The speed of light is 3 x 10 m/s.Calculate the frequency of light that is absorbed the most by the 100m length of fibre.
Give your answer in standard form.

Answers

Answer 1

Answer:

Answer:

$3 * 10 {}^6$

Related Questions

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The pressure in a section of horizontal pipe with a diameter of 2.5 cm is 139 kPa. Water ï¬ows through the pipe at 2.9 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should be the diameter of the constricted section? The acceleration of gravity is 9.81 m/s2 . Assume laminar nonviscous ï¬ow.

Answers

Answer:

d = 2*0.87 = 1.75 cm

Explanation:

by using flow rate equation to determine the speed in larger pipe

$\phi =\pi r^2 v$

$v = (\phi)/(\pi r^2)$

$= (2900 cm^3/s)/(3.14(1.25cm)^2)$

= 591.10 cm/s

= 5.91 m/s

by Bernoulli's EQUATION

$p1 +(1)/(2) \rho v1^2 = p2 +(1)/(2) \rho v2^2$

$139000+ (1)/(2)*1000*5.91^2 = 101000 +(1)/(2)*1000* v2^2$

solving for v2

v2 = 10.53 m/s

diameter can be determine by using flow rate equation

$q = v \pi r^2$

$r^2 = (q)/(\pi v)$

$= (2900)/(3.14*1053)$

r = 0.87 cm

d = 2*0.87 = 1.75 cm

An airplane is in level flight over Antarctica, where the magnetic field of the earth is mostly directed upward away from the ground. As viewed by a passenger facing toward the front of the plane, is the left or the right wingtip at higher potential? Does your answer depend on the direction the plane is flying?

Answers

Answer:

It does not depend on direction of plane and the left windtips more potential

Explanation:

Because if the Fleming right hand rule is applied the the right han is pointed in the direction of flight, and the fingers are curled in the direction of the magnetic lines. Thus , the lines are vertical and so are pointing down, thus by the right hand rule, the electrons move to the left hand side of the plane, although the potentials are equal and opposite in direction

Temperature°F = (9/5 * °C) + 32°
°C = 5/9 * (°F - 32°)
1 pt each. Using the table above as a guide, complete the following conversions. Be sure to show your work to the side:
1. 5 cm = ________ mm
2. 83 cm = ________ m
3. 459 L = _______ ml
4. .378 Kg = ______ g
5. 45°F = ________ °C
6. 80°C = _________ °F

Answers

$5cm = 50mm$
$2.83cm = 0.0283m$
$3.459l = 3459ml$
$4.378kg = 4378g$
$5.45f = - 47.79c$
$6.80c = 44.24f$

25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d, vcmax and vg for the time, tc, it takes the cheetah to catch the gazelle.

Answers

maximum speed of cheetah is

$v_1 = v_(max)$

speed of gazelle is given as

$v_2 = v_(g)$

Now the relative speed of Cheetah with respect to Gazelle

$v_(12) = v_1 - v_2$

$v_(12) = v_(max) - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_(12)* t$

so by rearranging the terms we can say

$t = (d)/(v_(12))$

$t = (d)/(v_(max) - v_g)$

so above is the relation between all given variable

Water flows at speed of 4.4 m/s through ahorizontal pipe of diameter 3.3 cm . The gauge
pressure P1 of the water in the pipe is 2 atm .
A short segment of the pipe is constricted to
a smaller diameter of 2.4 cm
(IMAGE)
What is the gauge pressure of the water
flowing through the constricted segment? Atmospheric pressure is 1.013 × 10^5 Pa . The density of water is 1000 kg/m^3
. The viscosity
of water is negligible.
Answer in units of atm

Answers

Answer:

1.75 atm

Explanation:

Mass is conserved, so the mass flow before the constriction equals the mass flow after the constriction.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = h₂:

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Writing v₂ in terms of v₁:

P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ d₁²/d₂²)²

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (d₁/d₂)⁴

P₁ + ½ ρ v₁² (1 − (d₁/d₂)⁴) = P₂

Plugging in values:

P₂ = 2 atm + ½ (1000 kg/m³) (4.4 m/s)² (1 − (3.3 cm / 2.4 cm)⁴) (1 atm / 1.013×10⁵ Pa)

P₂ = 1.75 atm

The temperature within a thin plate with thermal conductivity of 10 W/m/K depends on position as given by the following expression: TT=(100 K)????????−xx2/????????xx 2cos�yy/????????yy�+300 K Where, Lx = 1 m, and Ly = 2 m. At the point (0.4 m, 1 m), find: a. The magnitude of the heat flux b. The direction of the heat flux