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serious [3.7K]
3 years ago
6

The cells lie odjacent to the sieve tubes​

Physics
1 answer:
nignag [31]3 years ago
4 0

Answer:

<h2><em>Almost</em><em> </em><em>always</em><em> </em><em>adjacent</em><em> </em><em>to</em><em> </em><em>nucleus</em><em> </em><em>containing</em><em> </em><em>companion</em><em> </em><em>cells</em><em>,</em><em> </em><em>which</em><em> </em><em>have</em><em> </em><em>been</em><em> </em><em>produced</em><em> </em><em>as</em><em> </em><em>sister</em><em> </em><em>cells</em><em> </em><em>with</em><em> </em><em>the</em><em> </em><em>sieve</em><em> </em><em>elements</em><em> </em><em>from</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>mother</em><em> </em><em>cell</em><em>.</em><em> </em></h2>
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Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0
2 years ago
[]Answer the question below[]
marysya [2.9K]
Answer:

The answer is D. density.
8 0
2 years ago
Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay
kaheart [24]

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

mu=(m+M)v

where

m = 0.200 kg is the mass of the koala bear

u = 0.750 m/s is the initial velocity of the koala bear

M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

Solving the equation for v, we get

v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s

8 0
3 years ago
Find equivalent resistance. <br><br>Answer asap and please, please don't spam.​
Alinara [238K]

Answer:

R = 4.77 ohms

Explanation:

Four resistors are given such that,

R₁ = 2 ohms

R₂ = 3 ohms

R₃ = 5 ohms

R₄ = 10 ohms

Here, R₁ and R₂ in series. The equivalent is given by :

R₁₂ = R₁ + R₂

= 2 + 5

R₁₂ = 7 ohms

Similarly, R₃ and R₄ are in series. so,

R₃₄ = R₃ + R₄

= 10+5

R₃₄ = 15 ohms

Now, R₁₂ and R₃₄ are in parallel. So,

\dfrac{1}{R}=\dfrac{1}{R_{12}}+\dfrac{1}{R_{34}}\\\\\dfrac{1}{R}=\dfrac{1}{7}+\dfrac{1}{15}\\\\R=4.77\ \Omega

So, the equivalent resistance s 4.77 ohms.

4 0
3 years ago
Find the noise level of a sound having an intensity of 1.5x10^-14W/m^2 given I0=10^12 W/m2
Llana [10]

Answer:

Noise level will be -18.2 watt

So option (b) will be correct answer

Explanation:

We have given sound intensity I=1.5\times 10^{-14}w/m^2

And threshold intensity I_0=\times 10^{-12}w/m^2 ( in question it is given as 10^{12}w/m^2 but its standard value is 10^{-12}w/m^2 )

Now noise level  =10log\frac{I}{I_0}=10log\frac{1.5\times 10^{-14}}{10^{-12}}=10log0.015-18.23

So the noise level will be -18.2

So option (b) will be correct answer

5 0
3 years ago
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