Game must be about two objects or vehicles colliding (like a tank game)
Language must be C#\
Assignment a Create a video showing how your game runs, play the game and explain how it plays. (don't worry about code in video).

To change the clock set in Eastern Standard Time (EST) to Pacific Standard Time (PST) in Mastering Engineering, follow these steps:

Locate the clock or time settings in Mastering Engineering.

Adjust the clock to reflect the current time in EST.

Calculate the time difference between EST and PST, which is typically 3 hours.

Subtract the time difference from the current EST time.

Update the clock by setting it to the resulting PST time.

Verify that the clock now displays the correct time in Pacific Standard Time.

Remember to account for daylight saving time changes, if applicable.

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Incomplete question. The options read;

A. can change the story's ending

B. listens to the dialogue

C. hears the rock songs

D. feels more connected to the text.

Answer:

D. feels more connected to the text.

Explanation:

Remember, the underlying motive of an author when writing any text is to make his readers or audience more connected to the material been read or heard.

Hence, in "Great Rock and Roll  Pauses" we can conclude that the author's choice of medium was motivated by the same desire of making the audience feel more connected to the text.

Answer:

The audience can actually hear the music.

Explanation: dont listen to below average iq people :0

Thrust angle alignments can be performed on both rear wheel drive and front wheel drive vehicles. Therefore, the correct option is (OD) Neither Technician A nor B.

Neither Technician A nor B is correct.

Thrust angle alignments are not exclusive to either rear wheel drive or front wheel drive vehicles.

A thrust angle alignment is a procedure that ensures the rear wheels are properly aligned with the vehicle's centerline, regardless of the drivetrain configuration.

It is performed to correct any misalignment that can cause the vehicle to pull to one side.

The thrust angle can be affected by various factors, including rear axle issues, frame damage, or improper wheel alignment.

Therefore, both technicians' statements are inaccurate, and a thrust angle alignment can be necessary for vehicles of any drivetrain type.

Therefore, the correct option is (OD) Neither Technician A nor B.

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For the MOS transistor shown in Fig. 2, assume that it is sized and biased so that gm = 1mA/V and ro= 100 k2. Using the small-signal model and assigning RL = 10 k2, R₁ = 500 k2, and R2 = 1 MS2, find the following.

Draw the equivalent small-signal circuit.(b) The overall voltage gain vo / V sig(c) The input resistance R..The equivalent small-signal circuit is shown below :Equivalent small-signal circuit The voltage gain can be calculated as follows: Since the value of the current source is equal to gmVgs:Vgs = igm / gm Thus, Vgs = 0.001/1 mV = 1VThe output voltage is given by:Vo = -gm * ro * Vgs * RLVsig = Vo / igm = -ro * RL * VgsInput resistance, R, is given by:R = Rsig = R1 || R2 || rπwhere rπ = 1/gm = 1 kΩThen,R = 357 ΩThe main answer to each part of the question is as follows: a) The equivalent small-signal circuit for the MOS transistor is shown in the diagram above.

The voltage gain can be calculated as:Vgs = 1mV, Vo = -100V and igm = 1mARL = 10kΩThe voltage gain is given as:Av = Vo / Vsig = (-100V) / (1mV) = -100000V/Vc) The input resistance is given as:R = Rsig = R1 || R2 || rπR1 = 500kΩ, R2 = 1MΩ and rπ = 1/gm = 1kΩSo,R = 1 / (1/R1 + 1/R2 + 1/rπ)R = 1 / (1/500kΩ + 1/1MΩ + 1/1kΩ)R = 357ΩTherefore, the voltage gain is -100000 V/V and the input resistance is 357 Ω.

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Answer:

The condition in which a problem exists

Explanation:

Context is the situation surrounding information, e.g. if you don't understand a word in a passge you can look for "context clues" where you use surrounding words to infer the meaning of the unknown one

Answer:

b

Explanation:

Micropulse arc welding is a well-known tool for speeding up simple bench operations during the welding process like ring size, making connections near thermally sensitive gemstones, and mending porosity castings.

The practice performs effectively on gold, silver, copper, and stainless steel whether joining tiny jewelry materials or bigger pieces for miniature metal sculptures.

The welding arc temperatures typically range from 6000°C to 8000°C, which translates to about 10000F to 15000F.

This increased range of thermal heat is high enough to turn Gold and silver rings into a molten state. A molten state is a phase at which a solid element is being melted and turns into a liquid state under the influence of a high temperature.

Therefore, from the above explanation, we can conclude that it is possible and true for gold and silver rings to receive an arc and turn molten.

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The probability that the player gets doubles less than three times in 5 attempts is; 0.9645.

The formula for binomial probability distribution is;

P(X = x) = ⁿCₓ × pˣ × (1 - p)⁽ⁿ ⁻ ˣ⁾

where;

x = total number of successes

p = probability of a success on an individual trial

n = number of trials

We are given;

x = 3

p = 1/6 = 0.1667

n = 5

Thus;

P(X < 3) = (⁵C₂ * 0.1667 * (1 - 0.1667)⁽⁵ ⁻ ²⁾) + (⁵C₁ * 0.1667 * (1 - 0.1667)⁽⁵ ⁻ ¹⁾) +  (⁵C₀ * 0.1667 * (1 - 0.1667)⁽⁵ ⁻ ⁰⁾)

Thus;

P(X < 3) = 0.9645

Thus, the probability that the player gets doubles less than three times in 5 attempts is 0.9645.

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Explanation:

1 meter ≈ 3.281 ft.

so,

1 m² = 1m×1m ≈ 3.281 × 3.281 = 10.764961 ft²

now we have 22.3 ft² of skin.

that would be then

22.3/10.764961 = 2.071535605 ≈ 2.1 m² or even more rounded ≈ 2 m²

Answer:

a) the mass flow rate of the steam is  \(\mathbf{m_1 =6.92 \ kg/s}\)

b) the exit velocity of the steam  is \(\mathbf{V_2 = 562.7 \ m/s}\)

c) the exit area of the nozzle is  \(A_2\) = 0.0015435 m²

Explanation:

Given that:

A steam with 5 MPa and 400° C enters a nozzle steadily

So;

Inlet:

\(P_1 =\) 5 MPa

\(T_1\) = 400° C

Velocity V = 80 m/s

Exit:

\(P_2 =\) 2 MPa

\(T_2\) = 300° C

From the properties of steam tables  at \(P_1 =\) 5 MPa and \(T_1\) = 400° C we obtain the following properties for enthalpy h and the speed v

\(h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg\)

From the properties of steam tables  at \(P_2 =\) 2 MPa and \(T_1\) = 300° C we obtain the following properties for enthalpy h and the speed v

\(h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg\)

Inlet Area of the nozzle = 50 cm²

Heat lost Q = 120 kJ/s

We are to determine the following:

a) the mass flow rate of the steam.

From the system in a steady flow state;

\(m_1=m_2=m_3\)

Thus

\(m_1 =\dfrac{V_1 \times A_1}{v_1}\)

\(m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}\)

\(m_1 =\dfrac{0.4 }{0.057838 }\)

\(\mathbf{m_1 =6.92 \ kg/s}\)

b) the exit velocity of the steam.

Using Energy Balance equation:

\(\Delta E _{system} = E_{in}-E_{out}\)

In a steady flow process;

\(\Delta E _{system} = 0\)

\(E_{in} = E_{out}\)

\(m(h_1 + \dfrac{V_1^2}{2})\) \(= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})\)

\(- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})\)

\(- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})\)

\(- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})\)

\(- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})\)

\(V_2^2 = 316631.29 \ m/s\)

\(V_2 = \sqrt{316631.29 \ m/s\)

\(\mathbf{V_2 = 562.7 \ m/s}\)

c) the exit area of the nozzle.

The exit of the nozzle can be determined by using the expression:

\(m = \dfrac{V_2A_2}{v_2}\)

making \(A_2\) the subject of the formula ; we have:

\(A_2 = \dfrac{ m \times v_2}{V_2}\)

\(A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}\)

\(A_2\) = 0.0015435 m²

Dumping fuel is a procedure used by pilots to reduce aircraft weight in emergency situations or when there is a need to land the aircraft immediately after takeoff.

However, fuel dumping should be done with caution and in accordance with specific procedures to ensure safety and minimize environmental impact. Here are some precautions that pilots must take when dumping fuel:

1. Follow aircraft manufacturer procedures: Pilots should be familiar with the specific procedures and guidelines provided by the aircraft manufacturer for fuel dumping. These procedures outline the correct methods and limitations for fuel dumping.

2. Consult with air traffic control (ATC): Before initiating fuel dumping, the pilot should establish communication with ATC to inform them about the situation and seek guidance. ATC can provide instructions on the most suitable location and altitude for fuel dumping, taking into account air traffic and environmental considerations.

3. Select appropriate altitude and location: Pilots should choose a safe altitude for fuel dumping to ensure the fuel disperses and evaporates before reaching the ground. The altitude should be above populated areas, bodies of water, and environmentally sensitive regions whenever possible.

4. Dump fuel in approved areas: Pilots should follow regulations and guidelines regarding approved fuel dumping areas. Some airports have designated fuel dumping areas to minimize the impact on populated areas and the environment.

5. Consider weather conditions: Pilots should take into account weather conditions, such as wind direction and speed, when determining the appropriate location for fuel dumping. Dumping fuel against the wind can help disperse the fuel more effectively.

6. Monitor fuel dump rate: Pilots should closely monitor the fuel dump rate to ensure it is within the aircraft's limitations. Excessive fuel dumping can lead to safety concerns and environmental damage.

7. Inform air traffic control and emergency services: Pilots should inform ATC and emergency services about the fuel dumping operation, including the estimated quantity and location of the dumped fuel. This information helps authorities assess any potential risks and take appropriate actions.

It's important to note that fuel dumping should only be conducted when necessary and in compliance with applicable regulations and procedures. Pilots should prioritize the safety of passengers, crew, and the environment when considering fuel dumping as an emergency measure.

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A common application that is used for the rod and tube type of control is in a duct-mounted limit control.

A control system can be defined as a type of input and output system that is designed and developed to typically maintain stable (constant) conditions and processes, usually without excessive human interference.

In Control engineering, a duct-mounted limit control is a control system that is commonly used for the operation of the rod and tube type of control.

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To complete this challenge in Python, you will need to define a function called print_popcorn_time() that takes a parameter called bag_ounces. The function should first check if the value of bag_ounces is less than 3 using an if statement.

If it is, print the string "Too small". If it is not, check if the value of bag_ounces is greater than 10 using another if statement. If it is, print the string "Too large". If it is not, calculate the time it takes to pop the popcorn using the formula 6 * bag_ounces and print the result followed by the string "seconds". Finally, end with a newline character.
Here is some example code that demonstrates how to define the print_popcorn_time() function in Python:
```python
def print_popcorn_time(bag_ounces):
   if bag_ounces < 3:
       print("Too small")
   elif bag_ounces > 10:
       print("Too large")
   else:
       time = 6 * bag_ounces
       print(str(time) + " seconds")
   print()
```
To use the function, you can simply call it with a value for the bag_ounces parameter:
```python
print_popcorn_time(7)
```
This will output the following string:
```text
42 seconds
```
I hope this helps you complete this Python challenge. Let me know if you have any further questions!

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An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car  to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both  cars which is equal from instant point of abreast.

So; we can say :

\(D_{pursuit} =D_{police}\)

By using the second equation of motion to find the distance S;

\(S= ut + \dfrac{1}{2}at^2\)

\(D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)\)

\(D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)\)

\(D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)\)

\(D_{police} = ut _P + \dfrac{1}{2}at_p^2\)

where ;

u  = 0

\(D_{police} = \dfrac{1}{2}at_p^2\)

\(D_{police} = \dfrac{1}{2}*(1.96)*(t+12)^2\)

\(D_{police} = 0.98*(t+12)^2\)

\(D_{police} = 0.98*(t^2 + 144 + 24t)\)

\(D_{police} = 0.98t^2 + 141.12 + 23.52t\)

Recall that:

\(D_{pursuit} =D_{police}\)

\((187.8+15.65 \ t-1.525 t^2)= 0.98t^2 + 141.12 + 23.52t\)

\((187.8 - 141.12) + (15.65 \ t - 23.52t) -( 1.525 t^2 - 0.98t^2) = 0\)

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

t = -6.16 OR  t = 3.02

Since we can only take consideration of the value with a  positive integer only; then t = 3.02 secs

From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;

Therefore ; the total time  required for the police car  to over take the automobile = 12 s + 3.02 s

Total time  required for the police car  to over take the automobile = 15.02 sec

Answer:

True

Explanation:

They have to be sure to keep cement smooth and keep an eye out for imperfections

The Mo-99 activity produced in 10 days Irradiation is 2.40599 × 10³² disintegrations.

The given data is as follows: Intensity of Neutrons: 4.18×10¹⁷ neutron/sec

Energy of Neutrons: 0.1 eV

Sphere shell of Uranium-235 Density of Uranium-235: 19 g/cm³

Inner radius: 10 cm

Outer radius: 12 cm

Duration of Irradiation: 10 days

1. Calculation of Mo-99 activity in 10 days Irradiation:

Using the FLUKA code, the Mo-99 activity in 10 days Irradiation can be calculated.

2. Calculation of Mo-99 activity in 10 days Irradiation:

Considering the yield of Mo-99 is 0.062 per fission, and fission is 2.09 fission/primary neutron.

Therefore, the number of Mo-99 produced per neutron can be calculated as follows:

Number of Mo-99 produced per primary neutron = Yield of Mo-99 × Fission per primary neutron= 0.062 × 2.09 = 0.12958

Therefore, the activity of Mo-99 produced per primary neutron can be calculated as follows:

Activity of Mo-99 produced per primary neutron= (Number of Mo-99 produced per primary neutron) × (Disintegration rate of Mo-99)= 0.12958 × 1.44 × 10⁶= 1.8656 × 10⁵ disintegrations/sec

Now, the intensity of neutron is 4.18 × 10¹⁷ neutron/sec.

Hence, the number of neutron will be:

N = Intensity of neutron × Time= 4.18 × 10¹⁷ × 10 × 24 × 3600= 1.284288 × 10²⁴

The total number of Mo-99 produced in the given duration can be calculated by the following formula:

Number of Mo-99 produced in the given duration= (Number of Mo-99 produced per primary neutron) × (Number of primary neutron)× (Duration of Irradiation)= 0.12958 × 1.284288 × 10²⁴ × 10= 1.66992 × 10²⁶

The total activity of Mo-99 produced in 10 days Irradiation can be calculated by the following formula:

Activity of Mo-99 produced in 10 days Irradiation= (Number of Mo-99 produced in the given duration) × (Disintegration rate of Mo-99)= 1.66992 × 10²⁶ × 1.44 × 10⁶= 2.40599 × 10³² disintegrations.

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Answer:

2 1/3 A

Explanation:

power in = power out

V (14) =  6V a  

    14 = 6a

      a = 14/6 =

To prevent overfilling of recovery cylinders the service technician must manually close the valves on the tank and the unit.

To prevent overfilling of recovery cylinders, it is crucial for service technicians to follow proper procedures, which include manually closing the valves on both the tank and the unit.

This safety measure ensures that the recovery cylinders do not exceed their maximum capacity and helps prevent potential hazards.

By closing the valves on the tank and the unit, the technician effectively isolates the recovery cylinder from the refrigerant source. This prevents any additional refrigerant from entering the cylinder once it reaches its capacity limit.

Overfilling a recovery cylinder can lead to increased pressure, which may cause the cylinder to rupture, resulting in a hazardous situation.

Furthermore, closing the valves on the tank and the unit allows the technician to safely disconnect the recovery cylinder without any refrigerant leakage.

This step is crucial for maintaining a safe working environment and complying with environmental regulations.

It is important for service technicians to be vigilant and knowledgeable about proper procedures to prevent overfilling of recovery cylinders. By adhering to these guidelines and ensuring valves are manually closed, technicians can mitigate the risk of accidents, protect themselves and others, and contribute to a safe and sustainable working environment.

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Answer: true

Explanation:

because flushness means your dead   so you are dead now

Answer:

Explanation:

Your 3-B Annual Credit Reports & Scores with Enhanced Credit Report Monitoring.

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The credit building options that I think  is the easiest method that you can see yourself doing is Secured credit cards.

Secured credit cards are known to be a type of card that are said to be backed by a specific refundable security deposit.

This kind of deposit is known to limit or lower the lender's risk and thus makes it better for people who do not have credit or when they do not  qualify for that type of credit. This credit option is easier to get when compared to unsecured cards, as it does not need a lot of requirement.

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Answer:

20.833 ohms

Explanation:

V = IR    re-arrange

V/I = R

250 / 12 = 20.833    ( 20  5/6)

Answer:

You'd need to know things like the weight of the car, the spring constant of the suspension springs.

Explanation:

Driving over the curb slowly does not have much of an "impact" because of the small potential energy it has but to find the force that results from going over the curb at a much higher speed, you would need to know the weight of the car since it directly impact the force that it applies. You also would need to know about the car's suspension, the spring material, how stiff the springs are and their spring constant to calculate the force they absorb.

I hope this answer helps.

The bearings and shaft experience additional loads as a result of this increased tension. Bearings' lifespans might be shortened by excessive stress.

The inside of the Poly-V Belt, also known as the Ribbed Belt, is lined with longitudinal v-shaped ribs. These ribs give the pulley and the belt a bigger surface area to contact, which increases friction force. Compared to a flat belt or rubber v-belt of the same size, the belt can convey more power as a result.

The Poly-V Belt's design ensures superior protection for all reinforcing cords, resulting in a more even load distribution. The Poly-V Belt also serves as a guide for a belt tensioner or idler pulley. Less space is needed for Poly-V Belt drives, and the Belt can be stretched under higher tension.

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Explanation:

Ohm's law applies: V = IR, and variations. The current in a series circuit is the same through every element. That current is given as 1.1 A.

Rt = (8 V)/I3 = (8 V)/(1.1 A) = 80/11 Ω = 7 3/11 Ω

R3 = Rt -R1 -R2 = 7 3/11 -2 -5 = 3/11 Ω

It = I3 = 1.1 A

I1 = I2 = I3 = 1.1 A

V1 = (I1)(R1) = (1.1 A)(2 Ω) = 2.2 V

V2 = (I2)(R2) = (1.1 A)(5 Ω) = 5.5 V

V3 = (I3)(R3) = (1.1 A)(7/11 Ω) = 0.3 V