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3 years ago

At which point is the kinetic energy of the roller coaster at its maximum, and the potential energy at its minimum?

Physics

2 answers:

sesenic [268]3 years ago

7 0

A point b cause at a it’s gaining kinetic energy and at b it has the most kinetic energy

vitfil [10]3 years ago

5 0

It is point c because at this point all potential energy has turned into kinetic energy.( do not be mistake and say b. The answer is C)

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3 years ago

You are at the edge of a diving board that is 9 meters above the water. If you weigh 500 Newtons, what is your potential energy?

Semenov [28]

Answer:

4500 J

Explanation:

First, let's define some equations and derivations.

Our potential energy formula is:

$\displaystyle U = mgh$

Where <em>m </em>is mass (in kg), <em>g</em> is the gravitational constant (in m/s²), and <em>h</em> is height (in m).

We also know that <em>mg</em> is equal to the weight of an object (in N), from Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration).

Therefore, we can simply substitute force into the equation:

$\displaystyle U = Fh$

Where <em>F</em> is the force (in N) and <em>h</em> is still height (in m).

Now we can calculate the amount of potential energy in our system, measured in joules.

Substitute in the given variables, F = 500 N and h = 9 m:

$\displaystyle U = (500 \ N)(9 \ m)$

Using simple Pre-Algebra rules, we find that:

$\displaystyle U = 4500 \ J$

This tells us that the we have 4500 joules of potential energy when I am 9 meters above the water on the edge of the diving board.

6 0

3 years ago

Read 2 more answers

A baseball is batted from a height of 1.09 m with a speed of

kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0

2 years ago