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slavikrds [6]
4 years ago
11

Think about your displacement at three different times throughout your day and compare it with the distance you traveled.

Physics
1 answer:
Paul [167]4 years ago
6 0

-- The first thing I do when I wake up is go STRAIGHT to the bathroom.   Up to that time, my displacement is equal to distance I traveled from my bed.

-- Once I'm relaxed and back in my room, dithering around and getting dressed, the distance I've traveled since I woke up is growing and growing, but my displacement is staying pretty steady, because I'm still hanging right around my bed.

-- I walk to school, walk between classes, maybe run around the track a couple times, walk to the lunchroom and back to classes, then walk home.  By dinner time, my distance traveled during the day might be 3 or 4 MILES, but my displacement is only one floor down from my bedroom to the kitchen.

-- After my homework is done, I slide back into my warm bed and turn out the light.  My displacement for the day is now zero !  The straight-line distance from the place I started to the place I finished is zero.

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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;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

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