Answer:
see answer below
Explanation:
the equation is given by
k=ko* e^(-Q/RT)
then replacing the known values
k=1200 min⁻¹ e^[-8000 cal/mole/ (1.987 cal/mole K *T)]
then replacing values of T every 50 K from 100 to 500 K we get the series of values
![\left[\begin{array}{ccc}T&k\\100&3.924*10^{-15} min ^{-1} \\150&2.643*10^{-9} min ^{-1} \\200&2.17*10^{-6} min ^{-1} \\250&1.216*10^{-4} min ^{-1} \\300&1.781*10^{-3} min ^{-1} \\350&0.012 min ^{-1} \\400&0.051 min ^{-1} \\450&0.156 min ^{-1} \\500&0.382 min ^{-1} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DT%26k%5C%5C100%263.924%2A10%5E%7B-15%7D%20min%20%5E%7B-1%7D%20%5C%5C150%262.643%2A10%5E%7B-9%7D%20min%20%5E%7B-1%7D%20%5C%5C200%262.17%2A10%5E%7B-6%7D%20min%20%5E%7B-1%7D%20%5C%5C250%261.216%2A10%5E%7B-4%7D%20min%20%5E%7B-1%7D%20%5C%5C300%261.781%2A10%5E%7B-3%7D%20min%20%5E%7B-1%7D%20%5C%5C350%260.012%20min%20%5E%7B-1%7D%20%5C%5C400%260.051%20min%20%5E%7B-1%7D%20%5C%5C450%260.156%20min%20%5E%7B-1%7D%20%5C%5C500%260.382%20min%20%5E%7B-1%7D%20%5Cend%7Barray%7D%5Cright%5D)
Explanation:
Both station will retract their data. They will wait a random amount of time before sending out data again. This lessens the chance of collision again.
Answer: Time division multiplexing
Explanation: because a slot is equivalent to 125ms
And each position inside a slot is for each signal and a single bit frim each voice conversation is sent during each 125ms slot
Answer:
see explaination
Explanation:
import random
def number_guess(num):
n = random.randint(1, 100)
if num < n:
print(num, "is too low. Random number was " + str(n) + ".")
elif num > n:
print(num, "is too high. Random number was " + str(n) + ".")
else:
print(num, "is correct!")
if __name__ == '__main__':
# Use the seed 900 to get the same pseudo random numbers every time
random.seed(900)
# Convert the string tokens into integers
user_input = input()
tokens = user_input.split()
for token in tokens:
num = int(token)
number_guess(num)
Answer:
1. A high level algorithm for cooking a cheeseburger could be:
- Heat fry pan
- Cook one side of the hamburger
- Wait
- Turn hamburger upside down
- Put cheese over hamburger
- Wait
- Cut hamburger bread in half
- Put cooked hamburger inside bread
- End (eat)
2. A detailed algorithm for cooking a cheeseburger could be:
- Place fry pan over the stove heater
- Turn on heater (max temp)
- IF fry pan not hot: wait, else continue
- Place raw hamburger on fry pan
- IF hamburger not half cooked: Wait X time then go to line 5, else continue
- Turn hamburger upside down
- Put N slices of cheese over hamburger
- IF hamburger not fully cooked: Wait X time then go to line 8, else continue
- Turn off heater
- Cut hamburger bread in half horizontally
- Put cooked hamburger on one of the bread halves.
- Put second bread half on top of hamburger
- End (eat)
Explanation:
An algorithm is simply a list of steps to perform a defined action.
On 1, we described the most relevant steps to cook a simple cheeseburger.
Then on point 2, the same steps were taken and expanded with more detailed steps and conditions required to continue executing the following steps.
In computational terms, we used pseudo-code for the algorithm, since this is a list of actions not specific to any programming language.
Also we can say this is a structured programming example due to the sequential nature of the cooking process.
