Answer: D) All of the above
Explanation:
Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.
Answer:
23.3808 kW
20.7088 kW
Explanation:
ρ = Density of oil = 800 kg/m³
P₁ = Initial Pressure = 0.6 bar
P₂ = Final Pressure = 1.4 bar
Q = Volumetric flow rate = 0.2 m³/s
A₁ = Area of inlet = 0.06 m²
A₂ = Area of outlet = 0.03 m²
Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s
Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s
Height between inlet and outlet = z₂ - z₁ = 3m
Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation
Work done by pump
∴ Power input to the pump 23.3808 kW
Now neglecting kinetic energy
Work done by pump
∴ Power input to the pump 20.7088 kW
Answer:
Complete question is:
write the following decorators and apply them to a single function (applying multiple decorators to a single function):
1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <strong> and </strong> to the argument of the decorator. The return value of the wrapper should look like: return “<strong>” + func() + “</strong>”
2. The decorator will return the wrapper per usual.
3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.
4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.
5. Apply both decorators (by @ operator to greetings()).
6. Invoke the greetings() function and capture the result.
Code :
def strong_decorator(func):
def func_wrapper(name):
return "<strong>{0}</strong>".format(func(name))
return func_wrapper
def em_decorator(func):
def func_wrapper(name):
return "<em>{0}</em>".format(func(name))
return func_wrapper
@strong_decorator
@em_decorator
def Greetings(name):
return "{0}".format(name)
print(Greetings("Hello"))
Explanation:
Answer:
The correct answer is At least two of the stimulants will have different effects on the mean time spent awake.
Explanation:
The null hypothesis exposed leads to consider an alternative hypothesis that differs from the behavior of the 4 stimulants. In this case it is related to the effects on patients, since in practice at least one will present differences compared to the others, influencing the sleep time of consumers.
Answer:
Multiplying impulse response by t ( option D )
Explanation:
We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .
When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .
