Answer:

Explanation:
Given that:
The direction of the applied tensile stress =[001]
direction of the slip plane = [
01]
normal to the slip plane = [111]
Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:
![cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]](https://tex.z-dn.net/?f=cos%20%5Clambda%20%3D%20%5CBig%20%5B%5Cdfrac%7Bd_1d_2%2Be_1e_2%2Bf_1f_2%7D%7B%5Csqrt%7B%28d_1%5E2%2Be_1%5E2%2Bf_1%5E2%29%2B%28d_2%5E2%2Be_2%5E2%2Bf_2%5E2%29%20%7D%7D%20%5CBig%5D)
where;
= directional indices for tensile stress
= slip direction
replacing their values;
i.e
= 0 ,
= 0
= 1 &
= -1 ,
= 0 ,
= 1
![cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]](https://tex.z-dn.net/?f=cos%20%5Clambda%20%3D%20%5CBig%20%5B%5Cdfrac%7B%280%5Ctimes%20-1%29%2B%280%5Ctimes%200%29%20%2B%20%281%5Ctimes%201%29%20%7D%7B%5Csqrt%7B%280%5E2%2B0%5E2%2B1%5E2%29%2B%28%28-1%29%5E2%2B0%5E2%2B1%5E2%29%20%7D%7D%20%5CBig%5D)

Also, to find the angle
between the stress [001] & normal slip plane [111]
Then;
![cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]](https://tex.z-dn.net/?f=cos%20%5C%20%20%5Cphi%20%3D%20%5CBig%20%5B%5Cdfrac%7Bd_1d_3%2Be_1e_3%2Bf_1f_3%7D%7B%5Csqrt%7B%28d_1%5E2%2Be_1%5E2%2Bf_1%5E2%29%2B%28d_3%5E2%2Be_3%5E2%2Bf_3%5E2%29%20%7D%7D%20%5CBig%5D)
replacing their values;
i.e
= 0 ,
= 0
= 1 &
= 1 ,
= 1 ,
= 1
![cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]](https://tex.z-dn.net/?f=cos%20%5C%20%20%5Cphi%3D%20%5CBig%20%5B%20%5Cdfrac%7B%20%280%20%5Ctimes%201%29%2B%280%20%5Ctimes%201%29%2B%281%20%5Ctimes%201%29%7D%20%7B%5Csqrt%20%7B%280%5E2%2B0%5E2%2B1%5E2%29%2B%281%5E2%2B1%5E2%20%2B1%5E2%29%7D%20%7D%20%5CBig%5D)

However, the critical resolved SS(shear stress)
can be computed using the formula:

where;
applied tensile stress
13.9 MPa
∴


Answer:
1) 4.361 x 10 raised to power 8 revolutions
2) 1.744 x 10 raised to power 9 firings
3) 2.18 x 10 raised to power 8 intake strokes
Explanation:
The step by step explanation is as shown in the attachment
Answer:
1. You have the courage to help without expecting a reward.
2. Because actions are more eloquent than words. Actions are far more valuable and counted than words, and that's how she inspired me.
3. Doing simple things that can make someone grateful and happy without knowing that someone is inspired and motivated by your good deeds, and also doing some interesting things By.
Explanation:
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: