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irakobra [83]
3 years ago
8

A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

e surface?
Physics
1 answer:
saveliy_v [14]3 years ago
6 0

Answer :

Velocity will be 3.28\times 10^{-11}m/sec

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere  = =1.6+\frac{1.5}{2}=2.35mm = 0.00235m

Force on electron is given by F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}, here q is charge on sphere and e is charge on electron

F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N

This force work as centripetal force

So F=\frac{mv^2}{r}

4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}

v = =0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec

   

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Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

a=2.92164g

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s

Angular velocity is 1.3823 rad/s

Linear velocity is given by

v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g

a=2.92164g

Centripetal force is given by

F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N

The centripetal force is 2006.29095 N radially outward

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3 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

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V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

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Answer:

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Answer:

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6 0
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