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larisa [96]
3 years ago
12

ListenA bicycle and its rider have a combined mass of 80. kilograms and a speed of 6.0 meters per second. What is the magnitude

of the average force needed to bring the bicycle and its rider to a stop in 4.0 seconds?
a.1.2 × 102 N
b.3.2 × 102 N
c.4.8 × 102 N
d.1.9 × 103 N
Physics
1 answer:
Setler [38]3 years ago
3 0

Answer:

a) 1.2\times 10^2\ N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-6}{4}\\\Rightarrow a=-1.5\ m/s^2

The acceleration of the bicycle and rider is -1.5 m/s²

Force

F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N

The magnitude of the average force needed to bring the bicycle and its rider to a stop is 1.2\times 10^2\ N

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First of all, let's convert the radii of the inner and the outer sphere:
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r_B = 16.5 cm=0.165 m
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C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
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Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
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Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
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E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
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Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
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