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Nesterboy [21]
2 years ago
6

1) what is the potential energy at the end of the loop?

Physics
1 answer:
Agata [3.3K]2 years ago
8 0

At a roller coaster, energy conversions from potential to kinetic energy and back repeatedly several times across the course of a trip. Kinetic energy is the force that an object possesses as a consequence of its movement.

All moving objects possess kinetic energy, which is defined with the mass and speed of the object. Potential energy is the energy that is deposited in an object due to its position corresponding to some zero point. Gravitational potential energy is highest at the top of the looping point of a roller coaster and smallest at the deepest end.

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Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

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densk [106]

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<em><u>~ LadyBrain</u></em>

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3 years ago
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