What is the difference in moment and work?
1 answer:
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Answer:
the period of the physical pendulum is 0.498 s
Explanation:
Given the data in the question;
= 0.61 s
we know that, the relationship between T and angular frequency is;
T = 2π/ω ---------- let this be equation 1
Also, the angular frequency of physical pendulum is;
ω = √(mgL / ) ------ let this equation 2
where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and is moment of inertia of rod.
Now, moment of inertia of thin uniform rod D is;
=
mD²
since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.
we substitute equation 2 into equation 1
we have;
T = 2π/ω OR T = 2π/√(mgL/) OR T = 2π√(
/mgL)
so we can use =
mD² for moment of inertia of the rod
Since center of gravity of the uniform rod lies at the center of rod
so that L = D.
now, substituting these equations, the period becomes;
T = 2π/√(/mgL) OR T =
OR T = 2π√(2D/3g ) ----- equation 3
length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),
we have;
ω = 2π/
OR ω
= √(g/D) OR ω
= 2π√( D/g )
so we simple solve for D/g and insert into equation 3
so we have;
T = √(2/3) ×
we substitute in value of
T = √(2/3) × 0.61 s
T = 0.498 s
Therefore, the period of the physical pendulum is 0.498 s
Answer: Hello the missing piece of your question is attached
question : Determine mass of steam that has entered ( in kg )
answer : 0.206 kg
Explanation:
V1 = 0.1 m^3 ,
v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg
V2 = 0.2 m^3
using the steam tables
at ; P = 1000 kPa, v' = 0.167 m^3/kg
U1 = 2321 KJ/kg
at ; P = 1000 kPa , T2 = 280°C
v'2= 0.2481 m^3kg
U2 = 2760.6
at ; P = 5MPa , T = 500°C
h1 = 3434.7 KJ/Kg
calculate final mass ( m2 )
M2 = V2 / v'2
= 0.2 / 0.2481 = 0.806 kg
therefore the mass added = m2 - m1
= 0.806 - 0.6 = 0.206 kg
