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nasty-shy [4]
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.
Physics
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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Answer:

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Explanation:

8 0
3 years ago
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
babunello [35]
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.


Read more on Brainly.com - brainly.com/question/18743384#readmore
8 0
3 years ago
N discussing engines, the ratio of output work to input work expressed as a percentage is called
djyliett [7]

The ration of output work to input work expressed as a percentage is called <u>Efficiency</u>.

5 0
3 years ago
On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the
IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = \frac{1}{2}\varepsilon _{0}E^{2}

          =\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119

          = 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = \frac{B^{2}}{2\mu _{0}}

=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0
3 years ago
A car enters a horizontal, curved roadbed of radius 50 m. the coefficient of static friction between the tires and the roadbed i
UkoKoshka [18]
Previous results tell us the speed (v) is given in terms of the coefficient of friction (k) and the radius of the curve (r) as
  v = √(kgr)
  v = √(0.20·9.8 m/s²·50 m)
  = 7√2 m/s ≈ 9.90 m/s
6 0
3 years ago
Read 2 more answers
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