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Orlov [11]
4 years ago
15

A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. (a) A

t the moment contact is made with the battery the voltage across the capacitor is
Physics
1 answer:
maks197457 [2]4 years ago
6 0

Answer:

(a) D.  Zero.

(b) C.  Equal to the battery's terminal voltage.

Explanation:

The question is incomplete, see the complete question for your reference and information.

A resistor and a capacitor are connected in series across an ideal battery having a constant voltage

across its terminals. At the moment contact is made with the battery

(a) the voltage across the capacitor is

A) equal to the battery's terminal voltage.

B) less than the battery's terminal voltage, but greater than zero.

C) equal to the battery's terminal voltage.

D) zero.

(b) the voltage across the resistor is

A) equal to the battery's terminal voltage.

B) less than the battery's terminal voltage, but greater than zero.

C) equal to the battery's terminal voltage.

D) zero

A RC circuit is a circuit that is composed of both resistors and capacitors connect to a source of current or voltage.

basically when a voltage source is applied to an RC circuit, the capacitor, C charges up through the resistance, R

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What is x-rays ? and how it's formed ?​
tia_tia [17]

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3 years ago
A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

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3 0
3 years ago
A forklift raises a crate weighing 8.35 × 102 newtons to a height of 6.0 meters. What amount of work does the forklift do?
Vinil7 [7]
Answer:

<span>5010J</span>

Explanation:

Work is force times distance, or

<span>W=F⋅d</span>.

Substitute in values from the question to get

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6 0
3 years ago
Read 2 more answers
Two long, parallel wires each carry the same current I, but the two currents are anti-parallel. The two wires are a distance r a
Georgia [21]

Answer:

8 x 10⁻⁷ x  I / r

Explanation:

Two parallel long wires are carrying current I . Let the direction be towards the right in the farthest and towards the left in the nearest. Magnetic field due to current I  at a  distance d is given by the expression

B = μ₀ 2 I / 4π d

I the present case distance d = r/2  

Magnetic field due to one wire at point d = r/2  is

B₁ = μ₀ 2 I / (4π r / 2 )

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Magnetic field due to the other wire at point d = r/2  is

B₂ = μ₀ 2 I / (4π r / 2 )

= 10⁻⁷ x 4I / r

Direction of magnetic field due to both the wires at the mid  point P will be same . It will be in downward direction in the given scenario

So total magnetic field

B = B₁ + B₂

= 2 x  10⁻⁷ x 4I / r

= 8 x 10⁻⁷ x  I / r

6 0
3 years ago
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