To solve this problem it is necessary to apply the concepts related to internal torque, shear stress and the angle generated from the modulo of elasticity, torque, length and polar moment of inertia.
We will start by finding the angular velocity in units that allow us to find the total torque:
![\omega = 1700rev/min (\frac{2\pi rad}{1rev})(\frac{1min}{60s})](https://tex.z-dn.net/?f=%5Comega%20%3D%201700rev%2Fmin%20%28%5Cfrac%7B2%5Cpi%20rad%7D%7B1rev%7D%29%28%5Cfrac%7B1min%7D%7B60s%7D%29)
![\omega = 56.67\pi rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%2056.67%5Cpi%20rad%2Fs)
From the indicated variable of power, in English system this would be
![P = 2590HP (\frac{550ft \cdot lb/s}{1hp})](https://tex.z-dn.net/?f=P%20%3D%202590HP%20%28%5Cfrac%7B550ft%20%5Ccdot%20lb%2Fs%7D%7B1hp%7D%29)
![P = 1424500ft \cdot lb/s](https://tex.z-dn.net/?f=P%20%3D%201424500ft%20%5Ccdot%20lb%2Fs)
With the power and angular velocity found, the torque would then be:
![T = \frac{P}{\omega}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7BP%7D%7B%5Comega%7D)
![T = \frac{1424500}{56.67\pi}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B1424500%7D%7B56.67%5Cpi%7D)
![T = 8001.27lb\cdot ft](https://tex.z-dn.net/?f=T%20%3D%208001.27lb%5Ccdot%20ft)
The torsion formula will allow us to find maximum stress due to the shear force acting on the body:
![\tau_{max} = \frac{Tc}{J}](https://tex.z-dn.net/?f=%5Ctau_%7Bmax%7D%20%3D%20%5Cfrac%7BTc%7D%7BJ%7D)
![\tau_{max} = \frac{800.27(12)(4)}{\pi/4(4^4-3.625^4)}](https://tex.z-dn.net/?f=%5Ctau_%7Bmax%7D%20%3D%20%5Cfrac%7B800.27%2812%29%284%29%7D%7B%5Cpi%2F4%284%5E4-3.625%5E4%29%7D)
![\tau_{max} = 2.93ksi](https://tex.z-dn.net/?f=%5Ctau_%7Bmax%7D%20%3D%202.93ksi)
And finally the angle of twist
![\phi = \frac{TL}{JG}](https://tex.z-dn.net/?f=%5Cphi%20%3D%20%5Cfrac%7BTL%7D%7BJG%7D)
![\phi =\frac{(8001.27)(12)(100)(12)}{\pi/4(4^4-3.625^4)(4^4-3.625^4)(11)(10^6)}](https://tex.z-dn.net/?f=%5Cphi%20%3D%5Cfrac%7B%288001.27%29%2812%29%28100%29%2812%29%7D%7B%5Cpi%2F4%284%5E4-3.625%5E4%29%284%5E4-3.625%5E4%29%2811%29%2810%5E6%29%7D)
![\phi = 0.08002rad](https://tex.z-dn.net/?f=%5Cphi%20%3D%200.08002rad)
![\phi = 4.58\°](https://tex.z-dn.net/?f=%5Cphi%20%3D%204.58%5C%C2%B0)
Therefore the anle of twist in the shaft at full power is 4.58°
Answer:
y₀ = 10.625 m
Explanation:
For this exercise we will use the kinematic relations, where the upward direction is positive.
y = y₀ + v₀ t - ½ g t²
in the exercise they indicate the initial velocity v₀ = 8 m / s.
when the rock reaches the ground its height is zero
0 = y₀ + v₀ t - ½ g t²
y₀i = -v₀ t + ½ g t²
let's calculate
y₀ = - 8 2.5 + ½ 9.8 2.5²
y₀ = 10.625 m
Answer: a dark (absorption) line
Explanation:
This is as a result of absorption of electromagnetic a specific wavelength. The pattern followed by such lines is characteristic of specific atoms in the path of radiation.
It is important to pay attention to the fundamental units in the given information of physics problems; if you carefully arrange a system so the units cancel out appropriately, the answer is usually correct.
Force is mass x acceleration<span>, </span><span>and the units of force are (Kg/m</span>·s²). In this system, a 10 Kg system is accelerated by 5 m/s². Simply multiply:
10 Kg x 5 m/s² = 50 Kg·m/s² = 50 Newtons
Answer: D Demetrius who sees distortions and colors when he looks at a face.
Explanation: