Answer:
number of pulses produced = 162 pulses
Explanation:
give data
radius = 50 mm
encoder produces = 256 pulses per revolution
linear displacement = 200 mm
solution
first we consider here roll shaft encoder on the flat surface without any slipping
we get here now circumference that is
circumference = 2 π r .........1
circumference = 2 × π × 50
circumference = 314.16 mm
so now we get number of pulses produced
number of pulses produced =
× No of pulses per revolution .................2
number of pulses produced =
× 256
number of pulses produced = 162 pulses
Answer:
41.5° C
Explanation:
Given data :
1025 steel
Temperature = 4°C
allowed joint space = 5.4 mm
length of rails = 11.9 m
<u>Determine the highest possible temperature </u>
coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C
Applying thermal strain ( Δl / l ) = ∝ * ΔT
( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )
∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6
hence : T2 = 41.5°C
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