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earnstyle [38]
2 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]2 years ago
5 0
Its like newtons 3rd law that once in motion a outer force has to stop it
ad-work [718]2 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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Preston tossed a red ball upward and it reaches a maximum height of 3.0. What is the final velocity when it returns to prestons
Leokris [45]
That will depend on the units of the 3.0. We need to know if it's 3 feet, 3 yards, 3 meters, or 3 miles. Each one will have a different answer.
5 0
2 years ago
NEED THIS ASAP
horrorfan [7]

Answer:

Energy is essentially work done by an object or on object.

From,

W = Fd

It's directly proportional to mass.

from,

K. E = 1/2mv²

Energy is directly proportional to mass.

P. E = mgh

Energy is directly proportional to mass.

H = mc∆T

Energy is directly proportional to mass.

Thus increasing mass will increase the energy also imparted on another object since all the above eqns show that relationship.

And for 2 moving bodies

K.Ei = K.Ef(energy conservation)

m1u²1 + m2u²2 = m1v²1 + m2v²2

The relationship is the same that the greater mass the greater the impact.

5 0
2 years ago
Help please :pensive:
tino4ka555 [31]

Answer:

0m/s²

Explanation:

Given parameters:

Initial velocity of the boat = 8m/s

Final velocity  = 8m/s

Time taken  = 4s

Unknown:

Acceleration of the boat = ?

Solution:

Acceleration is the rate of change of velocity with time.

It is mathematically expressed as;

        A = \frac{v - u}{t}

A is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

    Insert the parameters and solve;

  A = \frac{8-8}{4}   = 0m/s²

6 0
2 years ago
When water in a lake freezes, the ice that forms floats on top of any water that is still liquid. Why does the ice float?
ddd [48]

Answer:

it floats because the lake is cold at that moment so when part of the lake freezes it still remains solid and floats because of the lake and the surrounding of the lake is still cold

5 0
3 years ago
An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
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