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loris [4]
3 years ago
10

When a ray of light is incident normally on an air glass interface what is its angle of refraction​

Physics
1 answer:
DiKsa [7]3 years ago
3 0

Answer:

When a ray of light falls on a glass slab normally, the angle between normal and incident ray is zero.

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A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr
aleksandr82 [10.1K]

Answer:

Explanation:

We shall apply Doppler's effect to solve the problem .

Formula for apparent frequency for a source of sound approaching an observer is as follows .

f₁ = f₀ V / (V - v )

where f₁ and f₀ are apparent and real frequency of source , V and v is velocity of sound and velocity of approaching source respectively .

Putting the given values and knowing that speed of sound is 340 m /s

f₁ =346x 340 / (340 - 39.6 )

f₁ = 391.6 Hz

In case of receding train , the formula is

f₂ = f₀ V / (V + v )

Putting the values

f₂ = 346x 340 / (340 + 39.6 )

= 309.9 Hz

Change in frequency =  391.6 - 309.9

= 81.7 Hz .

4 0
3 years ago
Suppose you apply a force of 75 N to a 25-kg object. What will the acceleration of the object b? (Remember a=F/m)
Vlada [557]

I would choose the option B.

F = ma

a = 75 / 25 = 3 m/s^2

8 0
2 years ago
Write short letters.
Svetradugi [14.3K]

Answer:

No. 67

Peter Street

12th Road

Chennai

24th June 201_

Dear Amrish

I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.

It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?

I am anxiously waiting for your reply.

Yours affectionately

your name

4 0
3 years ago
Read 2 more answers
A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh
n200080 [17]

Answer:

Part a)

T_L = 155.4 N

Part b)

T_R = 379 N

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

T_L cos65 = T_R cos80

T_L = 0.41 T_R

now we will have force balance in Y direction

mg = T_L sin65 + T_Rsin80

514 = 0.906T_L + 0.985T_R

Part a)

so from above equations we have

514 = 0.906T_L + 0.985(\frac{T_L}{0.41})

514 = 3.3 T_L

T_L = 155.4 N

Part b)

Now for tension in right string we will have

T_R = \frac{T_L}{0.41}

T_R = 379 N

3 0
3 years ago
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro
pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0
3 years ago
Read 2 more answers
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