The magnitude of shear stress (kPa) on an element a lying at a distance of 20 mm from neutral axis. (6 pts) are given, p = 2kn e = 210gpa distance from constant cease = 20mm = 0.2m.
<h3>What is the cantilever formula?</h3>
The cantilever beam equations (deflection) w = load. l = member length. e = young's modulus. i = the beam's moment of inertia.
- theta= 2000 2*210*10^ 9 * (0.006)^ four 12 [(0.1)^ 2 -(0.08)^ 2 ] =0.1587 rad
- ø at q i.e. x = 100 - 20 = 80mm = 0.08m from loose cease.
- theta = p/(2ei) * (l ^ 2 - x ^ 2) phase of facet a. i = (a ^ four)/12 for square.
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Answer:
El principio de arquimedes es un principio de la hidroestática que explica lo que experimenta un cuerpo ... la teoría y después como es costumbre pasaremos a ver problemas resueltos sobre éste principio. ... Si la magnitud del peso del cuerpo es menor a la magnitud de empuje. ... Alguien me ayuda con este Ejercicio :.
Explanation:
Answer:
Explanation:
Given that:
From process 1 → 2
![P_1 = 10 bar \\ \\ V_1 = 1 m^3 \\ \\ V_2 = 4 m^3](https://tex.z-dn.net/?f=P_1%20%3D%2010%20bar%20%20%20%5C%5C%20%20%5C%5C%20V_1%20%3D%201%20m%5E3%20%20%5C%5C%20%5C%5C%20%20V_2%20%3D%204%20m%5E3)
![PV^{1.5} = \ constant](https://tex.z-dn.net/?f=PV%5E%7B1.5%7D%20%3D%20%5C%20constant)
![\gamma = 1.5](https://tex.z-dn.net/?f=%5Cgamma%20%3D%201.5)
Process 2 → 3
The volume is constant i.e ![V_2 =V_3 = 4m^3](https://tex.z-dn.net/?f=V_2%20%3DV_3%20%3D%204m%5E3)
![P_3 = 10 \ bar](https://tex.z-dn.net/?f=P_3%20%3D%2010%20%5C%20bar)
Process 3 → 1
P = constant i.e the compression from state 1
Now, to start with 1 → 2
![P_1V_1^{1.5} = P_2V_2^{1.5}](https://tex.z-dn.net/?f=P_1V_1%5E%7B1.5%7D%20%3D%20P_2V_2%5E%7B1.5%7D)
![P_2 = P_1 (\dfrac{V_1}{V_2})^{1.5}](https://tex.z-dn.net/?f=P_2%20%3D%20P_1%20%28%5Cdfrac%7BV_1%7D%7BV_2%7D%29%5E%7B1.5%7D)
![P_2 = 10 \times (\dfrac{1}{4})^{1.5}](https://tex.z-dn.net/?f=P_2%20%3D%2010%20%5Ctimes%20%20%28%5Cdfrac%7B1%7D%7B4%7D%29%5E%7B1.5%7D)
![P_2 =1.25](https://tex.z-dn.net/?f=P_2%20%3D1.25)
The work-done for the process 1 → 2 through adiabatic expansion is:
![W = \dfrac{1}{1-\gamma}[P_2V_2-P_1V_1]](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7B1%7D%7B1-%5Cgamma%7D%5BP_2V_2-P_1V_1%5D)
We know that 1 bar = ![10^5 \ N/m^2](https://tex.z-dn.net/?f=10%5E5%20%5C%20N%2Fm%5E2)
∴
![W = \dfrac{1}{1-1.5}[1.25 \times 10^5 \times 4- 10 \times 10^5 \times 1]](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7B1%7D%7B1-1.5%7D%5B1.25%20%5Ctimes%2010%5E5%20%5Ctimes%204-%2010%20%5Ctimes%2010%5E5%20%5Ctimes%201%5D)
![W =1000000 \ J](https://tex.z-dn.net/?f=W%20%3D1000000%20%5C%20J)
![W_{1 \to 2} = 1000 kJ](https://tex.z-dn.net/?f=W_%7B1%20%5Cto%202%7D%20%3D%201000%20kJ)
For process 2 → 3
Since V is constant
Thus:
W = PΔV = 0
![W_{2 \to 3} = 0](https://tex.z-dn.net/?f=W_%7B2%20%5Cto%203%7D%20%3D%200)
For process 3 → 1
W = PΔV
![W _{3 \to 1} = P_3(V_1-V_3)](https://tex.z-dn.net/?f=W%20_%7B3%20%5Cto%201%7D%20%3D%20P_3%28V_1-V_3%29)
![W _{3 \to 1} = 10 \times 10^5 (1-4)](https://tex.z-dn.net/?f=W%20_%7B3%20%5Cto%201%7D%20%3D%2010%20%5Ctimes%2010%5E5%20%281-4%29)
![W _{3 \to 1} = 10 \times 10^5 (-3)](https://tex.z-dn.net/?f=W%20_%7B3%20%5Cto%201%7D%20%3D%2010%20%5Ctimes%2010%5E5%20%28-3%29)
![W _{3 \to 1} = -3 \times 10^6 \ J](https://tex.z-dn.net/?f=W%20_%7B3%20%5Cto%201%7D%20%3D%20-3%20%5Ctimes%2010%5E6%20%5C%20J)
![W _{3 \to 1} = -3000 \ kJ](https://tex.z-dn.net/?f=W%20_%7B3%20%5Cto%201%7D%20%3D%20-3000%20%20%5C%20kJ)
The net work-done now for the entire system is :
![W_{net} = W_{1 \to 2} + W_{2 \to 3 } + W_{ 3 \to 1 }](https://tex.z-dn.net/?f=W_%7Bnet%7D%20%3D%20W_%7B1%20%5Cto%202%7D%20%2B%20W_%7B2%20%5Cto%203%20%7D%20%2B%20W_%7B%203%20%5Cto%201%20%7D)
![W_{net} = (1000 + 0 + (-3000)) \ kJ](https://tex.z-dn.net/?f=W_%7Bnet%7D%20%3D%20%281000%20%2B%200%20%2B%20%28-3000%29%29%20%5C%20kJ)
![W_{net} =-2000 \ kJ](https://tex.z-dn.net/?f=W_%7Bnet%7D%20%3D-2000%20%5C%20kJ)
The sketch of the processes on p -V coordinates can be found in the image attached below.
Answer:
v ≈ 22.2 mph
θ ≈ 7.8° west of north
Explanation:
vₓ = -3 mph
vᵧ = 22 mph
Finding the speed (magnitude of velocity):
v = √(vₓ² + vᵧ²)
v = √((-3)² + (22)²)
v ≈ 22.2 mph
Finding the direction
θ = atan(vᵧ / vₓ)
θ = atan(22 / -3)
θ ≈ -82.2° or 97.8°
θ is in the second quadrant, so θ ≈ 97.8°, or 7.8° west of north.