Technician A is correct. Technician B is wrong because a gear's transmission is used to increase or decrease torque.
The relation torque is relying on multiplying the circumferential detail with the resource of the usage of the radius; massive gears experience a greater amount of torque, at the same time as smaller gears experience a great deal much less torque. Similarly, the torque ratio is equal to the ratio of the gears' radii. A gear's transmission torque modifications as it will boom or decreases speed. Commonly, with the resource of the usage of lowering the speed, a small torque on the doorway issue is transferred as a massive torque at the output issue. The calculation of torque is quantified with the resource of the usage of an extensive form of teeth.
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The question is asking whether that statement is true or false. Options are;
A) True
B) False
This is about usage of Swing arm restraints.
<em><u>B) False</u></em>
There are different safety features that people employ when a vehicle is lifted. However, for this question, we will only talk about swing arm restraints.
- Swing arm restraints are lifting restraint devices that are used to prevent a cars arms from shifting or going out of position after that car has been lifted and mounted.
- This swing arm restraint does not prevent a vehicle from falling off a lift as it just helps to ensure that the swing arms that are unloaded basically maintain their position.
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Here’s some of them
6. J
7. I
10. O
13. F
14. E
15. N
Answer:
I forget the word for it, but probably the guys who set up the power lines in the city.
Explanation:
Answer:
the minimum shaft diameter is 35.026 mm
the maximum shaft diameter is 35.042mm
Explanation:
Given data;
D-maximum = 35.020mm and d-minimum = 35.000mm
we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6
so From table, Selection of International Trade Grades metric series
the grade tolerance are;
ΔD = IT7(0.025 mm)
Δd = IT6(0.016 mm)
Also from Table "Fundamental Deviations for Shafts" metric series
Sf = 0.026
so
D-maximum
Dmax = d + Sf + Δd
we substitute
Dmax = 35 + 0.026 + 0.016
Dmax = 35.042 mm
therefore the maximum diameter of shaft is 35.042mm
d-minimum
Dmin = d + Sf
Dmin = 35 + 0.026
Dmin = 35.026 mm
therefore the minimum diameter of shaft is 35.026 mm
