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Bess [88]
2 years ago
6

Abigail runs one complete lap (400m) around the track, while Gabi runs a 50 meter dash in a straight line. Which runner had a gr

eater displacement?
Physics
2 answers:
MissTica2 years ago
5 0

Answer:

The distance of one lap on this track from the start line to the finish line is 400 meters. Two laps around the track is 800 meters, half a lap is 200 meters, and so on.

    Remember, displacement is the direction from the starting point and the length of a straight line from the starting point to the ending point. If a runner was going to run 400 meters (one lap around the track), they would start and stop at the same place on the track. Therefore, their displacement would be 0.

 

Explanation:

hope this helped

Ipatiy [6.2K]2 years ago
4 0

Answer:

\boxed {\boxed {\sf Gabi \ had \ greater \ displacement}}

Explanation:

Displacement is the change in the position of an object.

Abigail runs a complete lap around the track, which is 400 meters. Even though she ran 400 meters, she has <u>no displacement.</u> If she starts and ends at the same spot, there is no change in position.

Gabi runs a 50 meter dash in a straight line. Gabi has <u>50 meters of displacement</u>. She runs in a straight line and is 50 meters away from where she began.

While Abigail ran the farther distance, <u>Gabi had the greater displacement.</u>

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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:

1)

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

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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,

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2)

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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}

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

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