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3 years ago

For which of these scenarios is total internal reflection possible?

1 answer:

3 0

TIR is possible whenever light is travelling from a more optically dense medium to a less optically dense medium. You didnt provide any scenarios but just look up refractive indexes of the mediums the light is travelling through and if it is travelling from a more dense to a less dense, TIR is possible

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A 9800-N automobile is pushed along a level road by four students who apply a total forward force of 600 N. Neglecting friction,

Alexandra [31]

Answer:

c). $a = 0.60 m/s^2$

Explanation:

As we know that weight of the automobile is given here

so weight = mass times gravity

$W = mg$

$9800 = m(9.8)$

$m = 1000 kg$

now from Newton's law

$F = ma$

$600 = 1000 a$

$a = \frac{600}{1000}$

$a = 0.60 m/s^2$

4 0

3 years ago

A solenoid inductor has an emf of 0.80 V when the current through it changes at the rate 10.0 A/s. A steady current of 0.20 A pr

barxatty [35]

Answer:

The number of turns of the inductor is 2000 turns.

Explanation:

Given;

emf of the inductor, E = 0.8 V

the rate of change of current with time, dI/dt = 10 A/s

steady current in the solenoid, I = 0.2 A

flux per turn, Ф = 8.0 μWb per

Determine the inductance of the solenoid, L

E = L(dI/dt)

L = E / (dI/dt)

L = 0.8 / (10)

L = 0.08 H

The inductance of the solenoid is given by;

$L = \frac{\mu_o N^2 A}{l}$

Also, the magnetic field of the solenoid is given by;

$B = \frac{\mu_o NI}{l}$

I is 0.2 A

$B = \frac{\mu_oN(0.2)}{l} = \frac{0.2\mu_o N}{l}$

$\frac{B}{0.2 } = \frac{\mu_o N}{l}$

$L = \frac{\mu_o N^2 A}{l} \\\\L = \frac{\mu_o N }{l} (NA)\\\\L = \frac{B}{0.2} (NA)\\\\L = \frac{BA}{0.2} (N)$

But Ф = BA

$L = \frac{\phi N}{0.2} \\\\\phi N = 0.2 L\\\\N = \frac{0.2 L}{\phi} \\\\N = \frac{0.2 *0.08}{8*10^{-6}}\\\\N = 2000 \ turns$

Therefore, the number of turns of the inductor is 2000 turns.

5 0

4 years ago

What is the ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 125 km/h speed limit (about

nikitadnepr [17]

An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let θ denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,

• the normal force with magnitude n

• the car's weight with magnitude w

and the net force points toward the center of the circle made by the turn, with centripetal acceleration

a = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²

Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,

∑ F (⟂) = N + W (⟂) = m a (⟂)

and

∑ F (//) = W (//) = m a (//)

Let the direction of N be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle θ with the banked curve, and W makes the same angle with the negative perpendicular axis, so that the equations above reduce to

N - m g cos(θ) = m a sin(θ)

and

m g sin(θ) = m a cos(θ)

The second equation is all we need at this point to find the ideal θ. The mass m cancels out, and we can solve for θ to get

tan(θ) = a/g ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615

→ θ ≈ 3.52°