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zubka84 [21]
4 years ago
8

A tank contains liquid nitrogen at -190℃ is suspended in a vacuum shell by three stainless steel rods 0.80 cm in diameter and 3

meters long with a thermal conductivity of 16.3 W/m^2-℃. If the ambient air outside the vacuum shell is 15℃, calculate the magnitude of the conductive heat flow in watts along the support rods. A. 0.168 B. 0.0587 C. 0.182 D. 0.176
Engineering
1 answer:
Svetach [21]4 years ago
7 0

Answer:

C) 0.182 W

Explanation:

The effective conductive section is:

3 * A = 3 * \frac{\pi * d^2}{4} = 3 * \frac{\pi * (0.008 m)^2}{4} = 1.5e-4 m^2

The thermal conductivity of stainless steel is

k = 18 w/(m * K)

Heat conduction in a rod follows this equation:

q = k * A * \frac{(t2 - t1)}{l} = 18 * 1.5e-4 * \frac{(15 - (-190))}{3} = 0.182 W

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#198. Moment of inertia about center of a segmented bar A bar of width is formed of three uniform segments with lengths and area
zaharov [31]

Complete Complete

The complete question is shown on the first uploaded image

Answer:

The moment of inertia of the bar about the center of mass is

I_r = 1888.80  \  kg m^2

Explanation:

The free body diagram  is shown on the second uploaded image

From the diagram we see that is

The mass of each segment is

          m_1 = \rho_1  l_1 w = 1 * 6 * 2 = 12

          m_1 = \rho_2  l_2 w = 8 * 6 * 2 = 96

          m_1 = \rho_2  l_2 w = 5 * 5 * 2 = 50

The distance from the origin to the center of the segments i.e the center of masses for the individual segments

   x_2 = \frac{6}{2} + 6 = 9 m

   x_3 = \frac{4}{2} + 12 = 14 m

           

The  resultant center of mass is mathematically evaluated as

              x_r = \frac{m_1 * x_1 + m_2 *x_2 + m_3 *x_3}{m_1 + m_2 + m_3}    

        =   \frac{12 * 3 + 96 *9 + 50 *14}{12+ 96 + 50}

                      x_r = 10.13m        

The moment of Inertia of each segment of the bar is mathematically evaluated

             I_1 =\frac{m_1}{12}(l_1^2 + w^2) =    \frac{12}{12}(1^2 + 2^2)        

                   I_1 = 4 \ kgm^2

             I_2 =\frac{m_2}{12}(l_2^2 + w^2)  =    \frac{96}{12}(6^2 + 2^2)

                 I_2 = 320 \ kgm^2

             I_3 =\frac{m_3}{12}(l_3^2 + w^2)  =    \frac{50}{12}(4^2 + 2^2)        

                   I_2 = 83.334 \ kgm^2        

According to parallel axis theorem the moment of inertia about the center (x_r) is mathematically evaluated as

           I_r = (I_1 + m_1 r_1^2) + (I_2 + m_2 r_2^2) +(I_3 + m_3 r_3^2)

   I_r = (I_1 + m_1 |x_r - x_1|^2) + (I_2 + m_2 |x_r - x_2|^2) +(I_3 + m_3 |x_r - x_3|^2)

   I_r = (4  + 12 |10.13 - 3|^2) + (320 + 96 |10.13 - 9|^2) +(83.334 + 50 |10.13 - 14|^2)        

      I_r = 1888.80  \  kg m^2

6 0
3 years ago
A new approval process is being adapted by Ursa Major Solar. After an opportunity has been approved, the contract is sent to the
BigorU [14]
Install an app :] i think lol
4 0
3 years ago
What is required when setting up a smart phone as a WIFI hotspot?
emmainna [20.7K]
How to create a personal hot spot on an iPhone?

Go to Settings | Cellular | Personal Hotspot.

Tap the slider next to Allow Others to Join. ...

Your Wi-Fi Password will be shown right underneath the Allow Others to Join option. ...

Now, on another device, such as a laptop, go to the Wi-Fi section and search for nearby networks.
5 0
3 years ago
A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 × 106 psi) and an original diameter of 3.7
Keith_Richards [23]

Answer:

the maximum length of specimen before deformation is found to be 235.6 mm

Explanation:

First, we need to find the stress on the cylinder.

Stress = σ = P/A

where,

P = Load = 2000 N

A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4

A = 1.0752 x 10^-5 m²

σ = 2000 N/1.0752 x 10^-5 m²

σ = 186 MPa

Now, we find the strain (∈):

Elastic Modulus = Stress / Strain

E = σ / ∈

∈ = σ / E

∈ = 186 x 10^6 Pa/107 x 10^9 Pa

∈ = 1.74 x 10^-3 mm/mm

Now, we find the original length.

∈ = Elongation/Original Length

Original Length = Elongation/∈

Original Length = 0.41 mm/1.74 x 10^-3

<u>Original Length = 235.6 mm</u>

5 0
3 years ago
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
3 years ago
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