A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum gage pressure in the tank. Mark that point at the interior bottom of the tank. Draw the free surface at this acceleration.

Answer:

T = 438.87 N.m

Explanation:

The power required to raise the 4961 N load in 10 meters for 2 minutes is:

[tex]P = \dfrac{4961*10}{2*60}\\ \\ P = 413.42 Nm/sec[/tex]

P = Torque  × W

[tex]413.42 = T \times \dfrac{2 * \pi*9}{60}[/tex]

[tex]413.42 = T \times0.942 \\ \\ T = \dfrac{413.42}{0.942}[/tex]

[tex]T = \dfrac{413.42}{0.942}[/tex]

T = 438.87 N.m

Answer:

shaft diameter = [tex]\sqrt[3]{0.3512}[/tex] mm = 0.7055 mm

Explanation:

Ratio of inside diameter to outside diameter ( i.e. d/D )= 0.6

Shear stress of material ( Z ) ≤ 500 KPa

power transmitted by shaft ( P ) = 1.5MW of mechanical power

Revolution ( N ) = 1500 rev/min

Calculate shaft Diameter

Given that: P = [tex]\frac{2\pi NT}{60}[/tex]  ---- 1

therefore; T = ( 1.5 *10^3 * 60 ) / ( 2[tex]\pi[/tex] * 1500 )  = 9.554 KN-M

next

[tex]\frac{T}{I_{p} } = \frac{Z}{R}[/tex]

hence ; T = Z[tex]_{p} *Z[/tex]

attached below is the remaining part of the solution

Answer:

Explanation:

From the information given:

[tex]E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN[/tex]

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

[tex]P = P_1+P_2 --- (1)[/tex]

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

[tex]\delta_1=\delta_2[/tex]

[tex]\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}[/tex]

[tex]\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}[/tex]

[tex]P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})[/tex]

[tex]P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}[/tex]

[tex]P_2 = 6.6212 \ P_1[/tex]

Replace [tex]P_2[/tex] into equation (1)

[tex]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[/tex]

Finally, to determine the normal stress in aluminum rod:

[tex]\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}[/tex]

[tex]\sigma_1 = - 23.523 \ MPa}[/tex]

Thus, the normal stress = 23.523 MPa in compression.

Answer:

Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.

Explanation:

Hi sorry I need points I'm New

Answer: My favorite color is Red favorite letter C favorite fruit watermelon.

Where are your answer options?

Answer:

Implement a hazard communication program

Explanation: i took the quiz

Answer:

Gas turbines in Helicopters require lesser space.

Explanation:

[1] In terms of Space Requirements:

The gas used in helicopters requires lesser space as compared to Automotive gas turbines. The gas in automobile have higher thermal efficiency.

[2]. In terms of Environmental impact:

The occurrence of environmental solution is very slim when  used in helicopters' engines.

[3]. In terms of power-to-weight ratio:

The vibrations in engines of helicopters make it to have lesser efficiency as compared to automobile.

[4]. In terms of Fuel availability:

Fuel is available. Automobile can make use of gas as fuel.

Answer:

Attached below is the derived next state equations

Explanation:

Attached below is the derived next state equations

used for the solution of the given problem.

Answer:

a. 2.30

b. decreases with increasing velocity.

c. 0.179 kg/s.

Explanation:

Without mincing let's dive straight into the solution to the question above.

                                                         [a].

The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:

The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.

While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.

Array of slot noozle = [10 × (2  × 0.004)]/ 20.92  × 10^-6] = 3824.

where A = 4/56 =0.714.

And Ar = [ 60 + 4 (40/2  × 4) - 2 ]^2 ]-1/2 = 0.1021.

N = 2/3 (0.1021)^3/4 [ 2  ×  3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.

h = 24.3  ×  0.030/0.004 = 91.1 W/m^2k.

Therefore; 659.6  × 0.030/0.5 = 39.0 W/m²k.

The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.

The slot noozle = 91.1  ×  0.5  ×  0.5 [ 140 -15] = 2846.87W.

The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.

                                                     [b].

2.3 [ (2^2/3)/ 2^4/5] = 2.1

Thus, it decreases with increasing velocity

                                                      [c].

The  air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.

Answer:

Explanation:

This is Answer....

Explanation:

Nested piezometers indicate an upward flow if the elevation of the top of the water in the piezometer tube that penetrates the aquifer to the deeper point is greater than the elevation of the water in the shallower tube

def digit_sum(number):

 sumOfDigits = 0

 while number > 0:

     sumOfDigits += number % 10

     number = number // 10

 return sumOfDigits

or you can use,

In [2]: digit_sum(10)

Out[2]: 1

In [3]: digit_sum(434)

Out[3]: 11

or you can use,

digits = "12345678901234567890"

digit_sum = sum(map(int, digits))

print("The equation is: ", '+'.join(digits))

print("The sum is:", digit_sum)

Have a great day <3

Answer:

it is indeed C

Explanation:

Answer:

c

Explanation: