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Answer: So you are dealing with maximum and minimum weights and you want to know what MINIMUM number of supporting strands for this block and tackle system are needed I believe. If so you are dealing with economic imbalances Though we are not worrying about money Right? Right we need physics which Physics study matter and how it moves You would need 8 STRANDS
Explanation: Step By Step
Answer:
(a)
<em>d</em>Q = m<em>d</em>q
<em>d</em>q =
<em>d</em>T
=
(T₂ - T₁)
From the above equations, the underlying assumption is that
remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let
be heat constant of calorimeter
Q₂ =
ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m
ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m
' ΔT
= (16.73 - 6.14) / (15.085 x 3.10)
= 0.22646 KJ mol⁻¹ k⁻¹
Answer:
MRR = 1.984
Explanation:
Given that
Depth of cut ,d=0.105 in
Diameter D= 1 in
Speed V= 105 sfpm
feed f= 0.015 ipr
Now the metal removal rate given as
MRR= 12 f V d
d= depth of cut
V= Speed
f=Feed
MRR= Metal removal rate
By putting the values
MRR= 12 f V d
MRR = 12 x 0.015 x 105 x 0.105
MRR = 1.984
Therefore answer is -
1.944