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Stolb23 [73]
3 years ago
12

Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 44.0mph and half the

distance at 65.0 . On her return trip, she drives half the time at 44.0mph and half the time at 65.0mph .
1)What is Julie's average speed on the way to Grandmother's house?

2)What is her average speed on the return trip?



*Express your answer with the appropriate units.
Physics
1 answer:
sashaice [31]3 years ago
4 0

Answer:

52.47706 mph

54.5 mph

Explanation:

The average speed is given by

V_{av}=\dfrac{Distance}{Time}

V_{av}=\dfrac{100}{\dfrac{50}{44}+\dfrac{50}{65}}\\\Rightarrow V_{av}=52.47706\ mph

Julie's average speed on the way to Grandmother's house is 52.47706 mph

V_{av}=\dfrac{44+65}{2}\\\Rightarrow V_{av}=54.5\ mph

Average speed on the return trip is 54.5 mph

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0.661 s, 5.29 m

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In the x direction:

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A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
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Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

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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

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q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

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R = 1.29 cm

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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}

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The difference between the radial distance  and center can be expressed 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 }]^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

8 0
2 years ago
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