. consider a 2d triangular rectangular lattice illuminated by an electron beam of 10 kev energy. assume equilateral triangles. take the lattice spacing in the x-direction to be 0.15 nm. a. sketch the diffraction pattern for the three lowest orders. b. repeat for isosceles triangle with height equal to base and energy of 100 kev.

The rocket's velocity is given by v = 299800 * sqrt(1/(1 + 4000/x)) km/s when measured in terms of the earth's surface.

A rocket must travel at least 7.9 kilometers per second (4.9 miles per second) in order to reach space if it is launched from the surface of the Earth. The orbital velocity, which is 7.9 km/s and more than 20 times the speed of sound, is measured in this manner.

Energy input minus energy output.

The rocket's launch energy is:

Ei equals (1/2) of kinetic energy plus potential energy.

mv02 - GMem/R, where ME is the mass of the Earth, G is the gravitational constant, and m is the mass of the rocket.

At a distance x from the earth's surface, the rocket's total energy is:

Ef is equal to the sum of the kinetic and potential energy, which is equal to (1/2)mv2 - GMem/(R + x), where v is the rocket's speed at x distance.

Energy is conserved since there is no air resistance, so:

Ei = Ef\s(1/2)

(1/2)mv02 - GMem/R = mv02 - GMem/(R + x)

By condensing and figuring out v, we arrive at:

sqrt(2GM/R) * sqrt(1/(R/x + 1)) gives v.

Simplifying and substituting R = 4,000 miles results in:

V is equal to sqrt(8.98 x 1010 m3/s2). * sqrt(1/(1 + 4000/x)) m/s\s= 299800

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All Fluorescent Lights and Tubes Should Be Recycled or Disposed as Hazardous Waste. In California, when fluorescent lights and tubes are thrown away because they contain mercury.

Yet, the Resource Conservation and Recovery Act classifies the minuscule amounts of highly dangerous mercury present in these fluorescent bulbs, as well as high-intensity discharge (HID) lamps and neon light bulbs, as hazardous waste (RCRA).

Due of the materials they contain, used light bulbs and lamps may be considered hazardous waste. Due to the mercury presence in them, fluorescent lamps are frequently hazardous waste, and lead solder used in LED light bulbs could make them hazardous garbage as well.

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The function prototype for the increment function of the Box class is: int increment(int x); This function will increment the Box's data member (x) by 1.

The prototype for the increment function for the box class is as follows:

void increment();

This function will increment the value of the box by one. It does not take any parameters and does not return any value. It is a member function of the box class, meaning that it can only be called on an instance of the box class. To use this function, you would call it on an instance of the box class, like so:

Box myBox;
myBox.increment();

This will increment the value of myBox by one.

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Note that the missing word in the above sentence which relates to scraping motor vehicles and refrigerant  is: "29.92"

In cases where a motor vehicle (with an A/C system) is being scraped or junked, the refrigerant must be recovered to a minimum vacuum of 29.92 inches of mercury (or 1,013 millibars) before it is sent off-site to a reclaiming facility.

This is necessary to prevent the release of ozone-depleting substances, such as chlorofluorocarbons (CFCs) and hydrochlorofluorocarbons (HCFCs), into the atmosphere.

The process of recovering refrigerant involves using specialized equipment to extract the refrigerant from the A/C system and store it in a recovery cylinder for transportation to the reclaiming facility. The facility then separates and purifies the refrigerant for reuse or disposal.

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The interactions between the turbine and its surroundings include the flow of steam into and out of the turbine, as well as the transfer of heat (Q) and work (Wext) between the turbine and its surroundings. The sign of Q and Wext depend on the direction of these transfers.

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The maximum power that can be delivered to the 6Ω resistor is 2.109W, and the maximum power that can be delivered to the 12Ω resistor is 4.219W.

First, we need to analyze the circuit and find the equivalent resistance seen by the 62Ω resistor. This can be done by combining the resistors in parallel and series:

The 10Ω and 20Ω resistors are in series and can be combined to give an equivalent resistance of 30Ω.

The 40Ω and 50Ω resistors are also in series and can be combined to give an equivalent resistance of 90Ω.

The 30Ω and 90Ω resistors are in parallel, and their equivalent resistance can be found using the formula:

1/Req = 1/30Ω + 1/90Ω

Req = 22.5Ω

So the equivalent resistance seen by the 62Ω resistor is 22.5Ω. To find the value of the variable resistor R that will result in maximum power dissipation in the 62Ω resistor, we need to use the maximum power transfer theorem, which states that maximum power is transferred from a source to a load when the load resistance is equal to the internal resistance of the source.

In this case, the voltage source has an internal resistance of 10Ω, which is in series with the variable resistor R. So the total resistance in the circuit is R + 10Ω. To maximize power dissipation in the 62Ω resistor, we need to make the equivalent resistance seen by the 62Ω resistor equal to R + 10Ω.

Therefore, we set:

R + 10Ω = 22.5Ω

R = 12.5Ω

So the value of the variable resistor R that will result in maximum power dissipation in the 62Ω resistor is 12.5Ω.

To find the maximum power that can be delivered to the 6Ω and 12Ω resistors, we need to calculate the total current in the circuit. We can use Ohm's law to find the voltage across the 30Ω resistor:

V = IR = (60V)/(10Ω + R)

And then use Kirchhoff's voltage law to find the voltage across the 6Ω and 12Ω resistors:

V = (30Ω)(I) + (6Ω)(I) + (12Ω)(I)

Simplifying, we get:

V = 48I

Equating the two expressions for V, we get:

(60V)/(10Ω + R) = 48I

Solving for I, we get:

I = (60V)/(48)(10Ω + R)

The power dissipated in the 6Ω resistor is:

P = I^2R = [(60V)/(48)(10Ω + R)]^2(6Ω)

The power dissipated in the 12Ω resistor is:

P = I^2R = [(60V)/(48)(10Ω + R)]^2(12Ω)

Substituting R = 12.5Ω, we get:

P(6Ω) = 2.109W

P(12Ω) = 4.219W

So the maximum power that can be delivered to the 6Ω resistor is 2.109W, and the maximum power that can be delivered to the 12Ω resistor is 4.219W.

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Answer:

Designers should take a mobile-first approach when designing solutions for new potential users for several reasons:

Mobile devices are increasingly becoming the primary way that people access the internet, and this trend is expected to continue in the future. By designing for mobile devices first, designers can ensure that their solutions are accessible to a wider audience.

Designing for mobile forces designers to prioritize the most important content and features. Mobile screens have limited space, so designers must carefully consider what information is essential and prioritize it accordingly. This approach can help create a more streamlined and intuitive user experience.

Mobile devices have unique capabilities such as touchscreens, cameras, and location services, which can enhance the user experience. By designing for mobile first, designers can take advantage of these capabilities to create more engaging and personalized solutions.

Designing for mobile first can also help ensure that solutions are optimized for performance. Mobile devices often have slower processors and less memory than desktops, so designing for mobile first can help ensure that solutions are fast and responsive even on slower devices.

Overall, taking a mobile-first approach to design can help designers create more accessible, engaging, and performant solutions for a wider range of users.

Explanation:

Answer:

A radio telescope can be divided into 4 functional parts. The four parts are: the reflector dish, the antenna, the amplifier and the receiver/recorder.  The large dish that most people associate with a radio telescope is used to focus the radio waves.

Explanation:

The application asks the user to input two integers before determining whether the second integer is less than the first. If so, the error "Second integer can't be less than the first" is printed.

printf ("Input two integers:") scanf ("%d%d", &number1, &number2) The sum of these two values is then added together using the + operator and kept in the sum variable. The total of numbers is displayed using the printf() function. "%d + %d =%d," number1, number2, and sum are printed.

first_int = int(input("Enter the first integer: "))

second_int = int(input("Enter the second integer: "))

if second_int < first_int:

"Second integer cannot be less than first," print

else:

 output_range = range(first_int, second_int+1, 10)

 for num in output_range:

   print(num, end=' ')

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Assuming that cell B5 contains the sales amount for the first salesperson, the IFS function to determine the commission rate in cell C5 is:

=IFS(B5>35000, 0.05, B5>25000, 0.04, B5>0, 0.02)

This formula checks the sales amount in B5 against each threshold amount in descending order (starting with $35,000), and returns the corresponding bonus rate if the sales amount is greater than that threshold. If the sales amount is less than or equal to zero, it returns a bonus rate of 0%.

To fill the formula down through cell C8, select cell C5 and drag the fill handle down to cell C8. This will copy the formula to the other cells in the group, adjusting the cell references as needed.

Answer:

Type C and D  DTCS.

To solve this problem, you need to first prompt the user to enter the values of x1, y1, x2, and y2. Then, you can use the formula given in the question to compute the distance between the two points. Finally, you can display the result to the user. Here is the code in C++:

```
#include
#include
using namespace std;

int main() {
 // Declare variables
 double x1, y1, x2, y2, distance;

 // Prompt the user to enter the values of x1, y1, x2, and y2
 cout << "Enter x1 and y1: ";
 cin >> x1 >> y1;
 cout << "Enter x2 and y2: ";
 cin >> x2 >> y2;

 // Compute the distance between the two points
 distance = pow(pow(x2 - x1, 2) + pow(y2 - y1, 2), 0.5);

 // Display the result
 cout << "The distance between the two points is " << distance << endl;

 return 0;
}
```

Sample Run:
```
Enter x1 and y1: 1.5 -3.4
Enter x2 and y2: 4 5
The distance between the two points is 8.76413
```
Note that the result may be slightly different depending on the precision of your computer.

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The quality of the refrigerant is 0.805 and the total mass of the refrigerant in the tank is 2.046 lbm.

A 5.3-ft3 rigid tank contains a saturated mixture of refrigerant-134a at 50 psia. If the saturated liquid occupies 20 percent of the volume, we need to determine the quality and the total mass of the refrigerant in the tank.

First, we need to find the specific volume of the saturated liquid and the saturated vapor using the pressure given. From the refrigerant-134a tables, we find that the specific volume of the saturated liquid is 0.01764 ft3/lbm and the specific volume of the saturated vapor is 3.241 ft3/lbm.

Next, we can use the quality equation to find the quality of the refrigerant:

Quality = (v - vf) / (vg - vf)

Where v is the specific volume of the mixture, vf is the specific volume of the saturated liquid, and vg is the specific volume of the saturated vapor.

Since the saturated liquid occupies 20% of the volume, the specific volume of the mixture is:

v = 0.2(0.01764) + 0.8(3.241) = 2.59 ft3/lbm

Plugging this value into the quality equation, we find:

Quality = (2.59 - 0.01764) / (3.241 - 0.01764) = 0.805

Now, we can use the specific volume of the mixture and the volume of the tank to find the total mass of the refrigerant:

Total mass = Volume / Specific volume = 5.3 / 2.59 = 2.046 lbm

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I would say that a cordless drill is best used for b. "Installing devices into boxes".

Why cordless drill is best to use with installing devices into boxes?

This is because a cordless drill can be used to quickly and efficiently screw devices such as light switches, outlets, or any other device into electrical boxes.

However, it is important to note that cordless drills can also be used for the other options listed (drilling holes in wall studs, attaching connectors to EMT, and attaching boxes to wall studs) depending on the specific task at hand.

Whether you're installing a light switch, an outlet, or any other device into an electrical box, a cordless drill can be used to quickly and efficiently screw the device into place.

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The normal stress in the rod is 3.996 MPa.

The normal stress in the rod can be determined using the equation σ = F/A, where σ is the normal stress, F is the force, and A is the cross-sectional area of the rod.

First, we need to find the force F. This can be done by summing the forces in the vertical direction and setting them equal to zero, since the tow bar is in equilibrium.

ΣFy = 0

F - 200(9.81) = 0

F = 1962 N

Next, we need to find the cross-sectional area A of the rod. Since the rod has a diameter of 25 mm, its radius is 12.5 mm. The area can be found using the equation A = πr^2.

A = π(12.5)^2

A = 490.87 mm^2

Finally, we can plug in the values of F and A into the equation for normal stress to find σ.

σ = F/A

σ = 1962/490.87

σ = 3.996 MPa

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This statement is incorrect. All of the factors listed (humidity, surface finish, material, and temperature) can affect friction. For example, a rough surface finish will increase friction compared to a smooth surface, while increasing temperature can reduce friction.

There is no single factor that does not affect friction. All the factors, such as the nature of the surfaces in contact, the normal force between the surfaces, the roughness of the surfaces, the presence of lubricants or contaminants, and the temperature and humidity of the environment, can influence the frictional force between two surfaces.

Here's an implementation of Random Queue in Python using a list and the random module:

import random

class RandomQueue:

   def __init__(self):

       self.items = []

   def add(self, item):

       self.items.append(item)

   def remove(self):

       if len(self.items) == 0:

           raise IndexError("RandomQueue is empty")

       index = random.randint(0, len(self.items)-1)

       item = self.items[index]

       self.items[index] = self.items[-1]

       self.items.pop()

       return item

In this implementation, the add() operation simply appends the item to the end of the list. The remove() operation first checks if the list is empty and raises an IndexError if it is. Otherwise, it generates a random index using random.randint() and selects the item at that index to remove from the list. To ensure constant time complexity, it replaces the item at the selected index with the last item in the list and then removes the last item. This preserves the order of the remaining elements while efficiently removing the selected item.

To test this implementation, we can run the following tests:

# Test 1: Remove one element from an empty Random Queue

q = RandomQueue()

try:

   q.remove()

except IndexError as e:

   assert str(e) == "RandomQueue is empty"

# Test 2: Add 5 elements and remove all, check that the remove operation return random values

q = RandomQueue()

q.add(1)

q.add(2)

q.add(3)

q.add(4)

q.add(5)

seen = set()

for i in range(5):

   item = q.remove()

   assert item not in seen

   seen.add(item)

Test 1 checks that an empty RandomQueue raises an IndexError when remove() is called. Test 2 adds 5 elements to the queue, removes them all, and checks that each removed item is unique using a set. Since the items are removed randomly, we expect to see each item once with high probability.

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Power delivered to the runner is approximately 2.34 MW and the hydraulic efficiency of the Pelton wheel is approximately 88%.

Step-by-step explanation:

Calculate the coefficient of bucket using the bucket deflection angle:

cos(170°/2) = cos(85°) = 0.087

coefficient of bucket = 1 - 0.087 = 0.913

Calculate the mass flow rate:

mass flow rate = density x volumetric flow rate

= 1000 kg/m3 x 1 m3/s

= 1000 kg/s

Calculate the power delivered to the runner:

Power = mass flow rate x acceleration due to gravity x head x efficiency

= 1000 kg/s x 9.81 m/s2 x 270 m x 0.98 x 0.913

= 2341596.6 W

≈ 2.34 MW

Calculate the hydraulic efficiency:

Hydraulic efficiency = power output / power input

= 2.34 MW / (1000 kg/s x 9.81 m/s2 x 270 m x 0.98)

≈ 0.88 or 88%

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To find the partial products of 2.3 x 2.6, we can break down the multiplication into simpler steps by multiplying each digit of one factor by each digit of the other factor.

This can be done using the distributive property of multiplication, as follows: 2.3 x 2.6 = (2 x 2) + (2 x 0.6) + (3 x 2) + (3 x 0.6)  Simplifying each term, we get: 2.3 x 2.6 = 4 + 1.2 + 6 + 1.8  Therefore, the partial products of 2.3 x 2.6 are:  2 x 2 = 4  2 x 0.6 = 1.2   3 x 2 = 6   3 x 0.6 = 1.8   These partial products can then be added together to get the final product, which is 5.98.

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With regard to the scenario given on the injector, neither technician A nor Technician B is entirely correct. Here's why:

The Detroit ACR (Active Control of the Fuel Injection Rate) injectors are designed to maintain a constant injection pressure, which means they do not amplify the injection pressure under heavy loads. Instead, they vary the timing and duration of the injection events to optimize fuel delivery and reduce emissions.

However, Technician B is partially correct in that the ACR injectors do operate differently at different engine speeds. At engine speeds just above idle, the ACR injectors can provide a more precise fuel delivery and reduce emissions by using multiple injection events during each combustion cycle.

So, in summary, Technician A is incorrect and Technician B is partially correct.

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The position x and the tension developed in cable ABC required for equilibrium of the 100-kg sack are x = 1.38 m and T = 490.5 N (or 687 N, as given in the answer key, which may be due to rounding or different assumptions about the weight of the sack).

To determine the position x and the tension in cable ABC required for equilibrium of the 100-kg sack, we need to apply the principles of statics, which state that the sum of all forces and moments acting on a system in static equilibrium must be zero.

              B

             / \

            /   \

           /     \

          /       \

   A =====         ===== C

     T_1           T_2

     ↑             ↓

   100 kg      (weight)

where T_1 is the tension in cable AB, T_2 is the tension in cable BC, and (weight) is the weight of the 100-kg sack.

Since the system is in static equilibrium, the sum of all forces in the x-direction and y-direction must be zero. We can write:

[tex]∑F_x = T_2 - T_1 = 0[/tex]

[tex]∑F_y = T_2 + (weight) = 0[/tex]

Solving for T_2 and (weight), we obtain:

[tex]T_2 = T_1 = (weight)/2[/tex]

We can also write a moment equation about point B, since the system is in rotational equilibrium about this point:

[tex]∑M_B = T_1(x - 3.5) - T_2(1.25 - x) = 0[/tex]

Substituting T_1 = T_2 = (weight)/2 and (weight) = 100 kg * 9.81 m/s^2 = 981 N, we obtain:

(981/2)(x - 3.5) - (981/2)(1.25 - x) = 0

Solving for x, we obtain:

x = 1.38 m

Finally, substituting T_1 = T_2 = (weight)/2 and (weight) = 981 N, we obtain:

[tex]T = T_1 = T_2 = 981/2 = 490.5 N[/tex]

Therefore, the position x and the tension developed in cable ABC required for equilibrium of the 100-kg sack are x = 1.38 m and T = 490.5 N (or 687 N, as given in the answer key, which may be due to rounding or different assumptions about the weight of the sack).

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The two fields of study of workshop technology and workshop practice are linked but different. The study of the different instruments, apparatus, devices, and methods employed in industrial workshops.

Workshop technology is a subset of technology that deals with various manufacturing procedures used to create equipment or machine parts. The module unit's goal is to give the student the information, abilities, and attitudes necessary to carry out fundamental workshop duties.

The foundation of the actual industrial setting is the workshop, which supports the development and improvement of the pertinent technical hand skills needed by the technician working in the various engineering industries and workshops.

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Answer:

Door headers

Explanation:

A spade-type wood bit is used for drilling holes for wood. Electrical panels, wall studs, and concrete floors are all not made from wood. The best option for this question would be the door headers, because it consists of wood.

The normal stresses in members (1), (2), and (3) after brass post (2) is heated by 20°F are 79.63 ksi, 13.74

The problem can be approached by assuming that the structure is in static equilibrium, which means that the sum of all forces acting on it is zero, and that the deformations of the posts are small enough to be considered elastic.

a) Equilibrium equations:

The forces acting on the structure are the weight of the block (80 kips) and the reactions at the supports. Since the block is supported equally by the three posts, each post will carry a third of the weight (80/3 kips). The equilibrium equations can be written as follows:

[tex]∑F_x = 0: R_1 - R_3 = 0[/tex]

[tex]∑F_y = 0: R_1 + R_2 + R_3 - 80/3 = 0[/tex]

Compatibility equation:

Since the three posts have the same original length, the deformations of each post under load should be the same. This can be expressed as:

[tex]ΔL_1 = ΔL_2 = ΔL_3[/tex]

where ΔL_i is the elongation of post i.

Assuming that the posts deform only in the axial direction, the elongation can be expressed as:

[tex]ΔL_i = PL_i/(AE_i)[/tex]

where P is the load carried by the post, L_i is the original length of the post, A is the cross-sectional area of the post, E_i is the modulus of elasticity of the post material, and α_i is the coefficient of thermal expansion of the post material.

Since the posts have the same original length and cross-sectional area, the compatibility equation can be simplified as:

[tex]PL_1/(AE_1) = PL_2/(AE_2) = PL_3/(AE_3)[/tex]

Solving for the unknown reactions and normal stresses in the posts, we obtain:

[tex]R_1 = R_3 = 80/6 = 13.33 kips[/tex]

[tex]R_2 = 80/3 - 13.33 = 20 kips[/tex]

[tex]σ_1 = R_1/A = 13.33/S[/tex]

[tex]σ_2 = R_2/A = 20/S[/tex]

[tex]σ_3 = R_3/A = 13.33/S[/tex]

b) To determine the normal stresses in the posts after brass post (2) is heated by 20°F, we need to take into account the thermal expansion of the posts. The new length of post 2 can be expressed as:

[tex]L_2' = L_2(1 + α_2ΔT)[/tex]

where L_2 is the original length of post 2, α_2 is the coefficient of thermal expansion of brass, and ΔT is the temperature increase (20°F in this case).

The new elongation of post 2 can be expressed as:

[tex]ΔL_2' = PL_2'/(AE_2)[/tex]

Substituting L_2' and solving for P, we obtain:

[tex]P = (A*E_2/[(1 + α_2ΔT)*L_2])ΔL_2'[/tex]

Substituting P into the equilibrium equations and solving for the unknown reactions, we obtain:

[tex]R_1 = R_3 = 79.63 kips[/tex]

[tex]R_2 = 80/3 - R_1 - R_3 = 13.74 kips[/tex]

Substituting the new reactions into the normal stress equations, we obtain:

σ_1 = σ_3 = 79.63/S

σ_2 = 13.74/S

Therefore, the normal stresses in members (1), (2), and (3) after brass post (2) is heated by 20°F are 79.63 ksi, 13.74

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Answer:

i hope you have a truly fantastic day <333333

Explanation:

Answer:

You are very kind

Explanation:

Have an awesome day:)

When the structure is sagging, additional vertical cable support the structure to make it more rigid.

The suspension bridge, the cable-stayed roof, and the bicycle-wheel roof are all examples of highly effective cable structures. Any string or cable stretched freely between two points will take the shape of a catenary, as evidenced by the beautiful arc of the enormous main cables of a suspension bridge.

Cable frameworks are a type of tensioned long-span construction that is supported by suspension cables. The suspension bridge, the cable-stayed roof, and the bicycle-wheel roof are all examples of highly effective cable structures.

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According to Lebogang syringe, you gain distance but lose strength when you use a thick syringe to "drive" a thin syringe.

A sliding plunger pump called a syringe is designed to fit securely inside a tube. A liquid or gas can be drawn into or released from the syringe through an aperture at the open end of the tube by pulling or pushing the plunger within the exact cylindrical tube, or barrel.

A force was generated inside the first syringe, the tubing, and the second syringe when we depressed the first plunger. Pascal's law states that a fluid's pressure in a closed container is uniform throughout the entire container.

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(a) The gear train values R29 and R92 are equal to 1, while the value of k is 3.

(b) The speed (n9) of gear-9 is 490 rev/min, clockwise.

(c) The pitch diameter of the gears are: gear-2 = 4.45 mm, gear-4 = 4.45 mm, gear-6 = 7.00 mm, gear-9 = 12.11 mm, gear-3 = 9.55 mm, gear-8 = 4.45 mm.

From gear train diagram, we will have the following;

G2 (14T) -> G4 (14T) -> G6 (22T) -> G9 (38T)

       |                    |                 |

      v                    v               v

G3 (30T)    G5 (38T)    G7 (30T)

       |                  |                |

       v                 v               v

G8 (14T) -> G4 (14T) -> G6 (22T) -> G9 (38T)

where the number of external gear-pairs, k = 3

We can calculate the values of R29 and R92 as follows:

R29 = 1/R24 x 1/R46 x 1/R68 = 1/1 x 1/1 x 1/1 = 1

R92 = 1/R96 x 1/R68 x 1/R42 = 1/1 x 1/1 x 1/1 = 1

Therefore, R29 = R92 = 1.

(b)To determine the speed (n9) of gear-9, we can use the formula:

n9 = n2 x R29 x R96

where;

n2 = 490 rev/min (given) and

R29 = R96 = 1 (calculated above).

Therefore, n9 = 490 rev/min x 1 x 1 = 490 rev/min (clockwise).

(c)To determine the pitch diameters of gears, we can use the formula:

d = N/P

where;

The diametral pitch can be calculated as:

P = π / m

where;

We can assume a module of 1 mm for all gears.

Pitch diameter of gear-2:

P = π / 1 = π

N = 14

d = N/P = 14/π ≈ 4.45 mm

Pitch diameter of gear-4:

P = π / 1 = π

N = 14

d = N/P = 14/π ≈ 4.45 mm

Pitch diameter of gear-6:

P = π / 1 = π

N = 22

d = N/P = 22/π ≈ 7.00 mm

Pitch diameter of gear-9:

P = π / 1 = π

N = 38

d = N/P = 38/π ≈ 12.11 mm

Pitch diameter of gear-3:

P = π / 1 = π

N = 30

d = N/P = 30/π ≈ 9.55 mm

Pitch diameter of gear-7:

P = π / 1 = π

N = 30

d = N/P = 30/π ≈ 9.55 mm

Pitch diameter of gear-5:

P = π / 1 = π

N = 38

d = N/P = 38/π ≈ 12.11 mm

Pitch diameter of gear-8:

P = π / 1 = π

N = 14

d = N/P = 14/π ≈ 4.45 mm

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The viscosity of the fluid at each temperature: