Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque,    \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.  
Maximum Torque, ζ= τ ×  \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 ×  \frac{ π}{16} × d³
800= 11.78 ×  d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
 
        
                    
             
        
        
        
Answer:
i think yes it could make the color go lighter
Explanation:
 
        
                    
             
        
        
        
A torque wrench tool is a tool that ensures that a fastener has the proper amount of tightness.
<h3>What is the torque wrench used for?</h3>
The torque wrench tool is used to ensure screws and bolts are properly tightened. When performing home repairs and maintenance of equipment it is quite important that a torque wrench is used in other to prevent a scenario where a fastener (screws and bolts) does not become loose leading to equipment failure or damage. Because of its many advantages, this tool is often found in the possession of construction workers.
You can learn more about the benefits of a torque wrench tool here 
brainly.com/question/15075481
#SPJ1
 
        
             
        
        
        
Answer:
a) v = +/- 0.323 m/s
b) x = -0.080134 m
c) v = +/- 1.004 m/s
Explanation:
Given:
                              a = - (0.1 + sin(x/b))
b = 0.8
v = 1 m/s @ x = 0
Find:
(a) the velocity of the particle when x = -1 m
(b) the position where the velocity is maximum
(c) the maximum velocity.
Solution:
- We will compute the velocity by integrating a by dt.
                            a = v*dv / dx =  - (0.1 + sin(x/0.8))
- Separate variables:
                            v*dv = - (0.1 + sin(x/0.8)) . dx
-Integrate from v = 1 m/s @ x = 0:
                           0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5
                           0.5v^2 =  0.8cos(x/0.8) - 0.1x - 0.3
- Evaluate @ x = -1
                           0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3
                           v = sqrt (0.104516)
                           v = +/- 0.323 m/s
- v = v_max when a = 0:
                            -0.1 = sin(x/0.8)
                              x = -0.8*0.1002
                              x = -0.080134 m
- Hence,
                             v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134
                             v = sqrt (0.504)
                             v = +/- 1.004 m/s