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Alex_Xolod [135]

3 years ago

9

For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the

length is Δp1. If the length of the pipe is then doubled, what is the relation of the new pressure drop Δp2 to the original pressure drop Δp1 at the original mass flow rate?
delta P2=?

1 answer:

Goshia [24]3 years ago

8 0

Answer:

$\Delta p_{2} = 2\cdot \Delta p_{1}$

Explanation:

The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.

$\Delta p_{2} = 2\cdot \Delta p_{1}$

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A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N

soldier1979 [14.2K]

Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, $P_{t} = 8 kW = 8000 W$

Now, the force, F is given by:

$P_{t} = Force(F)\times velocity(v) = Fv$ (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, $F_{c}$ is provided by the gravitational force, $F_{G}$ :

$F_{c} = F_{G}$

$\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}$

Thus from the above, velocity comes out to be:

$v = \sqrt{\frac{GM_{e}}{R}}$

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s$

where

R = $R_{e} + H$

R = $\sqrt{GM_{e}(\frac{T}{2\pi})^{2}}$

where

G = Gravitational constant

T = Time period of rotation of Earth

R is calculated as 42166 km

Now, from eqn (1):

$8000 = F\times 3075.36$

F = 2.60 N

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3 years ago

A batch of 1000 is split into 10 smaller batches of equal size 100. The processing time of each unit is 2

Vika [28.1K]

The lead time of the actual batch will be in

2950 in minutes

<h3>What is Processing Time?</h3>

This refers to the amount of time which is taken for a processor to run a procedure and return a result.

We can see that a batch of 1000 is split so that they each have 10 smaller batches which has an equal size of 100 each, then if the processing time is 2 mins per machine and the set up time is 30 mins.

Hence, when this batch is processed over a serial line of 5 machines, then the lead time of the actual batch would be 2950 in minutes

Read more about processing time here:

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4 0

1 year ago

The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod

Vsevolod [243]

Answer:

Explanation:

From the information given:

$E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN$

The total load is distributed across both the rod and tube:

$P = P_1+P_2 --- (1)$

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

$\delta_1=\delta_2$

$\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}$

$\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}$

$P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})$

$P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}$

$P_2 = 6.6212 \ P_1$

Replace $P_2$ into equation (1)

$P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN$

Finally, to determine the normal stress in aluminum rod:

$\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}$

$\sigma_1 = - 23.523 \ MPa}$

Thus, the normal stress = 23.523 MPa in compression.

8 0

3 years ago

A parallel plates capacitor is filled with a dielectric of relative permittivity ε = 12 and a conductivity σ = 10^-10 S/m. The c

monitta

Answer:

t = 1.06 sec

Explanation:

Once disconnected from the battery, the capacitor discharges through the internal resistance of the dielectric, which can be expressed as follows:

R = (1/σ)*d/A, where d is is the separation between plates, and A is the area of one of the plates.

The capacitance C , for a parallel plates capacitor filled with a dielectric of a relative permittivity ε, can be expressed in this way:

C = ε₀*ε*A/d = 8.85*10⁻¹² *12*A/d

The voltage in the capacitor (which is proportional to the residual charge as it discharges through the resistance of the dielectric) follows an exponential decay, as follows:

V = V₀*e(-t/RC)

The product RC (which is called the time constant of the circuit) can be calculated as follows:

R*C = (1/10⁻¹⁰)*d/A*8.85*10⁻¹² *12*A/d

Simplifying common terms, we finally have:

R*C = 8.85*10⁻¹² *12 / (1/10⁻¹⁰) sec = 1.06 sec

If we want to know the time at which the voltage will decay to 3.67 V, we can write the following expression:

V= V₀*e(-t/RC) ⇒ e(-t/RC) = 3.67/10 ⇒ -t/RC = ln(3.67/10)= -1

⇒ t = RC = 1.06 sec.

3 0

3 years ago

Write a C program that asks the user to enter two numbers, obtains the two numbers from the user and prints the sum, product, di

Bogdan [553]

Answer:

View Image

Explanation:

Initialize your variable as a float or double since you're going to be using fractions in your answer.

User scanf() to get user input.

Print out the sum, product, quotient, and difference between the two numbers.

8 0

3 years ago