Answer:
Explanation:
Time period is the reciprocal of frequency
T = 1/F
F = 1/T
but angular frequency w = 2πF
F = w/2π
The detailed steps is as shown in the attached file
Answer: 71.7 KJ
Explanation:
The rotational kinetic energy of a rotating body can be written as follows:
Krot = ½ I ω2
Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:
Fc = m. ac
It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:
ac = ω2 r
We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.
As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.
Replacing in the expression for the Krot, we have:
Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ
There would be very less percentage loss<span> of the kinetic energy during </span>the conversion<span> to internal energy, assuming that there is less air in the </span>surroundings<span>. Also, the friction will contribute to the conversion where if it is, the percentage loses is negligible.</span>
Acceleration of the both masses tied together= 6m/s²
Explanation:
The force is given by F= ma
so 5= m1 (8)
m1=0.625 Kg
for m2
5=m2 (24)
m2=0.208 kg
now total mass= m1+m2=0.625+0.208
Total mass=M=0.833 Kg
now F= ma
5= 0.833 (a)
a= 5/0.833
a=6m/s²
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
