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ratelena [41]
3 years ago
9

A computational model predicts the maximum kinetic energy a roller coaster car can have given its mass, the speed at the highest

point of its journey, and the height of the highest point. The mass is 250 kg, and the highest point of the journey is at 30m, where the cart is at rest. What is the maximum kinetic energy?
A. 113500 J
B. 468750 J
C. 0 J
D. 73500 J
Physics
1 answer:
Ymorist [56]3 years ago
3 0

Answer:

D

Explanation:

The decrease in potential energy is equal to the increase in kinetic energy.

mgh

250 x 9.8 x 30

=73, 500

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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
nikitadnepr [17]

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

6 0
3 years ago
Introduction Team Problem 2 You jump into the deep end of Legion Pool and swim to the bottom with a pressure gauge. The gauge at
Kaylis [27]

Answer:

Depth of the pool, h = 4.004 cm

Explanation:

Pressure at the bottom, P = 39240 N/m²

The density of water, d = 1000 kg/m³

The pressure at the bottom is given by :

P = dgh

We need to find the depth of pool. Let h is the depth of the pool. So,

h=\dfrac{P}{dg}

h=\dfrac{39240\ N/m^2}{1000\ kg/m^3\times 9.8\ m/s^2}

h = 4.004 m

So, the pool is 4.004 meters pool. Hence, this is the required solution.

5 0
4 years ago
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