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3 years ago

Please help me out !!!!!!

Physics

1 answer:

nikdorinn [45]3 years ago

8 0

Answer:

Explanation:

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A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac

vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

$n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence

${\theta_r}$ is the angle of refraction = 90°

${n_2}$ is the refractive index of the refraction medium

${n_1}$ is the refractive index of the incidence medium

Thus,

$n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}$

The formula for the calculation of critical angle is:

${sin\theta_{critical}}=\frac {n_2}{n_1}$

Where,

${\theta_{critical}}$ is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

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3 years ago

express each of the following aa indicated a. 2 dm expressed in millimeter b. 2h 10min expessed in sec c. 16g expressed in centi

Alex787 [66]

I thinks or know it is A or B cuz magnets and metal attract

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Vladimir [108]

Answer:

Explanation:

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Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. S

dedylja [7]

The concept that we need here to give a proper solution is mutual inductance.

The mutual inductance is given by the expression

$M=\frac{N\Phi}{I}$

Where,

I = current

N = Number of turns

$\Phi =$ Flux through the solenoid.

Part A) Then we have in our values that,

$I=6.6A$

$\Phi= 3.50*10^{-2}Wb$

$N=450$

Replacing in the equation,

$M = \frac{450*350*10^{-2}}{6.60}$

$M = 2.39H$

Part B) Here is required the Flux, then using the same expression we have that

$\Phi = \frac{IM'}{N}$

We conserve the same value for the Inductance but now we have a current of 2.6, then

$\Phi = \frac{2.6*2.39}{690}$

$\Phi = 9*10^{-3}Wb$

Therefore the flux in Solenoid 1 is $9*10^{-3}Wb$

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