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:
26.7 min
Explanation:
First, we will find the <u>time required to drill each hole</u>:
- N = 300 x 12/0.75 = 1527.7 rev/min
- fr = 1527.7 (0.015) = 22.916 in/min
Formula for <u>distance per hole</u>: 0.5 + A + 1.75
- A = 0.5 (0.75) tan (90-100 / 2) = 0.315 in
- Tm = (0.5 + 0.315 + 1.75) / 22.916 = 0.112 min
Now, we will calculate the <u>time required to draw back the drill form hole</u>:
= 0.112 / 2 = 0.056 min
Time to move between holes = 1.5 / 15 = 0.1 min
For 100 holes, the number of moves between holes = 99
Total time required to drill 100 holes (t):
t = 100 (0.112 + 0.056) + 99 (0.1) = 26.7 min
Answer:
The correct answer is "20.8 kN" and "31 kN". A further explanation is given below.
Explanation:
- The angular touch bearing seems to be a fine replacement while accommodating radial and even some displacement pressures.
- You may receive static as well as dynamic scores from either the manufacturer's collections.
The load ratings should be for the SKF bearing including its predetermined distance:
Static load
= 20.8 kN
Dynamic load
= 31 kN
Answer:
B
Explanation:
A robot's work envelope is its range of movement. It is the shape created when a manipulator reaches forward, backward, up and down. These distances are determined by the length of a robot's arm and the design of its axes. ... A robot can only perform within the confines of this work envelope.