Answer:
a. 1.91 b. -8.13 mm
Explanation:
Modulus =stress/strain; calculating stress =F/A, hence determine the strain
Poisson's ratio =(change in diameter/diameter)/strain
The following scenarios are pertinent to driving conditions that one may encounter. See the following rules of driving.
<h3>What do you do when the car is forced into the guardrail?</h3>
Best response:
- I'll keep my hands on the wheel and slow down gradually.
- The reason I keep my hands on the steering wheel is to avoid losing control.
- This will allow me to slowly back away from the guard rail.
- The next phase is to gradually return to the fast lane.
- Slamming on the brakes at this moment would result in a collision with the car behind.
Scenario 2: When driving on a wet road and the car begins to slide
Best response:
- It is not advised to accelerate.
- Pumping the brakes is not recommended.
- Even lightly depressing and holding down the brake pedal is not recommended.
- The best thing to do is take one foot off the gas pedal.
- There should be no severe twists at this time.
Scenario 3: When you are in slow traffic and you hear the siren of an ambulance behind
Best response:
- The best thing to do at this moment is to go to the right side of the lane and come to a complete stop.
- This helps to keep the patient in the ambulance alive.
- It also provide a clear path for the ambulance.
- Moving to the left is NOT recommended.
- This will exacerbate the situation. If there is no place to park on the right shoulder of the road, it is preferable to stay in the lane.
Learn more about rules of driving. at;
brainly.com/question/8384066
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the required documents is 3000
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:
the torque capacity is 30316.369 lb-in
Explanation:
Given data
OD = 9 in
ID = 7 in
coefficient of friction = 0.2
maximum pressure = 1.5 in-kip = 1500 lb
To find out
the torque capacity using the uniform-pressure assumption.
Solution
We know the the torque formula for uniform pressure theory is
torque = 2/3 × 
× coefficient of friction × maximum pressure ( R³ - r³ ) .....................................1
here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in
now put all these value R, r, coefficient of friction and maximum pressure in equation 1 and we will get here torque
torque = 2/3 × × 0.2 × 1500 ( 4.5³ - 3.5³ )
so the torque = 30316.369 lb-in
