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2 years ago

11

Develop a model (diagram) that shows how different amounts of gravitational potential energy (GPE) are stored in the earth-ball

system when the ball is raised to different heights on the ramp.

1 answer:

hoa [83]2 years ago

3 0

The gravitational potential energy of the ball can be modelled in

relation to the velocity attained by the ball rolling down the ramp.

A model of the GPE of the ball is as follows;

The height of ball is directly proportional to the square of the velocity the ball reaches while rolling down the ramp. h ∝ v²

<h3>Which is the method used to obtain the model for the GPE of the ball?</h3>

The gravitational potential energy, GPE, is given as follows;

GPE = m·g·Δh

Where;

m = The mass of the ball

g = The acceleration due to gravity

Δh = The change in elevation of the ball

The gravitational energy of the ball at the starting (lowest) part of the

ramp, where Δh = 0, is 0.

When a small amount of energy is applied to the ball by moving it with a

slow speed, the ball moves some distance up the ramp then stops.

When a higher amount of energy is applied to the ball, the ball moves to

a higher point on the ramp.

When energy is applied to the ball, the energy is converted into

gravitational potential energy such that the height the ball reaches,

indicates the amount of gravitational potential energy in the ball.

Based on the model, we have;

Kinetic energy K.E. applied = GPE gained

Which gives;

$\dfrac{1}{2} \cdot m \cdot v^2 = \mathbf{m \cdot g \cdot h}$

$v^2 = 2 \cdot g \cdot h$

$h \propto v^2$

<u>The height of the ball is directly proportional to the square of the velocity</u>

Learn more about gravitational potential energy here:

brainly.com/question/16116226

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vichka [17]

Answer:

Radius r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Explanation:

Area of a circle;

A = πr^2

A = area

r = radius

Making r the subject of formula;

r = √(A/π) ........1

Given;

A = 1300 cm^2

Substituting into the equation 1;

r = √(1300/π)

r = 20.34214472564 cm

r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

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3 years ago

Which of the following statements concerning the lab is TRUE?

Over [174]

Answer:

(b) To get m3 to slide, m1 must be increased, never decreased.

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Lab experiments require attentiveness. If there is one thing missed or not taken seriously whole experiment could go wrong. In this case to slide m3 there should be more weight at m1. If the weight of m1 is lesser than m3 then the object will not slide. It will remain at the point where there is more weight. To slide an object there must be less frictional surface and more weight placed at the desired end point.

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2 years ago

Which objects in space formed from the huge disk cisce and debris beyond the outer planets? Select two option

Dennis_Churaev [7]

Answer:

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3 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

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$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

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Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

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This is in the $3$ points range as given in the figure.

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$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

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3 years ago

Which is the force of repulsion between two positively-charged particles?

klio [65]

<span> answer>>>>electric force <<<<by the way i don't like physics but i answer this for you ^-^</span>

<span>This is because centripetal force is just the net force of a circular motion. There are no attractive or repulsive forces here. This is not the case here. </span>
<span>The gravitational force is a force reliant on mass and attraction of the masses. There are attractive forces here, but not really repulsive forces. </span>
<span>The electric force is the only one that would make sense because it has to do with a relationship between charges and includes both repulsive and attractive forces.</span>

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3 years ago

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