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3 years ago

Diagram B D c с Which car has: Ke = 100 PE=0? * 1 point A B C D

Physics

1 answer:

erma4kov [3.2K]3 years ago

6 0

Answer:

The car C has KE = 100, PE = 0

Explanation:

The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.

The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.

Kinetic energy is the energy that a body possesses due to its motion.

Potential energy is the energy a body possesses due to its position.

From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.

In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.

Therefore, the car C has KE = 100, PE = 0

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3 years ago

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2 years ago

The motion of a particle is defined by the relation x 5 t 3 2 9t 2 1 24t 2 8, where x and t are expressed in inches and seconds,

shutvik [7]

Answer:

a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m

Explanation:

This is an exercise in kinematics, the relationship of position and time is indicated

x = 5 t³ - 9t² -24 t - 8

a) ask when the velocity is zero

speed is defined by

v = $\frac{dx}{dt}$

let's perform the derivative

v = 15 t² - 18t - 24

0 = 15 t² - 18t - 24

let's solve the quadratic equation

$t = \frac{18 \pm \sqrt{18^2 + 4 \ 15 \ 24} }{2 \ 15}$

t = $\frac{18 \pm 42}{30}$

t1 = -0.8 s

t2 = 2.0 s

the time has to be positive therefore the correct answer is t = 2.0 s

b) the position and distance traveled for a = 0

acceleration is defined by

a = dv / dt

a = 30 t - 18

a = 0

30 t = 18

t = 18/30

t = 0.6 s

we substitute this time in the expression of the position

x = 5 0.6³ - 9 0.6² - 24 0.6 - 8

x = 1.08 - 3.24 - 14.4 - 8

x = -24.56 m

we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s

the position for t = 0

x₀ = -8 m

the position for t = 0.6 s

x_f = - 24.56 m

the distance

ΔX = x_f - x₀

Δx = | -24.56 -(-8) |

Δx = 16.56 m

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3 years ago

Diagram B D c с Which car has: Ke = 100 PE=0? * 1 point A B C D​

Diagram B D c с Which car has: Ke = 100 PE=0? * 1 point A B C D