Answer:
s= 20.4 m
Explanation:
First lets write down equations for each ball:
s=so+vo*t+1/2a_c*t^2
for ball A:
s_a=30+5*t+1/2*9.81*t^2
for ball B:
s_b=20*t-1/2*9.81*t^2
to find time deeded to pass we just put that
s_a = s_b
30+5*t-4.91*t^2=20*t-4.9*t^2
t=2 s
now we just have to put that time in any of those equations an get distance from the ground:
s = 30 + 5*2 -1/2*9.81 *2^2
s= 20.4 m
Answer:
a) P ≥ 22.164 Kips
b) Q = 5.4 Kips
Explanation:
GIven
W = 18 Kips
μ₁ = 0.30
μ₂ = 0.60
a) P = ?
We get F₁ and F₂ as follows:
F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips
F₂ = μ₂*Nef = 0.6*Nef
Then, we apply
∑Fy = 0 (+↑)
Nef*Cos 12º - F₂*Sin 12º = W
⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18
⇒ Nef = 21.09 Kips
Wedge moves if
P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º
⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º
⇒ P ≥ 22.164 Kips
b) For the static equilibrium of base plate
Q = F₁ = 5.4 Kips
We can see the pic shown in order to understand the question.
Answer:
The level of service of of compound grade freeway is LOSB.
Explanation:
Find the provided attachments for explanation
Answer:
1) The exergy of destruction is approximately 456.93 kW
2) The reversible power output is approximately 5456.93 kW
Explanation:
1) The given parameters are;
P₁ = 8 MPa
T₁ = 500°C
From which we have;
s₁ = 6.727 kJ/(kg·K)
h₁ = 3399 kJ/kg
P₂ = 2 MPa
T₂ = 350°C
From which we have;
s₂ = 6.958 kJ/(kg·K)
h₂ = 3138 kJ/kg
P₃ = 2 MPa
T₃ = 500°C
From which we have;
s₃ = 7.434 kJ/(kg·K)
h₃ = 3468 kJ/kg
P₄ = 30 KPa
T₄ = 69.09 C (saturation temperature)
From which we have;
h₄ =
+ x₄×
= 289.229 + 0.97*2335.32 = 2554.49 kJ/kg
s₄ =
+ x₄×
= 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)
The exergy of destruction,
, is given as follows;
= T₀ ×
= T₀ ×
× (s₄ + s₂ - s₁ - s₃)
= T₀ ×
×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)
∴
= 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW
The exergy of destruction ≈ 456.93 kW
2) The reversible power output,
=
+
≈ 5000 + 456.93 kW = 5456.93 kW
The reversible power output ≈ 5456.93 kW.