All categories

3 years ago

Two electrodes connected to a 9.0 v battery are charged to ±45 nc. What is the capacitance of the electrode?

Physics

1 answer:

mart [117]3 years ago

6 0

Answer:

$5\cdot 10^{-9} F$

Explanation:

The capacitance of the electrode is given by:

$C=\frac{Q}{V}$

where

C is the capacitance

Q is the charge on the electrode

V is the potential difference

In this problem, we have

$Q=45 nC=45\cdot 10^{-9} C$

V = 9.0 V

Substituting into the equation, we find

$C=\frac{45\cdot 10^{-9}C}{9.0 V}=5\cdot 10^{-9} F$ (5 nF)

You might be interested in

A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the

marin [14]

Answer:

0.395 m

2.4 m/s

Explanation:

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$x$ = initial position of spring from equilibrium position = 0.21 m

$v_{i}$ = initial speed of the cart = 2.0 ms⁻¹

$A$ = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

$(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m$

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$A$ = amplitude of the oscillation = 0.395 m

$v_{o}$ = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

$(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}$

5 0

3 years ago

Read 2 more answers

Unpolarized light with an intensity of 22.4 ????ux passes through a polarizer whose transmission axis is vertically oriented. (a

irina1246 [14]

Answer:

a) I₁ = 11.2 Lux , vertical direction , b) I₂ = 1.44 Lux

Explanation:

a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed

consequently the above the processed light has half of the incident intensity and the directional of the polarized

I₁ = I₀ / 2

I₁ = 22.4 / 2

I₁ = 11.2 Lux

is polarized in the vertical direction

b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus

I₂ = I₁ cos² θ

I₂ = 11.2 cos² 69

I₂ = 1.44 Lux

8 0

3 years ago

What happens when calcium reacts with chlorine?

choli [55]

Answer: They create calcuim chloride, CaCl2

7 0

3 years ago

by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one th

xenn [34]

Answer:

<em>The force is now 9 times the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

$\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}$

$\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}$

$\displaystyle F'=9k\frac{q_1q_2}{d^2}$

$\displaystyle F'=9F$

The force is now 9 times the original force

8 0

3 years ago

Sometimes balance point may not be obtained on the potentiometer wire why

scoundrel [369]

Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage

.
The balance point is not on the potentiometer wire - this statement means that

. In that case ,
l > L
V > E

8 0

3 years ago