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3 years ago

The current in resistor Y is..?

Physics

1 answer:

Aliun [14]3 years ago

4 0

(A)

Explanation:

We can see that the resistors are connected in parallel so all of them have the same voltage of 100 V. We also know that

$P = VI$

Since resistor Y dissipates 100 W of power, we can solve for the current as

$I = \dfrac{P}{V} = \dfrac{100\:\text{W}}{100\:\text{V}} = 1.0\:\text{A}$

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A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. The time it takes for the ball to fall halfway t

Naddik [55]

Answer:

1.697s

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8 0

3 years ago

Two parallel plates of area 7.34 x 10^-4 m^2 have 5.83 x 10^-8 C of charge placed on them. A 6.62 x 10^-6 C charge q1 is placed

inn [45]

Answer:

* if the two plates have the same charge sign F_net = 0

*if one plate is positive and the other is negative F_net = 59.4 N

Explanation:

The electric field created by a parallel plate is

E = $\frac { \sigma }{2 \epsilon_o }$

where sigma is the charge density

σ = Q / A

we substitute

E = $\frac{Q}{A \ 2 \epsilon_o }$

E = $\frac{ 5.83 \ 10^{-8} }{7.34 \ 10^{-4}\ 2 \ 8.5 \ 10^{-12} }$

E = 4.487 10⁶ N / C

the electric force is

F = E / q

in this force it is a vector, so if the charges are of the same sign they repel and if they are of the opposite sign they attract. In this case, the test load is between the two plates that have the same load sign, so the forces are in the opposite direction.

* if the two plates have the same charge sign

F_net = F₁ - F₂

F_net = q (E₁ -E₂)

since the electric field does not depend on the distance

E₁ = E₂

in consecuense

F_net = 0

In a more interesting case

*if one plate is positive and the other is negative

F_net = F₁ + F₂

F_net = q (E₁ + E₂) = 2 q E₁

F_net = 2 6.62 10⁻⁶ 4.487 10⁶

F_net = 59.4 N

net force goes from positive to negative plate

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