The stomach lining is made up of deep muscular grooves.How might these structures help the stomach to break down food?

Three polarizing filters are stacked, with the polarizing axis ofthe second and third filters at angles of 22.2^\circ and 68.0^\

andreev551 [17]

Answer:

I₂ = 25.4 W

Explanation:

Polarization problems can be solved with the malus law

I = I₀ cos² θ

Let's apply this formula to find the intendant intensity (Gone)

Second and third polarizer, at an angle between them is

θ₂ = 68.0-22.2 = 45.8º

I = I₂ cos² θ₂

I₂ = I / cos₂ θ₂

I₂ = 75.5 / cos² 45.8

I₂ = 155.3 W

We repeat for First and second polarizer

I₂ = I₁ cos² θ₁

I₁ = I₂ / cos² θ₁

I₁ = 155.3 / cos² 22.2

I₁ = 181.2 W

Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized

I₁ = I₀ / 2

I₀ = 2 I₁

I₀ = 2 181.2

I₀ = 362.4 W

Now we remove the second polarizer the intensity that reaches the third polarizer is

I₁ = 181.2 W

The intensity at the exit is

I₂ = I₁ cos² θ₂

I₂ = 181.2 cos² 68.0

I₂ = 25.4 W

8 0

4 years ago

Make a VIR chart. All voltage, resistance, and current.

KiRa [710]

$\fbox{Hope This Helps You .}$

5 0

3 years ago

Which object is not accelerating?

AleksAgata [21]

I would say d because the student isn’t moving, the Ferris wheel is.

8 0

3 years ago

Read 2 more answers

How much work must be done in the car to slow it from 100km/h to 50km/h

Brilliant_brown [7]

Answer:

$-96.465m$ joule

Explanation:

Let m = mass of the car and v1 = initial velocity and v2 = final velocity

Given.

Initial velocity = 100 km/h

final velocity = 50 km/h

What is work done in the car to slow it from 100km/h to 50km/h?

$v1=100km/h=27.78m/s$

$v2=50km/h=13.89m/s$

The work done in the car to slow it from v1 to v2.

w=Δk

$w=k2-k1$

$w= \frac{1}{2}m(v2)^{2}- \frac{1}{2}m(v1)^{2}$

$w=\frac{1}{2} m(v2-v1)^{2}$

$w=\frac{1}{2}\times m(13.89 -27.78)^{2}$

$w=\frac{1}{2}\times m(-13.89)^{2}$

$w=\frac{1}{2}\times m\times (-192.93)$

$w=-96.465m$ joule.

Therefore, the work done is $-96.465m$ joule

7 0

3 years ago

After three half-lives, one-ninth of an original radioactive parent isotope remains, and eight-ninths has decayed into the daugh

BartSMP [9]

Answer:

B. False

The true statement will be:

After three half-lives, one-eighth of an original radioactive parent isotope remains, and seven-eighth has decayed into daughter isotopes.

Explanation:

Half life means the time after which the amount of the given substance becomes half of its initial amount, by radiation.

Thus after each successive half life the material is halved or half of it is decayed, which leads to a general formula:

Remaining Amount = 1/2^n of parent isotope

where, n is the number of half lives passed. So, for three half-lives:

Remaining Amount = 1/2^3 of parent isotope

Remaining Amount = 1/8 of parent isotope

Thus, the decayed amount will be:

Decayed Amount = (1-1/8) of parent isotope

Decayed Amount = 7/8 of parent isotope

Thus, the given statement is B.False

8 0

3 years ago