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
Answer:
So the fraction of proeutectoid cementite is 44.3%
Explanation:
Given that
cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.
We know that
Fraction of proeutectoid cementite phase gievn as

Now by putting the values


So the fraction of proeutectoid cementite is 44.3%
Answer:
The Debye temperature for aluminum is 375.2361 K
Explanation:
Molecular weight of aluminum=26.98 g/mol
T=15 K
The mathematical equation for the specific heat and the absolute temperature is:

Substituting in the expression of the question:


Here

Replacing:

Answer:
drag coefficient for the parachute is Cd = 1.84177
Explanation:
given data
diameter = 8.0 m
average vertical speed = 3 m/s
total weight of load = 200 N
to find out
drag coefficient for the parachute
solution
we will apply here drag coefficient formula that is express as here
drag coefficient Cd =
......................1
here w is total load and A is area and v is speed
and here ρ = 1.22 kg/cu
put here value we get
Cd =
Cd = 1.84177
drag coefficient for the parachute is Cd = 1.84177