Answer:
Explanation:
Force of friction acting on the body = μ mg cosθ
= .4 x 70 x 9.8 x cos30
= 237.63 N
component of weight = mgsinθ
= 70 x 9.8 x sin30
= 343 N
Net upward force = 600 - mgsinθ - μ mg cosθ
= 600 - 343 - 237.63
= 105.37 N
acceleration in upward direction = 105.37 / 70
= 1.5 m /s²
s = ut + 1/2 a t²
= 0 + .5 x 1.5 x 3²
= 6.75 m .
Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 =
ω2 = 216.47
Helium, Neon, and Xenon are all part of the same column on the Periodic Table. Such a column is referred to as a Group, because they have the same number of valence electrons in their outermost shell. Hope this helps!
The capacitive reactance is reduced by a factor of 2.
<h3>Calculation:</h3>
We know the capacitive reactance is given as,

where,
= capacitive reactance
f = frequency
C = capacitance
It is given that frequency is doubled, i.e.,
f' = 2f
To find,
=?




Therefore, the capacitive reactance is reduced by a factor of 2.
I understand the question you are looking for is this:
A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?
- The capacitive reactance is doubled.
- The capacitive reactance is traduced by a factor of 4.
- The capacitive reactance remains constant.
- The capacitive reactance is quadrupled.
- The capacitive reactance is reduced by a factor of 2.
Learn more about capacitive reactance here:
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Answer:
By 16.7% or 0.167 IPM
Explanation:
Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.