1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lutik1710 [3]
3 years ago
11

Polarizers #1 and #3 are "crossed" such that their transmission axes are perpendicular to each other. Polarizer #2 is placed bet

ween the polarizers #1 and #3 with its transmission axis at 60◦ with respect to the transmission axis of the polarizer #1 (see the sketch). #1 #2 #3 60◦ After passing through polarizer #2 the intensity I2 (in terms of the intermediate intensity I1) is
Physics
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

Explanation:

The angle between the polariser 1 and 2 is 60°

If I₁ be the intensity of light after polariser 1 , its intensity after polariser 2 that is I₂ can be expressed as follows

I₂ = I₁ cos² 60

= I₁ x 1/4

=  I₁ x \frac{1}{4}

You might be interested in
A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with ne
gregori [183]

Answer:

The ball experiences the greater momentum change

Explanation:

The momentum change of each object is given by:

\Delta p = m \Delta v= m (v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

Both objects have same mass m and same initial velocity u. So we have:

- For the ball, the final velocity is

v=-u

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

\Delta p=m(v-u)=m((-u)-u)=-2mu

- For the clay, the final velocity is

v=0

since it sticks to the wall. So, the change in momentum is

\Delta p = m(v-u)=m(0-u)=-mu

So we see that the greater momentum change (in magnitude) is experienced by the ball.

3 0
3 years ago
Two blocks with rough surfaces are stacked on top of a slippery horizontal surface. The top block has a mass of m, and the botto
lapo4ka [179]

Answer:

Please refer to the attached schematic for free-body diagrams of the blocks.

The contact forces N1 = mg, N2 = 3mg.

The friction force fs = mg/2.

The tension force T = 3mg/2.

The distance y = (gt^2)/4

Explanation:

Starting with the third block which is declining a time t. Newton's Second Law implies that

F = (3m)a\\3mg - T = 3ma\\T = 3mg - 3ma

There are two unknowns in one equation: T and a. Therefore, we need to find more equations to solve these unknown variables.

Let's look at free-body diagram 1:

Newton's Second Law:

F = ma\\f_s = ma

That is another equation with one more unknown: fs.

Free-body diagram 2:

F = (2m)a\\T - f_s = 2ma

That is our third equation. We now have three equations with three unknowns. Combining equation 2 and 3 gives:

T - ma = 2ma\\T = 3ma

Plugging this equation into the first equation gives

T = 3mg - 3ma\\3ma = 3mg - 3ma\\3mg = 6ma\\a = g/2

Following this, we can find the other unknowns:

T = 3ma = 3m(g/2) = \frac{3mg}{2}

f_s = ma = mg/2

The contact forces are N1 and N2.

N_1 = mg\\N_2 = 2mg + N1 = 3mg

Finally, using the acceleration and equation of kinematics, we can find the distance the hanging weight drops in a time t.

y - y_0 = v_0t + \frac{1}{2}at^2\\y = \frac{1}{2}(\frac{g}{2})t^2 = \frac{gt^2}{4}

6 0
4 years ago
A cat chase a mouse across a 1.0 m high table. The mouse steps out of the way, and the car slides off the table and strikes the
Tema [17]

The cat fell 1.0 m from the ground.

Using the formula

h = v_{oy}t - \frac{gt^2}{2}

Here, v_{oy} = 0, \; h = 1.0 \; m.

Solving for t, the time it spent in the air is

t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.0 \; m)}{9.8 \; m/s^2}} = 0.451753951 \; s

The cat does not accelerate along the horizontal, so it has constant horizontal velocity. Since it strikes the floor 2.2 m from the table, then

v_x = \Delta x/t = \frac{2.2\; m}{0.451753951 s} = 4.869907597 \; m/s \Rightarrow 4.9 \; m/s


5 0
3 years ago
What coefficients would balance the following equation? __C2H6 + __O2 → __CO2 + __H2O A. 1C2H6 + 5O2 → 2CO2 + 3H2O B. 2C2H6 + 5O
sladkih [1.3K]
It's C, with the 2/7/4/6 in front of each reactant and product.
4 0
4 years ago
Planet A takes 1 year to go around its star at an average of 1 A.U. distance. Planet B is 4 A.U. from the star. Calculate how lo
pentagon [3]

Planet Geos in orbit a distance of 1 A.U. (astronomical unit) from the star Astra has an orbital period of 1 "year." If planet Logos is 4 A.U. from Astra, how long does Logos require for a complete orbit?

TB = <span>8</span> years

7 0
4 years ago
Read 2 more answers
Other questions:
  • An object takes 5s to reach the ground from a hight of 5m on planet.what is the value of g on planet?
    15·1 answer
  • A filled water tower sits on the top of the highest hill near a city. The cylindrical tower has a height of 55.0 m
    10·1 answer
  • A cylindrical cork (area A, length L, density pc) is floating in a liquid of density Pw with only a part of its length L submerg
    12·1 answer
  • It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of
    8·2 answers
  • If element X has 39 protons, how many electrons does it have?
    11·1 answer
  • Someone please help me
    5·1 answer
  • If you exert a force of 10.0 N to lift a box a distance of 0.75 m, how much work do you do?
    14·1 answer
  • 11 Design Imagine that a scientistdiscovered a way to make africtionless surface. What wouldbe some useful applications forthis
    8·1 answer
  • 2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
    9·1 answer
  • *please refer to photo attached* The figure below shows a small, charged sphere, with a charge of q = +44.0 nC, that moves a dis
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!