²/s

\frac{278-250}{300-250}= \frac{k-22.3(10^{-3}}{26.3(10^{-3}-22.3(10^{-3})} → k = 0.0245 W/m.K

\frac{278-250}{300-250}= \frac{\alpha - 15.9(10^{-6})}{22.5(10^{-6}-15.9(10^{-6})} → ∝ = 19.6(10⁻⁶)m²/s

\frac{278-250}{300-250}= \frac{Pr-0.720}{0.707-0.720} → Pr = 0.713

\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}

The Rayleigh number for vertical cavity

Ra_L  = \frac{g \beta (T_1-T_2)L^3}{\alpha v}

= \frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}

= 8.38*10^5

\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24

For the rectangular cavity enclosure , the Nusselt number empirical correlation:

Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}

NU_L= \frac{hL}{k}= 4.878

\frac{hL}{k}= 4.878

\frac{h*0.06}{0.0245}= 4.878

h = \frac{4.878*0.0245}{0.06}

h = 1.99 W/m².K

Finally; the rate of heat loss by convection across the air space;

q = hA(T₁ - T₂)

q = 1.99(1.4*0.96)(20-(-10))

q = 82.53 W

3 0
3 years ago
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