What happens to the resistance of two resistors when connected in series circuit

A ray of light travels across a liquid-to-glass interface. if the indices of refraction for the liquid and glass are, respective

prisoha [69]

When it comes to optics, Snell's law is the basic formula to be used. If you notice, when light hits the water, the light does not travel in the same direction. After, it hits the water, it changes in angle. Light becomes refracted. This is observed when your hands tend to become bigger if you place it underwater. The formula for Snell's Law is

n₁ sin θ₁ = n₂sin θ₂, where n is the index of refraction. This depends on the type of medium. For example, for air, n=1. The parameters θ₁ is the angle of incidence, and θ₂ is the angle of refraction. Critical angle is the incident angle needed so that the refract angle is 90°. So, modifying the equation:

n₁ sin θcrit = n₂sin 90°, since sin 90°=1,
sin θcrit = n₂/n₁
θcrit = sin ⁻¹ (n₂/n₁)

Since liquid comes first before glass, n₁=1.75 and n₂=1.52. Substituting,
θcrit = sin ⁻¹ (1.52/1.75)
θcrit = 60.29°

7 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

A motor has a rating power of 20 W and motor is used for 10 s. Calculate the

zepelin [54]

Explanation:

oii ay school ens?

6 0

3 years ago

The temperatures (in degrees Fahrenheit) in Long Island recorded by the weather bureau over a week were 42, 49, 53, 55, 50, 47,

lora16 [44]

Answer:

Inter Quartile Range

Explanation:

Quartile is a positional statistical average, which divided the data into 4 equal halves.

Q1 (Lower Quartile) has 25% data below it, 75% above it. Q3 (Upper Quartile) has 75% data below it, 25% above it.

Interquartile range is the measure used to calculate how far the lower & upper quartiles are.

7 0

3 years ago

A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision

BigorU [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$f= -75 \ cm = - 0.75 \ m$

b

$P = -1.33 \ diopters$

Explanation:

From the question we are told that

The image distance is $d_i = -75 cm$

The value of the image is negative because it is on the same side with the corrective glasses

The object distance is $d_o = \infty$

The reason object distance is because the object father than it being picture by the eye

General focal length is mathematically represented as

$\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}$

substituting values

$\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}$

=> $f= -75 \ cm = - 0.75 \ m$

Generally the power of the corrective lens is mathematically represented as

$P = \frac{1}{f}$

substituting values

$P = \frac{1}{-0.75}$

$P = -1.33 \ diopters$

7 0

3 years ago

What happens to the resistance of two resistors when connected in series​ circuit

What happens to the resistance of two resistors when connected in series circuit