Answer:
μ=0.329, 2.671 turns.
Explanation:
(a) ln(T2/T1)=μβ β=angle of contact in radians
take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.
T2=5000 lb and T1=80 lb
we have two full turns which makes total angle of contact=4π radians
μ=ln(T2/T1)/β=(ln(5000/80))/4π
μ=0.329
(b) using the same relation as above we will now compute the angle of contact.
take greater tension as T2 and smaller as T1.
T2=20000 lb T1=80 lb μ=0.329
β=ln(20000/80)/0.329=16.7825 radians
divide the angle of contact by 2π to obtain number of turns.
16.7825/2π =2.671 turns
Answer: um wuh anyways thxs for the points!
Explanation: ....:/
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:
The rate of entropy change of the air is -0.10067kW/K
Explanation:
We'll assume the following
1. It is a steady-flow process;
2. The changes in the kinetic energy and the potential energy are negligible;
3. Lastly, the air is an ideal gas
Energy balance will be required to calculate heat loss;
mh1 + W = mh2 + Q where W = Q.
Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;
Rate of Entropy Change = -Q/T
Where Q = 30Kw
T = Temperature of air = 25°C = 298K
Rate = -30/298
Rate = -0.100671140939597 KW/K
Rate = -0.10067kW/K
Hence, the rate of entropy change of the air is -0.10067kW/K
Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
