Answer:
5.1 Personnel Security. ...
5.2 Physical and Environmental Protection. ...
5.3 Production, Input and Output Controls. ...
5.4 Contingency Planning and Disaster Recovery. ...
5.5 System Configuration Management Controls. ...
5.6 Data Integrity / Validation Controls. ...
5.7 Documentation. ...
5.8 Security Awareness and Training.
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Answer:
(4.5125 * 10^-3 kg.m^2)ω_A^2
Explanation:
solution:
Moments of inertia:
I = mk^2
Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2
Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2
Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2
Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.
r_A=50 mm
r_1 =100 mm
r_2 =50 mm
r_C=150 mm
At the contact point between gears A and B,
r_1*ω_b = r_A*ω_A
ω_b = r_A/r_1*ω_A
= 0.5ω_A
At the contact point between gear B and C.
At the contact point between gears A and B,
r_C*ω_C = r_2*ω_B
ω_C = r_2/r_C*ω_B
= 0.1667ω_A
kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2
=(4.5125 * 10^-3 kg.m^2)ω_A^2