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scZoUnD [109]
3 years ago
8

If a roller coaster car had 40,000 J of gravitational potential energy when at rest on the top of a hill how much kinetic energy

does it have when it is 3/4 of the way down the hill
Physics
1 answer:
Inessa [10]3 years ago
6 0

Answer:

K.E = 30,000 J

Explanation:

Given,

The potential energy of the roller coaster car, P.E = 40000 J

The kinetic energy at height h/4, K.E = ?

According to the law of conservation of energy, the total energy of the system is conserved.

At height 'h', the total energy is,

                                    P.E = mgh

                                     K.E = 0

At height 'h/4', the total energy is

                                     P.E + K.E = mgh

                                     P.E = mgh/4

                                     K.E = 1/2 mv²

Therefore,

                                   mgh/4 + 1/2 mv² = mgh

                                    gh/4 + v²/2 = gh

Hence,

                                      v² = 3gh/2

Substituting in the K.E equation

                               K.E = 1/2 mv²

                                      = 1/2 m (3gh/2)

                                       = 3/4 mgh

                                        = 3/4 x 40000

                                         = 30000 J

Hence, the K.E of the roller coaster car is, K.E = 30000 J

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Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

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The graph is attached with the solution. The relation is linear as indicated by the line.

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                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

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Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

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Part b

Pressure variation is given as

                            \Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\

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P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

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