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ivann1987 [24]
4 years ago
7

Which of the following best represents potential energy being converted to kinetic energy? A. A man jogs and stops to drink an e

nergy drink. B. A drawn bow is released, causing an arrow to fly across a field. C. A roller coaster rounds a curve and climbs the next hill. D. A tree is struck by lightning, and then it is set on fire.
Physics
2 answers:
Schach [20]4 years ago
3 0
<span>So we want to know which of the following is the best representation of converting potential energy into kinetic energy. The correct answer is C. A roller coaster rounds a curve to climb the next hill. So before he climbed the hill, the roller coaster had kinetic energy which he used to climb to the hill. Then the potential energy he has on the hill can again be transformed into kinetic energy when he will go down hill. </span>
igomit [66]4 years ago
3 0

Answer;

B.  A drawn bow is released, causing an arrow to fly across the field.

Explanation;

Potential energy is the energy in a body due to its position. While kinetic energy is the energy in a body due to its motion. The formula for potential energy is mgh, where m stands for mass, g stands for gravitational acceleration and h stands for height.

Potential energy is the stored energy that can become kinetic energy. Kinetic energy is given by 1/2mv², where m is the mass and v is the velocity.

Potential energy = Stored energy, usually energy in bonds, energy in height.  

Kinetic energy = Energy of movement

As you draw the bow, the elasticity of the bow stores the energy (potential), and once you release, the arrow flies across the field (kinetic).

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Find the density, in g/cm3, of a metal cube with a mass of 50.3 g and an edge length (l) of 2.65 cm. the volume (v) of a cube is
tamaranim1 [39]
<span>Density is a physical property which describes the mass of a substance per unit of volume of the substance. It is expressed as Density = m / V. We calculate as follows:

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3 years ago
Can someone help me?​
Leviafan [203]

Car X traveled 3d distance in t time.  Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t,  the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

V_y.t = \frac{2d}{t} .t = 2d

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

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