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earnstyle [38]
3 years ago
9

What happens in terms of energy when a moving car hit a parked car causing the park card to move?

Physics
2 answers:
olga_2 [115]3 years ago
5 0
Its like newtons 3rd law that once in motion a outer force has to stop it
ad-work [718]3 years ago
5 0

Answer:

The moving car transfers Kinetic Energy to the parked car

Explanation:

Options are not supplied here but I have previously answered this question in a test and these options were supplied.

A) The moving car transfers kinetic energy to the parked car.

B) Kinetic energy in the moving car disappears.

C) Kinetic energy in the parked car is created.

D) The parked car transfers kinetic energy to the moving car.

A) This option provides us with the true outcome, the moving car transfers Kinetic energy to the parked car. This is in agreement with the law of conservation of energy. A transference of energy is the outcome of this event.

B) Kinetic energy cannot disappear, this is because according to the law of conservation of energy, energy can neither be created nor destroyed.

If energy disappears, this would imply that energy is destroyed which is not agreeable.

C) Kinetic energy in the parked car is created... Just like the option B above, and according to the above stated law, energy can neither be created nor destroyed. It is also not agreeable here that energy is created.

D) Energy can be transferred but, "a parked car indicates that the car only has potential energy" and as such cannot transfer kinetic energy to the moving car since it does not possess kinetic energy.

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kirill115 [55]

The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.

The given parameters;

  • time of fall of the first ball, t = 1 s
  • time of fall of the second ball, t = 3 s

The distance traveled by each ball is calculated using the second equation of motion as shown below.

The distance traveled by the first ball is calculated as follows;

h = u_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 1^2)\\\\h = 4.9 \ m

The distance traveled by the second ball is calculated as follows;

h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m

Thus, the second ball traveled a greater distance because it spent more time in motion.

Learn more here:brainly.com/question/5868480

3 0
2 years ago
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Marianna [84]
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8 0
3 years ago
A nasa spacecraft measures the rate r of at which atmospheric pressure on mars decreases with altitude. the result at a certain
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Answer:4.21 \times 10^{-10} J/cm^4

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\Rightarrow r= 0.0421 kPa/km= 0.0421 kPa/km \times \frac{10^{-8} J/cm^4}{1 kPa/km}= 0.0421 \times 10^{-8}J/cm^4=4.21 \times 10^{-10} J/cm^4

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Still go straight but would obviously go up in speed!!





Hope this helps plz mark as brainlist and 5 star
6 0
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alina1380 [7]

Answer:

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Explanation:

a- 9-7= 2cm³

b- 200 divided by 2= 100 g/cm³

Hope this helps... correct me if i'm wrong

7 0
3 years ago
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