The main purpose was for power. The vessel has come to symbolize Sweden's Great Power Period, when the nation became a major European power and controlled much of the Baltic.
Answer:
N = 38546.82 rpm
Explanation:
= 150 mm
![A_{1}= \frac{\pi }{4}\times 150^{2}](https://tex.z-dn.net/?f=A_%7B1%7D%3D%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Ctimes%20150%5E%7B2%7D)
= 17671.45 ![mm^{2}](https://tex.z-dn.net/?f=mm%5E%7B2%7D)
= 250 mm
![A_{2}= \frac{\pi }{4}\times 250^{2}](https://tex.z-dn.net/?f=A_%7B2%7D%3D%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Ctimes%20250%5E%7B2%7D)
= 49087.78 ![mm^{2}](https://tex.z-dn.net/?f=mm%5E%7B2%7D)
The centrifugal force acting on the flywheel is fiven by
F = M (
-
) x
------------(1)
Here F = ( -UTS x
+ UCS x
)
Since density, ![\rho = \frac{M}{V}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7BM%7D%7BV%7D)
![\rho = \frac{M}{A\times t}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7BM%7D%7BA%5Ctimes%20t%7D)
![M = \rho \times A\times t](https://tex.z-dn.net/?f=M%20%3D%20%5Crho%20%5Ctimes%20A%5Ctimes%20t)
![M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t](https://tex.z-dn.net/?f=M%20%3D%207100%20%5Ctimes%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Cleft%20%28%20D_%7B2%7D%5E%7B2%7D-D_%7B1%7D%5E%7B2%7D%20%5Cright%20%29%5Ctimes%20t)
![M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37](https://tex.z-dn.net/?f=M%20%3D%207100%20%5Ctimes%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Cleft%20%28%20250%5E%7B2%7D-150%5E%7B2%7D%20%5Cright%20%29%5Ctimes%2037)
![M = 8252963901](https://tex.z-dn.net/?f=M%20%3D%208252963901)
∴
-
= 50 mm
∴ F = ![763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}](https://tex.z-dn.net/?f=763%5Ctimes%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Ctimes%20250%5E%7B2%7D-217%5Ctimes%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Ctimes%20150%5E%7B2%7D)
F = 33618968.38 N --------(2)
Now comparing (1) and (2)
![33618968.38 = 8252963901\times 50\times \omega ^{2}](https://tex.z-dn.net/?f=33618968.38%20%3D%208252963901%5Ctimes%2050%5Ctimes%20%5Comega%20%5E%7B2%7D)
∴ ω = 4036.61
We know
![\omega = \frac{2\pi N}{60}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B2%5Cpi%20N%7D%7B60%7D)
![4036.61 = \frac{2\pi N}{60}](https://tex.z-dn.net/?f=4036.61%20%3D%20%5Cfrac%7B2%5Cpi%20N%7D%7B60%7D)
∴ N = 38546.82 rpm
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:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
Explanation:
Calculation to estimate the upper and lower bounds of the modulus of this composite.
First step is to calculate the maximum modulus for the combined material using this formula
Modulus of Elasticity for mixture
E= EcuVcu+EwVw
Let pug in the formula
E =( 110 x 0.40)+ (407 x 0.60)
E=44+244.2 GPa
E=288.2GPa
Second step is to calculate the combined specific gravity using this formula
p= pcuVcu+pwTw
Let plug in the formula
p = (19.3 x 0.40) + (8.9 x 0.60)
p=7.72+5.34
p=13.06
Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness
UPPER BOUNDS
Using this formula
Upper bounds=E/p
Let plug in the formula
Upper bounds=288.2/13.06
Upper bounds=22.07 GPa
LOWER BOUNDS
Using this formula
Lower bounds=EcuVcu/pcu+EwVw/pw
Let plug in the formula
Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3
Lower bounds=(44/8.9)+(244.2/19.3)
Lower bounds=4.94+12.65
Lower bounds=17.59 GPa
Therefore the Estimated upper and lower bounds of the modulus of this composite will be:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
A protective equipment which protects workers who are passing by from stray sparks or metal while another worker is welding is: E. Welding Screens.
A wielder refers to an individual who is saddled with responsibility of joining two or more metals together by wielding.
During the process of wielding, sparks and minute metallic objects are produced, which are usually hazardous to both the wielder and other workers within the vicinity.
Hence, the following protective equipment are meant to be worn or used directly by a wielder (worker) who is wielding:
However, a protective equipment which protects other workers who are passing by from stray sparks or metallic objects while wielder (worker) is welding is referred to as welding screens.
Find more information: brainly.com/question/15442363