why does the value of capacitance of a capacitor increases in parallel combination but not in series??

Answers

Answer 1

Answer: It is easiest to consider problems like this by thinking exclusively about parallel plate capacitors for which $C \equiv (Q)/(V) =\kappa \epsilon_0 (A)/(d)$ where Q is the charge separated (+Q on one plate, -Q on the other), V is the voltage difference between the plates, A is the area of each plate, and d is the separation between the plates.

When capacitors are connected in parallel, the voltage across each capacitor is the same. But with two capacitors, it will require more charge to reach the voltage V than it would with just one capacitor. In fact, if capacitor 1 requires charge

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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.80 N)k and the corresponding torque about the origin is vector tau = (3.40 N · m)i hat + (2.80 N · m)j + (0.800 N · m)k. Determine Fx.

Answers

Answer:

-4.40

Explanation:

explanation is in attachment

A circular loop of wire has radius 7.80cm . A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 2.03�10?2W/m2 , and the wavelength of the wave is 6.20m .What is the maximum emf induced in the loop? Express your answer with the appropriate units.I stumbled through the formulas I do know for EMF but cant seem to figure out how to get the right answer. Please help and provide explanation! Thanks

Answers

Answer:

fem = - 4.50 10²² V

Explanation:

For the solution of this problem we must use the equation of the induced electromotive force or Faraday's law

E = - d Φ $._(B)$ / dt = d (BA cos θ) dt

In this case they tell us that the magnetic field is perpendicular to the plane of the loop, as the normal to the surface of the loop is in the direction of the radius, the angle between the field and this normal is zero, so cos 0º = 1. The area of the loop is constant, with this the equation is

E = - A dB / dt (1)

To find field B, we have the relationships of electromagnetic waves

E = c B

The intensity or poynting vector for the wave is described by the equation

S = I = 1 / μ₀ E x B = 1 /μ₀ E B

We replace

I = 1 /μ₀ (cB) B = c /μ₀ B²

This is the instantaneous intensity.

B = √ (μ₀ I /c)

We substitute in equation 1

E = - A μ₀/c d I / dt

With the maximum value we are asked to change it derived from variations

E = -A c/μ₀ ΔI / Δt

It remains to find the time of the variation. Let's use the equation

c = λ f = λ / T

T = λ / c

T = 6.20 / 3 10⁸

T = 2.06 10⁻⁸ s

We already have all the values to calculate the fem

fem = - π r² c/μ₀ ΔI/Δt

fem = - (π 0.078²) (3 10⁸/(4π 10⁻⁷) (2.03 10² -0) / (2.06 10⁻⁸ - 0)

fem = - 4.50 10²² V

In a solution such as salt water, the component that does the dissolving is called the?

Answers

Answer:

Solvent

Explanation:

A solution is a substance that is made by dissolving one component in another.
It is made up of a solvent and a solute.
The component that dissolves the other is known as solvent. Examples of solvents include water, ethanol, milk, chloroform, etc.
The component that dissolves in the other is known as the solute. Examples of solute include salts, sugar, etc.
For example in an aqueous solution of sodium chloride(NaCl); sodium chloride salt is the solute while water is the solvent.

A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle of 60° with the plane of the coil. If the magnitude of this field increases by 6.0 mT every 10 ms, what is the magnitude of the emf induced in the coil?

Answers

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

$\epsilon=N(d\phi)/(dt)$

$(d\phi)/(dt)$ is the rate of change if magnetic flux.

$\phi=BA\ cos\theta$

$\theta$ is the angle between the magnetic field and the normal to area vector.

$\theta=90-60=30$

$\epsilon=NA(dB)/(dt)* cos30$

$\epsilon=2* (0.24\ m)^2* (6\ mT)/(10\ mT)* cos(30)$

$\epsilon=0.0598\ T$

$\epsilon=59.8\ mT$

EMF = 60 mT

So, the magnitude of emf induced in the coil is 60 mT. Hence, this is the required solution.

The mass of a string is 20 g and it has a length of 3.2 m. Assuming that the tension in the string is 2.5 N, what will be the wavelength of a travelling wave that is created by a sinusoidal excitation of this string with a frequency of 20 Hz. Provide the wavelength in units of m. Please note: You do not include the units in your answer. Just write in the number.

Answers

Answer:

The wavelength of the wave is 1 m

Explanation:

Given;

mass of the string, m = 20 g = 0.02 kg

length of the string, L = 3.2 m

tension on the string, T = 2.5 N

the frequency of the wave, f = 20 Hz

The velocity of the wave is given by;

$v = \sqrt(T)/(\mu) {}$

where;

μ is mass per unit length = 0.02 kg / 3.2 m

μ = 6.25 x 10⁻³ kg/m

$v = \sqrt{(T)/(\mu) } \n\nv = \sqrt{(2.5)/(6.25*10^(-3)) } \n\nv = 20 \ m/s$

The wavelength of the wave is given by;

λ = v / f

λ = (20 m/s )/ (20 Hz)

λ = 1 m

Therefore, the wavelength of the wave is 1 m

Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?