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4 years ago

Which term is defined as the change in direction of light when it goes from one medium into a different meduim?

Physics

2 answers:

Sauron [17]4 years ago

6 0

The answer is Refraction.

BTW, I just took the test, so I know that's the right answer!

I hope this helps. :)

icang [17]4 years ago

3 0

Refraction is the term

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Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m fr

Korolek [52]

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

Mathematically this can be described as:

$E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$

Where,

$\epsilon_0 =$ permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

$E= 49.000N/C$

$r= 4.1m$

$\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}$

Replacing we have,

$E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$

$49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}$

$q= 9.16*10^{-5} C$

$q= 91.6\mu C$

Therefore the amoun of charge on the outer surface of the larger shell is $91.6 \mu C$

6 0

4 years ago

An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic m

Sergeeva-Olga [200]

Answer:

a = 2 m/s2

Explanation:

we know from newtons 2nd law

F = ma.

we also know that from hookes law we have

F = kx

equate both value of force to get value of acceleration

kx = ma,

where,

k is spring constant = 8.0 N/m

x is maximum displacement 0.10 m

m is mass of object 0.40 kg

a = \frac{kx}{m}

= \frac{8 *0 .10}{0.40}

a = 2 m/s2

5 0

4 years ago

What are four metals other than iron that can be made to exhibit magnetic properties? Fill in the blank. Every magnet has ______

Morgarella [4.7K]

Answer:

Four metals other than iron that can be made to exhibit magnetic properties are nickel, cobalt, manganese, and chromium.

two

neutral region

north

magnetic

Explanation:

PennFoster

4 0

4 years ago

A satellite travels with a constant speed |v| as it moves around a circle centered on the earth. How much work is done by the gr

7nadin3 [17]

Answer:

W = 0

Explanation:

As the satellite moves in a circle the force is perpendicular to the path, therefore the work that is defined by

W = F. r = f r cos θ

Since the force and the radius are perpendicular, the angle θ = 90º and the cosine 90 = 0, therefore there is no work for the circular motion.

W = 0

3 0

3 years ago

suppose that the man pictured on the front side is orbiting the earth (mass=5.98 x 10^24kg at a distance of 310 miles above the

wlad13 [49]

Answer:

a = 9.81[m/s^2]; v = 18683.5[m/s]

Explanation:

The stament of the problem is:

Suppose that the man pictured on the front side is orbiting the earth (mass = 5.98 x 1024kg) at a distance of 310 miles (1600 meters = 1 mile) above the surface of the earth (radius = 4000 miles).

a. What acceleration does he experience due to the earth's pull?

b. What tangential velocity must he possess in order that he orbit safely (in m/s)?

First we need to convert all the initial data to units of the SI

Rs = 310 [mil] = 498897 [m]

RT= 400 [mil] = 643738 [m]

r = Rs + RT = 1142635 [m] "distance from the center of the earth to the man"

$F=G*\frac{M*m}{r^{2} } \\where:\\$

G = universal gravitational constant

M = mass of the earth [kg]

m = mass of the man [kg]

r = distance from the center of the earth to the man [m]

The acceleration he is experimenting is the same acceleration given by the gravity, therefore:

a = g = 9.81[m/s^2]

To find the tangential velocity, we must determinate the force exerted by the earth.

Now we will find the force exerted by the gravity when the man is orbiting the earth at distance r.

$G = 6.67*10^{-11} [\frac{N*m^{2} }{kg^{2} } ]\\M=5.98*10^{24}[kg]\\ m=100 [kg]\\Replacing:\\F = G*\frac{M*m}{r^{2} }$

$F = 6.67*10^{-11}*\frac{5.98*10^{24}*100 }{1142635^{2} } \\ F= 30550 [N]$

And this force will be equal to the following expression:

$F = m*\frac{v^{2} }{r} \\where:\\v= tangential velocity [m/s]\\v=\sqrt{\frac{F*r}{m} } \\v=\sqrt{\frac{30550*1142635}{100} } \\v=18683.5[m/s]$

8 0

3 years ago