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3 years ago

A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF.

2 answers:

N76 [4]3 years ago

8 0

Voltage across capacitor, V = 2.0V
Capacitance of Capacitor, C = 4.0 μF
Charge stored in Capacitor, Q = ?

The capacitance formula, C = Q/V
We can rewrite this as Q = CV

Therefore, the amount of charge stored in this capacitor = 2.0 * 4.0 μC
= 8.0 μC

Aleksandr-060686 [28]3 years ago

7 0

Answer:
Q = 8 μC

Explanation:
The relation between voltage, capacitance and charge can be expressed using the following rule:
Q = C * V
where:
Q is the amount of charge that we want to calculate
C is the capacitance = 4 * 10⁻⁶ F
V is the voltage applied = 2 V

Substitute with the givens in the above equation to get the amount of charge as follows:
Q = C * V
Q= 4 * 10⁻⁶ * 2
Q = 8 * 10⁻⁶ Coulumb
Q = 8 μC

Hope this helps :)

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Explanation:

Percent ionization is used for quantifying the number of ions present in the weak acid when dissolved in a solution. So it is similar to the pKa value. The percent ionization value can be determined as negative log of dissociation constant. Also the as the number of ions increases in weak acid, the concentration of acid will be decreasing . It can be calculated using the formula for percent ionization as follows:

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2 years ago

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3 years ago

Read 2 more answers

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

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2 years ago

For a fixed amount of gas at a fixed temperature, what will happen if the volume is doubled?.

mash [69]

Answer:

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nRT are all constant so they will equal 1

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2 years ago

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Neko [114]

Answer:

Balanced forces: When a number of forces acting on a body do not cause any change in its state of rest or of uniform motion along a straight line then the forces are said to be balanced forces. In other words, a body is said to be underbalanced forced when the resulting force acting on the body is zero.

The balanced forces:

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⋅ May change the shape and size of soft objects.

⋅ Cannot change the speed/velocity of a moving body.

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When the resultant of all the forces acting on a body is not zero, then forces are called unbalanced forces.

Example:

⋅ Game of tug of war: When the forces exerted by both the teams are equal, then the rope does not move. But, if the force applied by team A is greater than team B, then the rope, as well as members of the weaker team, i.e., B, will be pulled towards A. The unbalanced force can (a) Set a stationary body in motion.

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Explanation:

give me an one thanks please

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