Answer:
a) 5.197rev/s
b) Kf/Ki =2.28
Explanation:
a) Angular momentum of the system L = Iw
ButLi=Lf
Kiwi =Ifwf
wf = (Ii/If)will = (4.65/3.4)×3.8=5.197rev/s
b)Kinetic energy KE= 0.5Iw^2
Ki = 0.5Iiwi^2
Kf=0.5Ifwf^2
Kf/Ki = Ifwf/Iiwi
Kf/Ki = (4.65/3.4))(5.197/3.8)
Kf/Ki = 1.22(1.368)^2
Kf/Ki = 2.28
Answer: 6s
Explanation:
Vs=32m/s speed at beginning of slowing down
Vf=0m/s stop speed
a= -6 m/s² acceleration
----------------
Use equation for acceleration :
a=(Vf-Vs)/t
a*t=Vf-Vs
t=(Vf-Vs)/a
t=(0-36)/-6
t=-36/-6
t=6 s
Answer:
aaksj
Explanation:
a) the capacitance is given of a plate capacitor is given by:
C = \epsilon_0*(A/d)
Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:
The plates are squares so their area is given by:
A = L^2 = 0.19^2 = 0.0361 m^2
C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F
b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:
Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C
c) The electric field on a capacitor is given by:
E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]
E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m
d) The energy stored on the capacitor is given by:
W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J
105 miles because you have to use the gif arable
