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:
See explaination
Explanation:
2. 0-1 km shear value: taking winds at 1000mb and 850 mb
15 kts south easterly and 50 kts southerly
Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts
3. 0-6 km shear value: taking winds at 1000 mb and 500 mb
15 kts south easterly and 40 kts westerly
Vector difference 135/15 and 270/40 will be 281/51 kts
please see attachment
Answer:

Explanation:
Previous concepts
Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =
MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:
MO = H˙ O
Principle of Angular Impulse and Momentum
The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

Solution to the problem
For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is
".
If we analyze the staritning point we see that the initial velocity can be founded like this:

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

](https://tex.z-dn.net/?f=0%2B%5Csum%20%5Cint_%7B0%7D%5E%7B4%7D%2020t%20%280.15m%29%20dt%20%3D0.46875%20%5Comega%20%2B%2030kg%5B%5Comega%280.15m%29%5D%280.15m%29)
And if we integrate the left part and we simplify the right part we have

And if we solve for
we got:

I attached a photo that explains and gives the answer to your questions. Had to add a border because the whole picture didn’t fit.