A power winch is designed to raise a 4,961 N load 10 meter in 2 minutes. The winch is designed with a 283 mm diameter drum taking up the wire lifting the load. The drum will be connected to a 9 rpm motor through a gearbox. What is the minimum torque (Nm) that the motor shaft coupling should be designed to transmit

Answer:

it is indeed C

Explanation:

Answer:

c

Explanation:

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:

t2 = 256 hours

Explanation:

Given data:

Carbon concentration ( C ) at 1.2mm from surface, C = 0.38 wt%

Duration( t ) of heat treatment for 0.38wt% at 1.2mm =  9-hr

Estimate the time (in h) necessary to achieve the same concentration at a 6.4 mm

Assuming same concentration of 0.38wt%  we will apply Fick's second law for constant surface concentration

attached below is the remaining part of the solution

x1 = 1.2 mm

x2 = 6.4 mm

t1 = 9-hr

t2 = ?

t2 = [tex](\frac{6.4}{1.2} )^{2} * 9[/tex]  = 256 hours

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.

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:

3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O

Explanation:

It is all about unit conversions.

The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.

160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm

By the way, now we can determine the volume of MOS, in order to work with density.

1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³

But density is mg/m³, so we convert mm³ to m³

1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³

Now, we apply density to determine the mass of MOS

Density = mass /volume → Density . volume = mass

1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg

To make more easier the calculate, we convert mg to g.

3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g

To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol

We need to know moles of Si and O₂ in the MOS

Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles

1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:

5.86×10⁻¹⁸ mol of SiO₂ may have:

(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si

(5.86×10⁻¹⁸ .2) /1 =  1.17×10⁻¹⁷ moles of O₂

Let's count the atoms (1 mol of anything contain NA particles)

5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si

1.17×10⁻¹⁷ mol of O₂  . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O

The number of Si and O atoms present per mm² of the oxide layer are respectively; 3.52 × 10⁶ atoms of Si and 7.05×10⁶ atoms of O

We are given;

Area of MOS device; A = 1 mm²

Thickness of layer; t = 160 nm = 1.6 × 10⁻⁴ mm

Formula for volume is;

V = Area * thickness

V = 1.6 × 10⁻⁴ mm × 1 mm² = 1.6 × 10⁻⁴ mm³

Converting volume to m³ gives;

V = 1.6 × 10⁻¹³ m³

Now, to get the mass of MOS, we will use the formula;

Mass = Density * volume

we are given density = 2.20mg/m³

Thus;

Mass = 1.6 × 10⁻¹³ m³ × 2.20mg/m³

Mass = 3.52 × 10⁻¹³ mg = 3.52×10⁻¹⁶ g

From periodic table, the molar mass of SiO₂ = 60.08 g/mol

Then;

Number of moles of MOS = (3.52 × 10⁻¹⁶ g)/60.08 g/mol

Number of moles of MOS = 5.86 × 10⁻¹⁸ moles

Now, 1 mol of SiO₂  is formed from 1 mol of Si and 2 mol of O. Thus;

5.86×10⁻¹⁸ mol of SiO₂ will have;

5.86×10⁻¹⁸ moles of Si

and (5.86×10⁻¹⁸ .2) =  11.72 × 10⁻¹⁸ moles of O₂

From Avogadro's number that 1 mol equals 6.02×10²³ atoms , we can say that;

5.86 × 10⁻¹⁸ mol of Si × 6.02 × 10²³ atoms/mol = 3.52 × 10⁶ atoms of Si

11.72 × 10⁻¹⁸ mol of O₂ × 6.02 × 10²³ atoms/mol = 7.05×10⁶ atoms of O

Read more about number of atoms at; https://brainly.com/question/3157958

Answer:

The pavement of the carpool lanes is marked with the diamond symbol. These carpool lanes are reserved for buses and vehicles with a minimum of two or three people and that includes the driver

Explanation:

hope this helped

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:

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:

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.