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3 years ago

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Light traveling through air at 3.00 · 10^8 m/s reaches an unknown medium and slows down to 2.00 · 10^8 m/s. What is the index of

refraction of that medium? n =

2 answers:

nataly862011 [7]3 years ago

5 0

V 1= 3.00 · 10^8 m/s
v 2 = 2.00 · 10^8 m/s
The index of refraction:
n = v 1 / v 2 = 3.00 · 10^8 m/s : 2.00 · 10^8 m/s = 1.5
Answer:
The index of refraction of that medium is 1.5

lesya692 [45]3 years ago

4 0

Answer

The index of refraction of medium is 1.5 .

Explanation:

Equation for index of refraction is given by .

$n = \frac{c}{v}$

Where n is index of refraction in medium , v is speed of light in the medium and c is the speed of light in air .

As given

Light traveling through air at $3.00\times 10^8$ m/s reaches an unknown medium and slows down to $3.00\times 10^8$ m/s.

$c = 3.00\times 10^{8}$

$v = 2.00\times 10^{8}$

Putting the values in the formula

$n = \frac{3.00\times 10^{8}}{2.00\times 10^{8}}$

Solving the above

n = 1.5

Therefore the index of refraction of medium is 1.5 .

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A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c

puteri [66]

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

$B =\mu_o I(\frac{N}{L} )\\\\B = \mu_o I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m$

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

3 0

3 years ago

At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh

Anna35 [415]

We have the equation for electric field E = kQ/ $d^{2}$

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to $d^{2}$

So, $\frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}$

$E_{1}$ = 485 N/C

$d_{1}$ = 0.208 cm

$d_{2}$ = 0.620 cm

$E_{2}$ = ?

$\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}$

$E_{2}$ = $\frac{485*0.208^{2}}{0.628^{2}}$

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3 years ago

Read 2 more answers

A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. How long does it take to come to rest? How many

Darya [45]

Answer:

$t=45.7s$

$\alpha =116revolutions$

Explanation:

Since we have given values of ω₀=32.o rad/s ,ω=0 and α=-0.700 rad/s² to find t we use below equation

$w=w_{o}+at\\ 0=(32.0rad/s)+(-0.700rad/s^{2} )t\\t=\frac{-32.0}{-0.700} \\t=45.7s$

To find revolutions we use below equation

$w^{2}=w_{o}^{2}+2a\alpha$

Substitute the given values to find revolutions α

So

$0=(32.0rad/s)^{2}+2(-0.700rad/s^{2} )\alpha \\\alpha =\frac{(-32.0rad/s)^{2}}{2(-0.700rad/s^{2} )} \\\alpha =731rad$

To convert rad to rev:

$\alpha =(731rad)*(\frac{1rev}{2\pi rad} )\\\alpha =116revolutions$

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4 years ago

What will be the value of both charges if they are 5 cm apart and suffer a

Mrac [35]

Answer:

$\boxed{q = 1.2 \times {10}^{ - 6} C}$

Explanation:

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3 years ago

What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29.5 km/h and the μs

Yakvenalex [24]

Answer:

Radius=15.773 m

Explanation:

Given data

v=29.5 km/h=8.2 m/s

μs=0.435

To find

Radius R

Solution

The acceleration is a centripetal acceleration which is experienced by the bicycle given by

$a=v^{2}/R$

This acceleration is only due to static force which given as

$f=ma\\f=m(v^{2}/R )$

The maximum value of the static force is given as

$Fs_{max}=u_{s}F_{N}$

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FN is normal force equal to mass*gravity

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Therefore the minimum radius should be found by the bicycle move without sliding

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4 years ago