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MrRissso [65]
3 years ago
5

An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displace

ment at the end of 2 minutes 20 s?
Physics
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

Explanation:

Displacement is the shortest distance possible between initial and final position.

As the athlete is in circular motion , displacement is zero. [ THis is because , the initial and final position is the same ]

Distance covered in 1 round = Circumference of the circle

Circumference of the circle = 2πr

Diameter = 200 m

Radius = 100 m

Distance covered in 1 round[ 40 sec ] = 2 × 22/7 × 100 = 628.57 m

Distance covered in 1 sec = 628.57 ÷ 40 = 15.71 m

2min 20 sec = 140 sec

Distance covered in 140 sec = 15.71 × 140 = 2199.4 m

For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.

At the end of his motion, the athlete will be in the diametrically opposite position. That is, displacement = diameter = 200 m.

Hence, the distance covered is 2200 m and the displacement is 200 m.

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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
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}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

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}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

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
8 0
3 years ago
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it
fomenos

Answer:68.15m/s

Explanation:

<u><em>Given: </em></u>

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

<u><em>Formula:</em></u>

v₁²=v₁²+2a (x)

<u>Set up:</u>

=\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)

<h2><u><em>Solution:</em></u></h2><h2><u><em>68.15m/s</em></u></h2>

<u />

6 0
3 years ago
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