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3 years ago

Where does a rider on a roller coaster get energy?

Physics

2 answers:

Annette [7]3 years ago

5 0

Gravitational potential energy.

As the rider falls, the GPE is converted into kinetic energy.

liraira [26]3 years ago

5 0

He gets it from the huge motor that lifts the cars and people to the top of the first hill and then let's them go.

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Two forces are applied to a car in an effort to accel-

mestny [16]

Answer:

R=2F

Explanation:

As the forces are in same direction so the resultant force will be:

R=F+F

R=2F

4 0

3 years ago

Help! I don’t really know what it’s asking

Misha Larkins [42]

You have to do the math of each and see which one adds up to 66.5

6 0

2 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

It is dangerous to stand near the railway leak when train passing by,why?

Harman [31]

It is dangerous to stand close to railroad tracks when a rapidly moving commuter train passes. Explain why atmospheric pressure would push you toward the moving train. Water pressure inside a hose nozzle can be less than atmospheric pressure due to the Bernoulli effect.

Pls brainliest!? :)

6 0

3 years ago

Distance versus Displacement Worksheet

Umnica [9.8K]

when we find the distance we will add all the blocks so

distance = 6+6+4

distance = 14blocks

when we find the displacement we will add and minus too

As you can read he goes to the south 6 and to north 6 so he leave that place and back to the place again so the displacement is 0. and again he goes to the west 4 blocks so the displacement = 4blocks to the west

6 0

3 years ago

Read 2 more answers