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"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a

My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0

4 years ago

A flat, rigid object oscillates as a physical pendulum in simple harmonic motion with a frequency f. The mass of the pendulum is

lyudmila [28]

Answer:

I = mgd/4π²f²

Explanation:

The period of the physical pendulum T = 1/f where f = frequency,

T = 1/f = 2π√(I/mgd) where I = moment of inertia of the physical pendulum, m =mass of pendulum, g = acceleration due to gravity and d = distance of center of mass of pendulum from pivot point.

1/f = 2π√(I/mgd)

dividing both sides by 2π, we have

1/2πf = √(I/mgd)

squaring both sides, we have

(1/2πf) = [√(I/mgd)]²

1/4π²f² = I/mgd

multiplying both sides by mgd, we have

I = mgd/4π²f²

7 0

3 years ago

At what point in the swing of the pendulum is the potential energy completely converted into Kinetic energy

artcher [175]

Point a because point a is the highest at potential energy converting into the highest kinetic energy.

4 0

3 years ago

Read 2 more answers

A car slams on its brakes, coming to a complete stop in 4.0 s. The car was traveling south at 60.0 mph. Calculate the accelerati

Cloud [144]

Answer:

Acceleration, $a=-6.705\ m/s^2$

Explanation:

It is given that,

Final velocity of the car, v = 0

Initial velocity of the car, u = 60 mph = 26.82 m/s

It comes to rest in 4 s

We need to find its acceleration. The rate of change of velocity is called its acceleration. Mathematically, it is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{0-26.82\ m/s}{4\ s}$

$a=-6.705\ m/s^2$

So, the car is decelerating with a magnitude of 6.705 m/s^2. hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

If we push one cart (Cart A) into a second stationary cart (Cart B), v_Ai is nonzero and v_Bi=0, respectively. Assume no outside

Darya [45]

Answer:

$v_A=(\frac{m_A-m_B}{m_A+m_B})u_A$

$v_B=(\frac{2m_A}{m_A+m_B})u_A$

Explanation:

In an elastic collision, the total momentum of the system and the total kinetic energy of the system are conserved.

So:

- Conservation of momentum:

$m_A u_A+m_B u_B = m_A v_A + m_B v_B$ (1)

where

$m_A, m_B$ are the masses of cart A and B

$u_A,u_B$ are the initial velocities of cart A and B

$v_A,v_B$ are the final velocities of cart A and B

- Conservation of kinetic energy:

$\frac{1}{2}m_A u_A^2+\frac{1}{2}m_B u_B^2= \frac{1}{2}m_A v_A^2 +\frac{1}{2} m_B v_B^2$ (2)

From (1) we can write:

$m_A (u_A-v_A)=m_B(v_B-u_B)\\\frac{m_A (u_A-v_A)}{m_B(v_B-u_B)}=1$ (3)

From (2) we get:

$m_A (u_A^2-v_A^2)=m_B(v_B^2-u_B^2)\\m_A(u_A-v_A)(u_A+v_A)=m_B(v_B-u_B)(v_B+u_B)\\\frac{m_A(u_A-v_A)}{m_B(v_B-u_B)}(u_A+v_A)=(v_B+u_B)$ (4)

Substituting (3) into (4),

$1\cdot (u_A+v_A)=(v_B+u_B)\\v_B=u_A+v_A-u_B$ (5)

And substituting (5) into (1),

$m_A(u_A-v_A)=m_B (u_A+v_A-2u_B)$

And now we can solve to find the final expressions:

$v_A=(\frac{m_A-m_B}{m_A+m_B})u_A+(\frac{2m_B}{m_A+m_B})u_B$

$v_B=(\frac{2m_A}{m_A+m_B})u_A+(\frac{m_B-m_A}{m_A+m_B})u_B$

But in this problem

$u_B=0$

So the equations can be rewritten as

$v_A=(\frac{m_A-m_B}{m_A+m_B})u_A$

$v_B=(\frac{2m_A}{m_A+m_B})u_A$

4 0

3 years ago