4x + 4 < 4x + 3 (expand it)
4 < 3 (cancel 4x on both sides)
Since 4 < 3 is not true there is no solution.
Answer: NO SOLUTION.
Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance. Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement. Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m
Answer:
150.6 km
Explanation:
One mile is about 1.61 km so multiply 93.6 by 1.6 which gives you above 150.6
Answer:
Psm = 30.66 [Psig]
Explanation:
To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.
P * v = R * T
where:
P = pressure [Pa]
v = specific volume [m^3/kg]
R = gas constant for air = 0.287 [kJ/kg*K]
T = temperature [K]
<u>For the initial state</u>
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P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)
T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)
Therefore we can calculate the specific volume:
v1 = R*T1 / P1
v1 = (0.287 * 270.4) / 266.8
v1 = 0.29 [m^3/kg]
As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.
V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.
T2 = 43 + 273 = 316 [K]
P2 = R*T2 / V2
P2 = (0.287 * 316) / 0.29
P2 = 312.73 [kPa]
Now calculating the manometric pressure
Psm = 312.73 -101.325 = 211.4 [kPa]
And converting this value to Psig
Psm = 30.66 [Psig]
