Answer:
Tire rotation is the least likely cause of tire wear. So, the option D is correct.
Explanation:
Step1
Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.
Step2
On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.
Step3
Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.
Step4
Tire rotation has least amount of wear and tear due to no rubbing action. It has less amount surface contact with the surface in rotation.
Thus, tire rotation is the least likely cause of tire wear. So, the option D is correct.
Answer:
(i) 169.68 volt
(ii) 16.90 volt
(iii) 16.90 volt
(iv) 108.07 volt
(v) 2.161 A
Explanation:
Turn ratio is given as 10:1
We have given that input voltage 
(i) We know that peak voltage is give by 
(ii) We know that for transformer 
So 
So peak voltage in secondary will be 16.90 volt
(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary
So peak voltage of the rectifier will be 16.90 volt
(iv) Dc voltage of the rectifier is given by 
(v) Now dc current is given by 
Answer:
Negative feedback
Explanation:
In Biology, negative feedback refers to the counteraction of an effect by its own influence on the process producing it. For instance, the presence of a high level of a particular hormone in the blood may inhibit further secretion of that hormone.
In other words, in negative feedback, the result of a certain action may inhibit further performance of that action
The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².
<h3>What is normal stress?</h3>
If the direction of deformation force is perpendicular to the cross-sectional area of the body, the stress is called normal stress. Changes in wire length and body volume will be normal.
σ = P/A
Where, σ = Normal stress
P = Pressure
A = Area
1 Kg = 9.81 N
800 kg = 7848 N
Since the rod is half bronze and half steel
800 kg = 7848/2
= 3924 N
Pₙ = Fₙ = 3924 N [n = Bronze]
Pₓ = 3924 N [x = steel]
Given,
σₙ = 90MPa
σₓ = 120MPa
Aₙ = ?
Aₓ = ?
Aₙ = Pₙ/σₙ
Aₙ = 3924/90
Aₙ = 43.6 mm²
Aₓ = Pₓ/σₓ
Aₓ = 3924/120
Aₓ = 32.7 mm²
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Answer:
Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible
