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dlinn [17]
3 years ago
6

A DOHC V-6 has how many camshafts?

Engineering
1 answer:
viktelen [127]3 years ago
4 0

Answer:

two camshafts

Explanation:

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Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f
natita [175]

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:

5 0
4 years ago
Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

Δs_{air} =0.240kJ/kgK

4 0
3 years ago
When you learned about the Highway Transportation system you learned about 6 different types of people as
Musya8 [376]

Answer: 3 different types of people using Highway Transportation system are :

1. Pedestrians - should be given priority while driving.

2. Cyclists - should be given enough space on road.

3. Motorcyclists- should be given enough space on road.

Explanation:

1. Pedestrians - Most of pedestrians use to walk on the footpath along side road that keeps them on a safe side. But there are places without footpath along side road, in that case they have to walk on the road itself. Here, we need to take care for them. We need to wait in case they are crossing road and also check for them while taking a turn.

2. Cyclists - They travel on road but are tough to figure out. They travel at a slower pace compared to cars. To avoid any accident with them, we are supposed to give them enough space which should be equivalent to a car's space.

3. Motorcyclists - They can pass by very closely and also come between lanes. Most of the things to be considered here are same as that of cyclists. Here also, we need to check for them carefully while taking a turn. Also, need to give them enough space.

4 0
3 years ago
Air at T1 = 32°C, p1 = 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate
RUDIKE [14]

Answer:

4.5kg/min

Explanation:

Given parameters

T_1 = 32^0 C,  m_1 = 3 kg/min, T_2 = 7^0 C ,T_3 = 17^0

if we take  

The mass flow rate of the second stream = m_2(kg/min)

The mass flow rate of mixed exit stream = m_3 (kg/min)

Now from mass conservation

m_3 = m_2 + m_1

m_3 = m_2 + 3 (kg/min)

The temperature of the mixed exit stream given as

T_3m_3 = T_2m_2 +T_1m_1\\\\17 ( 3 + m_2) = 7 \times m_2 + 32 \times 3\\\\51 + 17 m_2 = 7 m_2 + 96\\\\10 m_2 = 96 - 51\\\\m_2 = 4.5 kg/min\\\\\\\\

Therefore the mass flow rate of second stream will be 4.5 kg/min.

7 0
3 years ago
What is didactic apparatus?​
Soloha48 [4]
Didactic apparatus is a method of teaching in which scientific approach is follow in order to present the information to the student. This method effectively teaches the student with the required theoretical knowledge .
8 0
3 years ago
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