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amid [387]
2 years ago
15

When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t

o I(t), where t represents the thickness of the medium (in feet). In clear seawater, the intensity 3 feet below the surface is 25% of the initial intensity I0 of the incident beam. What is the intensity of the beam 18 feet below the surface
Physics
1 answer:
antoniya [11.8K]2 years ago
3 0

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

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If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi
netineya [11]

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

F_s = \mu_s*N

Where F_s is the static friction force, \mu_s is the coefficient of static friction and N is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, N = 110\text{ N}, to make the puppy moving we need to use a force of 80 N, therefore, F_s = 80 \text{ N}, so we can solve for the coefficient as shown below:

80 = \mu_s*110\\\mu_s = \frac{80}{110} = 0.7273\\

The coefficient of static friction between the puppy and the floor is 0.7273.

5 0
2 years ago
Timmy drove 2/5 of a journey at an average speed of 20 mph.
mixer [17]

Answer:

4hr

Explanation:

5 0
2 years ago
Read 2 more answers
Assume you are driving 20 mph on a straight road. Also, assume that at a speed of 20 miles per hour, it takes 100 feet to stop.
Viefleur [7K]

Answer:900  feet

Explanation:

Given

Velocity \left ( V_1\right )=20 mph\approx 29.334 ft/s

it take 100 feet to stop

Using Equation of motion

v^2-u^2=2as

where

v,u=Final and initial velocity

a=acceleration

s=distance moved

0-\left ( 29.334\right )^2=2\left (-a\right )\left ( 100\right )

a=\frac{29.334^2}{2\times 100}=4.302 ft/s^2

When velocity is 60 mph\approx 88.002 ft/s

v^2-u^2=2as

0-\left ( 88.002\right )^2=2\left ( -4.302\right )\left ( s\right )

s=900.08 feet

8 0
2 years ago
Read 2 more answers
Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward rad
mote1985 [20]

Answer:

468449163762.0812 W

Explanation:

m = Mass = \rhoV

V = Volume =\dfrac{4}{3}\pi r^3

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

\rho = Density = 19 g/cm³

c = Speed of light = 3\times 10^8\ m/s

A = Area = \pi R^2

I = Intensity = \dfrac{P}{4\pi r^2}

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}

According to the question

F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W

The power required of the light source is 468449163762.0812 W

4 0
2 years ago
Not too long ago houses were protected from excessive currents by fuses rather than circuit breakers. sometimes a fuse blew out
olganol [36]

Answer: Resistance = 8.21 \times 10^{-8} \Omega

The approximate diameter of a penny is, <em>d</em> = 20 mm

thickness of penny is, <em>L = </em> 1.5×10^{-3} mm

The area of penny along circular face is,A = \frac{\pi d^2}{4} =\frac{\pi (20 mm\frac{1 m}{1000 m})^2}{4}

= 3.14×10^{-4} m²

The resistivity of copper is <em>ρ</em> = 1.72 x 10-8 Ωm.

Resistance,

R = \rho \frac{L}{A}  = (1.72 \times 10^{-8} \Omega m)\frac{1.5 \times 10^{-3} m}{3.14 \times 10^{-4} m^2} = 8.21 \times 10^{-8} \Omega

5 0
2 years ago
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