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3 years ago

Dispositivo que muestra tanto la dirección como la magnitud de la corriente eléctrica.

Physics

1 answer:

slamgirl [31]3 years ago

5 0

Answer:

Galvanómetro

Explanation:

Un galvanómetro es un dispositivo eléctrico utilizado para detectar la presencia de corriente y voltaje pequeños o para medir su magnitud. Los galvanómetros se utilizan principalmente en puentes y potenciómetros.

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The medical model of abnormality states that psychological disorders are __________.

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A. social issues!
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Which explanation best describes a wave?

Gekata [30.6K]

Answer:

C. a disturbance that travels through a medium with a transfer of energy and without a transfer of matter

Explanation:

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2 years ago

A car battery can deliver up to 75 amps of current at a voltage of 12 volts. How much power is this?

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3 years ago

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According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 × 103 m/s, the uncertai

GREYUIT [131]

Explanation:

It is given that,

Uncertainty in the speed of an electron, $\Delta v=3.5\times 10^3\ m/s$

According to Heisenberg uncertainty principle,

$\Delta x.\Delta p=\dfrac{h}{4\pi}$

$\Delta x$ is the uncertainty in the position of an electron

Since, $\Delta p=m\Delta v$

$\Delta x=\dfrac{h}{4\pi.m \Delta v}$

$\Delta x=\dfrac{6.6\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 3.5\times 10^3}$

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3 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

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3 years ago