Answer:
1 pulse rotate = 9 degree
Explanation:
given data
incremental encoder rotating = 15 rpm
wheel holes = 40
solution
we get here first 1 revolution time
as 15 revolution take = 60 second
so 1 revolution take = 
1 revolution take = 4 seconds
and
40 pulse are there for 1 revolution
40 pulse for 360 degree
so 1 pulse rotate is = 
1 pulse rotate = 9 degree
Answer:
In the acid processes, deoxidation can take place in the furnaces, leaving a reasonable time for the inclusions to rise into the sla*g and so be removed before casting. Whereas in the basic furnaces, deoxidation is rarely carried out in the presence of the sla*g, otherwise phosphorus would return to the metal.
Answer:
Engineers can design a train with a regenerative braking system
Explanation:
Assuming the point of the question is that the engineers want to focus on using energy efficiently when starting and stopping, they would likely want to consider a regenerative braking system. Such a system can store energy during braking so that it can be used during starting, reducing the amount of energy that must be supplied by an outside power source.
Answer:
2.379m
Explanation:
The width = 23m
The depth = 3m
The radius is denoted as R
The wetted area is = A
The perimeter perimeter = P
Hydraulic radius
R = A/P
The area of a rectangular channel
= Width multiplied by Depth
A = 23x3
A = 69m²
Perimeter = (2x3)+23
P = 6+23
P= 29
Hydraulic radius R = 69/29
= 2.379m
This answers the question
Thank you!
Answer:
a. Rotational speed of the drill = 375.96 rev/min
b. Feed rate = 75 mm/min
c. Approach allowance = 3.815 mm
d. Cutting time = 0.67 minutes
e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min
Explanation:
Here we have
a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min
b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min
c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm
d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes
e. R = 0.25πD²fr = 9525 mm³/min.
