Two speedboats can each travel at V= 5 m/s with respect to water. One boat crosses the
1 answer:
Consider the travel of the speedboat from A to B and back.
Refer to the first figure.
The boat travels at 5 m/s relative to the water, and the downstream speed of the water is 0.5 m/s.
Therefore,
V₁=5 m/s, u = 0.5 m/s, sinθ = 0.5/5 = 0.1 => θ = arcsin(0.1) = 5.74°.
The boat should travel upstream at 5 m/s, at an angle of 5.74°.
Similarly, return speed from B to A is 5 m/s, at 5.74° upstream.
The horizontal component of velocity from A to B (or vice versa) is
5cos(5.74°) = 4.975 m/s.
The time required to travel 1000 m is
1000/4.975 = 202.02 = 3.367 min.
The time for the return trip is
t₁ = 2*3.367 = 6.734 min.
Consider travel from C to D and back, as shown in the second figure.
The resultant velocity upstream from C to D is
V₁ = 5 - 0.5 = 4.5 m/s
The time required to travel 1000 m fromC to D is
1000/4.5 = 222.22 s
The resultant velocity downstream from D to C is
V₂ = 5 + 0.5 = 5.5 m/s
The time required to travel fro D to C is
1000/5.5 = 181.82 s
Total time for the return trip between C and D is
t₂ = 222.22 + 181.82 = 404.04 s = 6.734 min
Answer:
The first boat should travel upstream at an angle of 5.74°. The time for the return trip between A and B is t₁ = 6.734 min or 404.04 s.
The second boat travels upstream against the current, and downstream with the current. The time for the return trip between C and D is t₂ = 6.734 min or 404.04 s.
Refer to the first figure.
The boat travels at 5 m/s relative to the water, and the downstream speed of the water is 0.5 m/s.
Therefore,
V₁=5 m/s, u = 0.5 m/s, sinθ = 0.5/5 = 0.1 => θ = arcsin(0.1) = 5.74°.
The boat should travel upstream at 5 m/s, at an angle of 5.74°.
Similarly, return speed from B to A is 5 m/s, at 5.74° upstream.
The horizontal component of velocity from A to B (or vice versa) is
5cos(5.74°) = 4.975 m/s.
The time required to travel 1000 m is
1000/4.975 = 202.02 = 3.367 min.
The time for the return trip is
t₁ = 2*3.367 = 6.734 min.
Consider travel from C to D and back, as shown in the second figure.
The resultant velocity upstream from C to D is
V₁ = 5 - 0.5 = 4.5 m/s
The time required to travel 1000 m fromC to D is
1000/4.5 = 222.22 s
The resultant velocity downstream from D to C is
V₂ = 5 + 0.5 = 5.5 m/s
The time required to travel fro D to C is
1000/5.5 = 181.82 s
Total time for the return trip between C and D is
t₂ = 222.22 + 181.82 = 404.04 s = 6.734 min
Answer:
The first boat should travel upstream at an angle of 5.74°. The time for the return trip between A and B is t₁ = 6.734 min or 404.04 s.
The second boat travels upstream against the current, and downstream with the current. The time for the return trip between C and D is t₂ = 6.734 min or 404.04 s.
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Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
k = 1.4
Work done is given as;
inlet velocity is negligible;
Therefore, the exit velocity is 629.41 m/s