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3 years ago

A light wave travels through water (n=1.33) at an angle of 35º. What angle

Physics

2 answers:

san4es73 [151]3 years ago

6 0

Answer:

$49.7^{\circ}$

Explanation:

We can solve the problem by using Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

n1 = 1.33 is the index of refraction of the first medium (water)

n2 = 1.00 is the index of refraction of the second medium (air)

$\theta_1 = 35^{\circ}$ is the angle of incidence of the wave in water

$\theta_2$ is the angle of refraction of the wave in air

Solving for the angle of refraction,

$sin \theta_2 = \frac{n_1 sin \theta_1}{n_2}=\frac{(1.33) sin 35^{\circ}}{1.00}=0.763\\\theta_2 = sin^{-1} (0.763)=49.7^{\circ}$

fiasKO [112]3 years ago

3 0

Answer:

49.7

Explanation:

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Which radioisotope is used in dating geological formations?

ziro4ka [17]

<h2>Answer: U-238</h2>

Explanation:

Let's begin by explaining that for radioactive geological dating (also called radioisotope dating) in which radioactive impurities were selectively incorporated when the fossil materials were formed, it is very useful to compare it with a naturally occurring radioisotope having a known half-life.

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To understand it better:

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3 years ago

When a puddle of water evaporates, does the water change into a new kind of matter

PIT_PIT [208]

Evaporated water changes form into gas

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3 years ago

A hard-boiled egg of mass 46.0 gg moves on the end of a spring with force constant 25.6 N/mN/m . The egg is released from rest a

soldi70 [24.7K]

Answer:

0.022kg/s

Explanation:

We are given that

Mass of boiled egg=46 g= $\frac{46}{1000} kg=0.046 kg$

$1kg=1000 g$

Constant force=F=25.6 N/m

Initial displacement= $A_1=0.296 m$

Final displacement= $A_2=0.12 m$

Time=t=4.55 s

Damping force= $F_x=-bv_x$

We have to find the magnitude of damping constant b.

We know that the displacement of the oscillator under damping motion is given by

$x=Ae^{-\frac{b}{2m}t}cos(w't+\phi)$

For maximum displacement $cos(w't+\phi)=1$

Therefore , $x=A_2$

Substitute the values

$A_2=A_1e^{-\frac{-b}{2m}t}$

$e^{-\frac{b}{2m}t}=\frac{A_2}{A_1}$

$-\frac{b}{2m}t=ln\frac{A_2}{A_1}$

$lnx=y\implies x=e^y$

Substitute the values

$-\frac{b}{2\times 0.046}\times 4.55=ln\frac{0.12}{0.296}$

$\frac{2\times 0.046}{4.55b}=ln\frac{0.296}{0.12}$

$\frac{2\times 0.046}{4.55}=0.9b$

$b=\frac{2\times 0.46}{4.55\times 0.9}=0.022kg/s$

Hence,the magnitude of damping constant b=0.022kg/s

3 0

3 years ago

Which of the following statements concerning the lab is TRUE?

Over [174]

Answer:

(b) To get m3 to slide, m1 must be increased, never decreased.

Explanation:

Lab experiments require attentiveness. If there is one thing missed or not taken seriously whole experiment could go wrong. In this case to slide m3 there should be more weight at m1. If the weight of m1 is lesser than m3 then the object will not slide. It will remain at the point where there is more weight. To slide an object there must be less frictional surface and more weight placed at the desired end point.

3 0

3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth’s center. Satellite A is to o

Vera_Pavlovna [14]

Answer:

Explanation:

Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km

Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km

Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite

Orbital potential energy of a satellite A = - GMm / Ra

Orbital potential energy of a satellite B = - GMm / Rb

PE of satellite B /PE of satellite A

= Ra / Rb

= 12740 / 25480

= 1 / 2

b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same

KE of satellite B /KE of satellite A

= 1 / 2

c ) Total energy will be as follows

Total energy = - PE + KE

- P E + PE/2

= - PE /2

Total energy of satellite B / Total energy of A

= 1 / 2

Satellite B will have greater total energy because its negative value is less.

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3 years ago