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iris [78.8K]
2 years ago
5

VP 3.12.1 Part APart complete A cyclist going around a circular track at 10.0 m/s has a centripetal acceleration of 5.00 m/s2. W

hat is the radius of the curve? Express your answer with the appropriate units. R = 20.0 m Previous Answers Correct VP 3.12.2 Part B A race car is moving at 40.0 m/s around a circular racetrack of radius 265 m. Calculate the period of the motion. Express your answer in seconds. T = nothing s Request Answer Part C Calculate the car’s centripetal acceleration.
Physics
1 answer:
viktelen [127]2 years ago
4 0

Answer:

A) r = 20.0 m

B) T = 41.6 s

C) = 6.1 m/s²

Explanation:

A)

  • The centripetal acceleration is the one that explains that even though the cyclist is moving at a constant speed, his velocity is changing the direction all the time, keeping him around a circle.
  • This acceleration can be expressed as follows:

        a_{c} =\frac{v^{2}}{r} = \frac{(10.0m/s)^{2}}{r} = 5.00 m/s2  (1)

  • Solving for r:

       r = \frac{v^{2}}{a_{c} } = \frac{(10.0m/s)^{2}}{5.00m/s2} = 20.0 m  (2)

B)

  • We can apply the definition of linear velocity, remembering that the period is the time needed to complete an entire circle (T).
  • The arc around a circumference (the distance traveled) , is just 2*π*r, so applying the definition of linear velocity, we can write the following expression:

        v = \frac{\Delta s}{\Delta t} = \frac{2*\pi*r}{T} (3)

  • Solving  for T:

       T = \frac{\Delta s}{v} = \frac{2*\pi*r}{v} = \frac{2*\pi*265m}{40.0m/s} =41.6 s  (4)

C)

  • The centripetal acceleration of the car from B) can be found as follows:

        a_{c} =\frac{v^{2}}{r} = \frac{(40.0m/s)^{2}}{265m} = 6.1 m/s2   (5)

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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

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})

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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}

3)

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

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Ede4ka [16]

When acceleration is constant, the average velocity is given by

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where v and v_0 are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

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where x,x_0 are the final/initial displacements, and t,t_0 are the final/initial times, respectively.

Take the car's starting position to be at t_0=0\,\mathrm s. Then

\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t

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