All categories

3 years ago

As an object falls to the ground, its potential energy is being converted to kinetic energy.

Physics

1 answer:

alekssr [168]3 years ago

5 0

Answer:

<em>The statement is true</em>

Explanation:

<u>Energy Conversion </u>

When an object starts to fall in free air, it speeds up as it falls. The force of gravity acting on the object causes energy to be transferred from its gravitational potential energy to its kinetic energy. We can safely say the height converts to speed and vice-versa. If no external forces act on the system, we can easily calculate heights and speeds by knowing the total mechanical energy (gravitational potential plus kinetic) is conserved.

Answer:

$\boxed{\text{The statement is true}}$

You might be interested in

A skateboarder shoots off a ramp with a velocity of 7.3 m/s, directed at an angle of 60° above the horizontal. The end of the ra

erica [24]

Answer:

y maximum 3.54 m, value X 2.35 m

Explanation:

We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero

Vyf² = Vyo² - 2g (Y-Yo)

Where Yo is the initial height of the ramp 1.5 m

0 = Vyo² -2g (Y-Yo)

Y-Yo = Voy² / 2g

Y = Yo + Voy² / 2g

Let's calculate the velocity components using trigonometry

Voy = vo without T

Vox = Vo cost

Voy = 7.3 sin 60

Vox = 7.3 cos 60

Voy = 6.32 m / s

Vox = 3.65 m / s

Let's calculate the maximum height

Y = 1.5 +6.32²/2 9.8

Y = 3.54 m

This is the maximum height from the ground

b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point

Vfy = Voy - gt

0 = Voy - gt

t = Voy / g

t = 6.32 / 9.8

t = 0.645 s

Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time

X = Vox t

X = 3.65 0.645

X= 2.35 m

8 0

3 years ago

A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.

g100num [7]

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle $=34^{\circ}$

Range=240 m

We know that Range $=\frac{u^2sin2\theta }{g}$

$240=\frac{u^2sin(68)}{9.81}$

u=50.39 m/s

(b)maximum height of projectile is given by

$H=\frac{u^2sin^2\theta }{2g}$

$H=\frac{50.39^2(sin34)^2}{2\times 9.81}$

H=40.46 m

3 0

3 years ago

Read 2 more answers

Please look at the diagram and ans that question

TEA [102]

I can see that they are running away like my dad did

3 0

2 years ago

PLEASE HELP!!! I'm dumb lol.

Len [333]

Answer:

A theory or hypothesis does not necessarily provide an accurate scientific explanation to any topic but predicts what can happen.

Explanation:

7 0

2 years ago

Read 2 more answers

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$