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luda_lava [24]
3 years ago
9

Consider the following ways of handling deadlock: (1) banker’s algorithm, (2) detect

Engineering
1 answer:
Andrew [12]3 years ago
7 0

Answer:

b

Explanation:

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The concrete canoe team does some analysis on their design and calculates that they need a compressive strength of 860 psi. They
vlada-n [284]

Answer:

874 psi

Explanation:

Given a sample mean (x') = 900,

and a standard error (SE) = 10

At 99% confidence, Z(critical) = 2.58

That gives 99% confidence interval as,

x' ± Z(critical) x SE = 900 ± 2.58 x 10

The value of the lower limit is,

900 - 25.8 = 874.2

≈ 874 psi

8 0
3 years ago
Please solve part two
Burka [1]

Answer:

Wat part 2

Explanation:

7 0
3 years ago
Determine the combined moment about O due to the weight of the mailbox and the cross member AB. The mailbox weighs 3.2 lb and th
koban [17]

Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

Answer : moment of inertia = 186.7 Ib - in

Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

The origin is of mailbox and cross member is 0

The perpendicular distance from Y axis of centroid of the mailbox

= 4 + (25/2) = 16.5"

The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4  = 13"

therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)

                                     = 52.8 + 133.9 = 186.7 Ib-in

8 0
3 years ago
The base class Pet has attributes name and age. The derived class Dog inherits attributes from the base class Pet class and incl
Nonamiya [84]

Answer:

Explanation:

class Pet:

   def __init__(self):

       self.name = ''

       self.age = 0

   def print_info(self):

       print('Pet Information:')

       print('   Name:', self.name)

       print('   Age:', self.age)

class Dog(Pet):

   def __init__(self):

       Pet.__init__(self)

       self.breed = ''

def main():

   my_pet = Pet()

   my_dog = Dog()

   pet_name = input()

   pet_age = int(input())

   dog_name = input()

   dog_age = int(input())

   dog_breed = input()

   my_pet.name = pet_name

   my_pet.age = pet_age

   my_pet.print_info()

   my_dog.name = dog_name

   my_dog.age = dog_age

   my_dog.breed = dog_breed

   my_dog.print_info()

   print('   Breed:', my_dog.breed)

main()

3 0
3 years ago
What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?
Yakvenalex [24]

Answer:

Activation\ Energy=2.5\times 10^{-19}\ J

Explanation:

Using the expression shown below as:

N_v=N\times e^{-\frac {Q_v}{k\times T}

Where,

N_v is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

{Q_v} is the activation energy

T is the temperature

Given that:

N_v=2.3\times 10^{13}

N = 10 moles

1 mole = 6.023\times 10^{23}

So,

N = 10\times 6.023\times 10^{23}=6.023\times 10^{24}

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (425 + 273.15) K = 698.15 K  

T = 698.15 K

Applying the values as:

2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

Q_v=2.5\times 10^{-19}\ J

4 0
3 years ago
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