The force F has a magnitude of 500 lb and acts along the line AM where M is the midpoint of the vertical side OB of the parallel
epiped. Express F as its magnitude times the appropriate unit vector and then determine its x-, y-, and z-scalar components.
1 answer:
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The amount of Li present to start the reaction is 55.18g
<u>Explanation:</u>
2Li + Br₂ → 2LiBr
Molecular weight of Br₂ = 159.808 g/mol
Mass of Br₂ present = 225 g
Moles of Br₂ present during the reaction = 225 / 159.808
m = 1.4
Molecular weight of LiBr = 86.845 g/mol
Mass of LiBr formed = 690 g
moles of LiBr produced = 690 / 86.845
m(LiBr) = 7.95
According to the balanced equation, 2 molecules of Li reacts to for 2 molecule of LiBr
So, 7.95 moles of LiBr would require 7.95 moles of Li
The molecular weight of Li is 6.941 g/mol
Thus, the amount of lithium present to start the reaction is
Therefore, the amount of Li present to start the reaction is 55.18g
Answer:
(a)10.5 rad/s2
(b) 20.9 rev
(c) 47.27 m
Explanation:
As the block of mass 53 kg is falling and pulling on the rope. The tension force on the rope must be equal to the gravity acting on the block according to Newton's 3rd law
T = mg = 53*9.81 = 519.93 N
Since this tension force would rotate the cylinder freely without any friction. The torque created by this tension force is
To = TR = 519.93 * 0.36 = 187.17 Nm
This solid cylinder would have a moment of inertia around it's rotating axis of:
(a)We can use Newton's 2nd law to calculate the angular acceleration exerted by such torque on the solid cylinder
(b) With such constant angular acceleration, the angle it would make after 5s is
Since each revolution equals to of angle, we can calculate the number of revolution it makes
(c) Assume the thickness of the rope is negligible (and its wounded radius is unchanging), we can calculate the rope length unwinded after rotating 131.3rad