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3 years ago

5

A very thin oil film (n = 1.25) floats on water (n = 1.33). What is the thinnest film that produces a strong reflection for gree

n light with a wavelength of 500 nm?

1 answer:

mixer [17]3 years ago

6 0

Answer:

200 nm is the thinnest film that produces a strong reflection for green light with a wavelength of 500 nm

Explanation:

If two reflected waves interfere constructively ,strong reflection is produced. Two reflected waves will experience a phase change

For constructive interference

$2\times n\times t=m\lambda$

for thinnest film m=1

refractive index should be taken for film n=1.25

thickness of the thinnest film is

$t=\frac{m\lambda}{2n} \\t=\frac{1\times 500}{2\times 1.25} \\t=200 nm$

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Answer:

D. Ted expanded more power.

Explanation:

Given the following data;

For Ted.

Force = 10N

Height = 1.5m

Time = 1 seconds

To find Ted's power;

Power = workdone/time

But workdone = force * distance

Workdone = 10 * 1.5

Workdone = 15 Nm

Power = 15/1

Power = 15 Watts.

For Johnny.

Force = 10N

Height = 1.5m

Time = 2 seconds

To find Ted's power;

Power = workdone/time

But workdone = force * distance

Workdone = 10 * 1.5

Workdone = 15 Nm

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Therefore, from the calculations we can deduce and conclude that Ted expanded more power.

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Answer:

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Explanation:

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(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface

navik [9.2K]

Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

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1. Which of the following factors would best help identify a biome?

siniylev [52]

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