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erik [133]
3 years ago
9

Using poiseulli assumption derive an expression for the rate of flow of a liquid through a horizontal tube in terms of a, l, p a

nd n
Physics
1 answer:
Marizza181 [45]3 years ago
7 0

Answer:

im sorry im just trying this app please dont report

Explanation:

please im begging you maam/sir

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What types of crust are colliding between the North American Plate and the Pacific Plate?
svet-max [94.6K]

Answer: G00gle got you bro

Explanation:

Yea

4 0
3 years ago
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
In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.
Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

Q_H

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

Q_C>0

That is why the answer will be

Q_H ,Q_C>0

8 0
3 years ago
2. A car accelerates uniformly from +10.0 m/s to +50.0 m/s over a distance of 225 m. How long did it take to go that distance? S
artcher [175]
Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.

From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.

Hence P(t)=a*t^2/2+10t

Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T

Hence T=225/30=7.5

It took 7.5 seconds


7 0
2 years ago
For this trajectory, what would the vertical component of acceleration for the module be at time tm=t0−σ=325s? Recall that accel
Karo-lina-s [1.5K]
I think its half of pie so you divid that by 3.14 .
4 0
3 years ago
Read 2 more answers
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