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Elan Coil [88]
3 years ago
13

How to calculate critical angle?? Given 2 indexes of refraction??

Physics
1 answer:
never [62]3 years ago
4 0
Given refrelactive index n,
1/n = sinc
c = sin inverse (1/n)
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What mechanical layer lies below the lithosphere
joja [24]
The <span>asthenosphere is under the lithosphere.</span>
3 0
3 years ago
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Newton’s third law of motion explains the two forces namely ‘action’ and ‘reaction’ coming into action when the two bodies are i
SIZIF [17.4K]

(b) Always act on the different bodies in opposite directions

7 0
3 years ago
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A car drives past a pole at 40km/hr. Describe the motion from the point of view of a) the car, and b) the pole. Thanks in advanc
ki77a [65]
I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.

To put a finer point on it, let's give the car a direction.  Say it's driving North.

a).  From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .

b).  From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.

c).  A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.

The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.

Now follow me here . . .

The car and the pole are both seen to be moving south.  BUT ... Since the
pole is moving south faster than the car is, it easily overtakes the car, and
passes it . . . going south.

That's what everybody on the train sees.

==============================================

Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:

You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ? 

Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself !  Only of others.

The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else.  And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.

And now I believe that I have adequately milked this one for 50 points worth.


7 0
3 years ago
Need help
Bad White [126]

Answer:

12.5 km

Explanation:

Since 30.0 minutes is half of an hour and the average speed is 25.0 km/h we can multiply the time by the speed to find the distance travelled.

25.0km/h x 0.5 hours = 12.5 km

4 0
3 years ago
1. A step-up transformer increases 15.7V to 110V. What is the current in the secondary as compared to the primary? Assume 100 pe
Serjik [45]

1. I_2 = 0.14 I_1

Explanation:

We have:

V_1 = 15.7 V voltage in the primary coil

V_2 = 110 V voltage in the secondary coil

The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal

P_1 = P_2\\V_1 I_1 = V_2 I_2

where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find

\frac{I_2}{I_1}=\frac{V_1}{V_2}=\frac{15.7 V}{110 V}=0.14

which means that the current in the secondary coil is 14% of the value of the current in the primary coil.

2. 5.7 V

We can solve the problem by using the transformer equation:

\frac{N_p}{N_s}=\frac{V_p}{V_s}

where:

Np = 400 is the number of turns in the primary coil

Ns = 19 is the number of turns in the secondary coil

Vp = 120 V is the voltage in the primary coil

Vs = ? is the voltage in the secondary coil

Re-arranging the formula and substituting the numbers, we find:

V_s = V_p \frac{N_s}{N_p}=(120 V)\frac{19}{400}=5.7 V

5 0
3 years ago
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