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Marrrta [24]
2 years ago
5

Race cars A and B are driving on the same circular racetrack at the same speed of 45.0 m/s. At a given instant car A is on the n

orth side of the track moving eastward and car B is on the south side of the track moving westward. Find the magnitude of the velocity vector of car A relative to car B.

Physics
1 answer:
Radda [10]2 years ago
4 0

Answer:90 m/s

Explanation:

Given

Speed of both the car is v=45\ m/s

Suppose North side and East side as Positive Y and x axis

so Velocity of car A can be written as

v_a=45\hat{i}

Velocity of car B is

v_b=-45\hat{i}

Velocity of car A relative to car B is given by

v_{ab}=v_a-v_b

v_{ab}=45\hat{i}-(-45\hat{i})

v_{ab}=90\hat{i}

Magnitude of velocity is

|v_{ab}|=90\ m/s

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Answer:

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a) r = 0.76R

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Q = ρ*V(r) = ρ*\frac{4}{3} *\pi *r^{3}

where r = 0.760* R = 0.760* 0.295 m = 0.224 m, and ρ = -151 nC/m³

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Applying Gauss' Law to a spherical gaussian surface of r= 0.76R, as the electric field is radial, and directed inward, we can write the following equation:

E*A = Q/ε₀, where Q= -7.11 nC, A= 4*π*(0.76R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-7.11e-9C}{(0.76*0.295m)^{2}} =-1.27e3 N/C

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b) r= 3.90 R

As this distance falls inside the conducting shell, and no electric field can exist within a conductor in electrostatic condition, E=0

c) r = 2.8 R

As this distance falls between the sphere and the inner radius of the shell, we can calculate the electric field, applying Gauss' law to a gaussian surface of radius equal to r= 2.80 R.

First we need to find the total charge of the sphere, as follows:

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Q = -151e-9 *\frac{4}{3} *\pi *0.295m^{3} = -16.2e-9 C

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E*A = Q/ε₀, where Q= -16.2 nC, A= 4*π*(2.8R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-16.2e-9C}{(2.8*0.295m)^{2}} =-0.21e3 N/C

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As the sphere has a charge of -16.2 nC, and the total charge of the conducting shell is 66.7nC, in order to make E=0 inside the shell, the total charge enclosed by a gaussian surface with a radius larger than the inner radius of the shell and shorter than the outer one, must be zero, which means that a charge of +16.2 nC must be distributed on the inner surface of the shell.

This leaves an excess charge on the outer surface of the shell as follows:

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E*A = Q/ε₀, where Q= 50.5 nC, A= 4*π*(7.3R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{50.5e-9C}{(7.3*0.295m)^{2}} =0.1e3 N/C

⇒ E = 0.1*10⁻³ N/C

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