Was rachel and ross on a break

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 and to the argument of the decorator. The return value of the wrapper should look like: return “” + func() + “”

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 and to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “” + func() + “.

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 "{0}".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "{0}".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

5 0

3 years ago

A contractor is planning on including several skylights in each unit of a residential development. What type of worker would she

Sladkaya [172]

Answer:

Glazier

Explanation:

Glaziers are workers who specializes in cutting and installation of glass works.

They work with glass in various surfaces and settings, such as cutting and installing windows and doors, skylights, storefronts, display cases, mirrors, facades, interior walls, etc.

Thus, the type of worker the contractor will hire for this project is a Glazier

8 0

3 years ago

For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because (select all that are

kotegsom [21]

Answer:

B. The thickness of the heated region near the plate is increasing.

Explanation:

First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.

From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.

3 0

3 years ago

Saturated liquid water flows through 2 cm ID stainless steel tubes at 200 g/s. The water is at 80oC and the inside surface of th

EleoNora [17]

Answer:

steel

Explanation:

8 0

3 years ago

Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi

MariettaO [177]

Answer:

Inner surface temperature= 783K.

Outer surface temperature= 873K

Explanation:

Parameters:

Heat,e= 3×10^6 W/m^3

Inner radius = 0.25 m

Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

5 0

3 years ago