A _______ contact allows current to flow when the switch's operator is not activated.?
1 answer:
Answer:
Normally closed (NC)
Explanation:
Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.
Mathematically, Ohm's law is given by the formula;
Where;
V represents voltage measured in voltage.
I represents current measured in amperes.
R represents resistance measured in ohms.
A switch can be defined as an electronic component or device that is typically used to prevent or allow the flow of electric current (electrons) from one point to another in an electrical circuit.
There are two logical forms used for designing a switch and these includes;
I. Normally opened (NO).
II. Normally closed (NC).
A Normally closed (NC) contact allows current to flow when the switch's operator is not activated.
Basically, a Normally closed (NC) contact would always remained closed unless a certain (specific) condition is met. Thus, when the contacts are moved, the circuit becomes open and thus, prevents the flow of electric current in the circuit.
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This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
the forces required to keep the artery in place is 1.65 N
Explanation:
Given the data in the question;
Inlet velocity V₁ = 50 cm/s = 0.5 m/s
diameter d₁ = 15 mm = 0.015 m
radius r₁ = 0.0075 m
diameter d₂ = 11 mm = 0.011 m
radius r₂ = 0.0055 m
A₁ = πr² = 3.14( 0.0075 )² = 1.76625 × 10⁻⁴ m²
A₂ = πr² = 3.14( 0.0055 )² = 9.4985 × 10⁻⁵ m²
pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal
pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal
Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s
given that; blood density is 1050 kg/m³
mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s
Now, using continuity equation
A₁V₁ = A₂V₂
V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁
we substitute
V₂ = (0.015 / 0.011 )² × 0.5
V₂ = 0.92975 m/s
from the diagram, force balance in x-direction;
0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )
so we substitute in our values
0 - (12665.6 × 9.4985 × 10⁻⁵) × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )
0 - 0.6014925 + Rₓ = 0.043106929 - 0
Rₓ = 0.043106929 + 0.6014925
Rₓ = 0.6446 N
Also, we do the same force balance in y-direction;
P₁A₁ - P₂A₂ × sin(60°) + R = m'( V₂sin(60°) - 0.5 )
we substitute
⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R = 0.092728( 0.92975sin(60°) - 0.5 )
⇒ 1.5484 + R = 0.092728( 0.305187 )
⇒ 1.5484 + R = 0.028299
R = 0.028299 - 1.5484
R = -1.52 N
Hence reaction force required will be;
R = √( Rₓ² + R² )
we substitute
R = √( (0.6446)² + (-1.52)² )
R = √( 0.41550916 + 2.3104 )
R = √( 2.72590916 )
R = 1.65 N
Therefore, the forces required to keep the artery in place is 1.65 N
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:
the generator induced voltage is 60.59 kV
Explanation:
Given:
S = 150 MVA
Vline = 24 kV = 24000 V
the network voltage phase is
the power transmitted is equal to:
the line induced voltage is