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galben [10]
4 years ago
15

Write a python program to apply this:

Engineering
1 answer:
lakkis [162]4 years ago
8 0

Answer:

# Initialize a dictionary with the keys

contestants = {"Darci Lynne":0, "Angelica Hale":0, "Angelina Green":0};

# Repeatedly prompt the user for a contestant name to vote for

while True:

 

  # Prompting user for contestant name

  cName = input("Enter contestant name to vote: ");

 

  # Checking for Done

  if cName.lower() == "done":

      break;

 

  # Checking in dictionary

  if cName in contestants.keys():

      # Updating vote value

      contestants[cName] += 1

  # New entry

  else:

      contestants[cName] = 1

 

# Printing header

print("\n%-20s %-15s\n" %("Contestant Name", "Votes Casted"))

# Printing results

for contestant in contestants:

  print("%-23s %-15d" %(contestant, contestants[contestant]))

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9. Calculate the total resistance and current in a parallel cir-
Taya2010 [7]

Answer:

  d. 2.3 ohms (5.3 amperes)

Explanation:

The calculator's 1/x key makes it convenient to calculate parallel resistance.

  Req = 1/(1/4 +1/8 +1/16) = 1/(7/16) = 16/7 ≈ 2.3 ohms

This corresponds to answer choice D.

__

<em>Additional comment</em>

This problem statement does not tell the applied voltage. The answer choices suggest that it is 12 V. If so, the current is 12/(16/7) = 21/4 = 5.25 amperes.

5 0
3 years ago
What are the disadvantages of having a liquid cooled engine?
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One notable disadvantage of liquid cooling over air cooling is that it is considerably costly to set up. Cooling fans are prevalent in the market, and this overabundance of supply means they are cheap. The components of a liquid cooling system can be expensive.
5 0
3 years ago
Please answer the questions !
gizmo_the_mogwai [7]

Answer:

120

Explanation:

6 0
3 years ago
Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat
Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 =  0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = \frac{m}{\rho}

V = Av = \frac{\pi}{4} * d^2 * v1

V = \frac{\pi}{4} * 0.28^2 * 5

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = \frac{V}{v} = \frac{0.30787}{0.11418}

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = \frac{v2}{A2}

v2 = \frac{0.3705}{\frac{\pi}{4} * 0.28^2}

v2 = 6.017 m/s

7 0
3 years ago
Problem definition
LekaFEV [45]

Answer:

ummm thats alot

Explanation:

8 0
3 years ago
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