Ask question

Ask question

All categories

4 years ago

8

The amount of light that undergoes reflection or transmission is demonstrated by how bright the reflected or transmitted ray is.

Under what conditions is the amount of transmission maximized and the amount of reflection minimized?

1 answer:

melamori03 [73]4 years ago

8 0

If you subscribe I’ll answer QF Aotrx

You might be interested in

Explain the limitations of science that you learned about in this unit. Write 1-2 sentences. (2 points

Kaylis [27]

Explanation:

Two limits of science are 1. observations may faulty and 2. science cannot deal with values or morals. Hopes this helps!!

8 0

3 years ago

A car and its passengers have a mass of 1200kg it is travelling at 12m/s.

Natali [406]

Answer:

<em>The increase of kinetic energy is 108,000 J</em>

Explanation:

<u>Kinetic Energy </u>

Is the energy an object has due to its state of motion. It's proportional to the square of the speed and the mass.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

A car has a total mass of m=1,200 kg and travels at v1=12 m/s. Then it increases its speed at v2=18 m/s.

It's required to compute the increase of kinetic energy. We'll calculate both energies K1 and K2 and then subtract them.

$\displaystyle K_1=\frac{1}{2}1,200*12^2=86,400\ J$

$\displaystyle K_2=\frac{1}{2}1,200*18^2=194,400\ J$

The increase of kinetic energy is:

$\Delta K=K_2-K_1 =194,400\ J-86,400\ J$

$\Delta K=108,000\ J$

The increase of kinetic energy is 108,000 J

4 0

3 years ago

What happens when the moon faces one side of the earth?

algol [13]

When the moon faces earth a solar eclipse happens :-)

5 0

3 years ago

Given the indices of refraction n_1 and n_2 of material 1 and material 2, respectively, rank these scenarios on the basis of the

aniked [119]

Answer:

a) order of refraction is a, e, de, c , c) a f g

Explanation:

a) when lightning is refracted it must comply with the law of refraction

.n₁ sin θ₁ = n₂ sinθ₂

sin θ / sin δδa) when lightning is refracted it must comply with the law of refraction

.n1 sin θ₁ = n2 sin θ ₂

Sint θ1 / sin θ2 = n2 / n1

1 we see that the ray deviation is promotional to the ratio of the refractive indices

The order of refraction is a, e, de, c

.b) When a ray passes from a medium with less indicated refraction to a higher index the part of the reflected ray has a phase change of 180º

a) no phase change

b) reflected ray has a phase change of 180º

c) no phase change

d) no phase change

e) there is a phase change

f) these phase changeo

1 we see that the ray deviation is promotional to the ratio of the refractive indices

The order of refraction is a, e, de, c

.b) When a ray passes from a medium with less indicated refraction to a higher index the part of the reflected ray has a phase change of 180º

a) no phase change

b) reflected ray has a phase change of 180º

c) no phase change

d) no phase change

e) there is a phase change

f) ay phase vabio

7 0

4 years ago

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in va

WINSTONCH [101]

Answer:

Part A: 7500 V

Part B: 2.899×10⁻³ m²

Part C: 10.27 pF or 10.27×10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d................ Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed............. Equation 2

Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m

Substitute into equation 2

V = 3.0×10⁶ (2.5×10⁻³ )

V = 7.5×10³ V

V = 7500 V

Part B:

Using,

E = Q/(e₀A).................... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q/(e₀E).............. Equation 4

Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m

Constant: e₀ = 8.854×10⁻¹² F/m

Substitute into equation 4

A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)

A = 77×10⁻⁹/(26.562×10⁻⁶)

A = 2.899×10⁻³ m²

A = 2.899×10⁻³ m².

Part C:

Using,

Q = CV.................. Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V.............. Equation 6

Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77×10⁻⁹/7500

C = 10.27×10⁻¹² F

C = 10.27 pF

5 0

3 years ago