Answer:
John should detail his Scrum Master.
Explanation:
The Team Lead or Scrum Master coordinates the tasks of individual team members and supports the progress of the team. The Scrum Master usually receives instructions from the Product Owner and then ensures that the tasks are performed accordingly. She also coaches the Development Team and works with the Product Owner to carry out daily development activities. She also drives the Scrum Values and Principles, ensuring that the team members understand and practice them.
Answer:
we need to see what you're being asked about
Explanation:
Answer:
first is the parentheses, (3+2)=5 next is the exponent 5^2=25, next is the division 5 / 5 = 1, then the multiplication 4*1=4 and then you add 4+25=29. so the answer is 29.
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()
Answer:
Δw =7.95 kg/1000m^3
q = 62362.3 kg/1000m^3
Explanation:
To solve this problem we first need to use the psycrometric chart to determine the enthalpy h1, specific volume vi and absolute humidity col by using the given temperature T1 = 32°C and the relative humidity Ф1 = 95%.
h_1 = 106.5 kJ/kg
v_1 = 0.91 m^3/kg
w_1 = 0.02905
We will also need the enthalpy h2 and the absolute humidity w_2 at the exit point. We will again use the pyscrometric chart and the given temperature T_2 = 24°C. From the problem we also know that the exit relative humidity is = 60%.
h2 = 52.6 kJ/kg
w_2 = 0.01119
We need to express the final results in units per 1000 m^3. To do that we will need the mass m of this volume of air V and to calculate that we will use the given pressure p = 1 atm = 101.3 kPa.
m = R_a*T_1/V.p
m = 1000*101.3/0.287*305K
m = 1157 kg
Because it is a closed system, the amount of water removed Δw can be calculated as:
Δw =w_1 - w_2
Δw =0.02905- 0.01119
Δw =0.00687 kg/kg* 1157kg/1000m^3
Δw =7.95 kg/1000m^3
From the energy balance equation we can calculate the specific heat q removed from the air.
q = h_1 - h_2
q = 106.5 kJ/kg - 52.6 kJ/kg
q = 53.9 kJ/kg * 1157kg/1000m^3
q = 62362.3 kg/1000m^3
