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Eddi Din [679]
2 years ago
12

Current is the movement of positive charges called electrons.

Physics
2 answers:
spayn [35]2 years ago
7 0

Answer:

False -> negative charge

eimsori [14]2 years ago
3 0

Answer:

false

Explanation: electrons are negative charges

protons are positive charges

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miskamm [114]

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Astronomers use the gap between Earth and the sun, which is ninety-three million miles, as a new unit of measure called the Astronomical Unit.

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3 0
2 years ago
The magnetic field 0.0 2M from a wire is 0.1 T. What is the magnitude of the magnetic field of 0.0 1M from the same wire?
Anika [276]

Answer:

Explanation:

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5 0
2 years ago
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bagirrra123 [75]

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6 0
2 years ago
Considere que helicóptero para na altitude a que voava para largar uma caixa de 90 kg para o solo.
Firdavs [7]

Answer:

El peso es de 882 newton.

Explicación:

El helicóptero se detiene a la altura en la que volaba para dejar caer una caja de 90 kg al suelo, el peso de la caja llega hasta los 882 Newton porque el peso es una fuerza dirigida hacia el centro de la tierra y depende de la gravedad. La fórmula del peso es multiplicar la masa por la gravedad, de modo que cuando se multiplican 90 kg de masa por 9,8 metros / seg2, obtenemos el peso del objeto que es 882 newton.

7 0
3 years ago
Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml
Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is 2.41\times10^{3}\ V/m

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of  electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}

Put the value into the formula

E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}

E=2.41\times10^{3}\ V/m

Hence, The electric field due to the right ring at a location midway between the two rings is 2.41\times10^{3}\ V/m

3 0
3 years ago
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