Answer:
maximum isolator stiffness k =1764 kN-m
Explanation:
mean speed of rotation
=65.44 rad/sec
= 0.1*(65.44)^2
F_T =428.36 N
Transmission ratio
also
transmission ratio
SOLVING FOR Wn
Wn = 42 rad/sec
k = m*W^2_n
k = 1000*42^2 = 1764 kN-m
k =1764 kN-m
Answer:
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Explanation:
Answer:
0.2 kcal/mol is the value of for this reaction.
Explanation:
The formula used for is:
where,
= Gibbs free energy for the reaction
= standard Gibbs free energy
R =Universal gas constant
T = temperature
Q = reaction quotient
k = Equilibrium constant
We have :
Reaction quotient of the reaction = Q = 46
Equilibrium constant of reaction = K = 35
Temperature of reaction = T = 25°C = 25 + 273 K = 298 K
R = 1.987 cal/K mol
1 cal = 0.001 kcal
0.2 kcal/mol is the value of for this reaction.
Answer:
(a) 6.91 mm (b) 160 MPa
Explanation:
Solution
Given that:
E = 200 GPa
The rod length = 48 mm
P =P¹ = 6 kN
Recall that,
1 kN = 10^3 N
1 m =10^3 mm
I GPa = 10^9 N/m²
Thus
The rod deformation is stated as follows:
δ = PL/AE-------(1)
σ = P/A----------(2)
Now,
(a) We substitute the values in equation and obtain the following:
48 * 10 ^⁻3 m = (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]
Thus, we simplify
A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²
A =0.0375 * 10 ^⁻3 m²
A =37.5 mm²
A = π/4 d²
Thus,
d² = 4A /π
After inserting the values we have,
d = √37.5 * 4/3.14 mm
= 6.9116 mm
or d = 6.91 mm
Therefore, the smallest that should be used is 6.91 mm
(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)
Thus,
σ = P/A
σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²
σ= 160 MPa
Note: I MPa = 10^6 N/m²
Hence the the corresponding normal stress is σ= 160 MPa