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Galina-37 [17]
4 years ago
9

an athlete runs 300 m up a hill at a steady speed of 3.0 m/s. She then immediately runs the same distance at 6.0 m/s . What is h

er average speed for 600 m run?
Physics
1 answer:
mina [271]4 years ago
4 0

Answer:

4.0 m/s

Explanation:

In the first part of the run, the athlete runs a distance of

d_1 = 300 m

at a speed of

v_1 = 3.0 m/s

So, the time he/she takes is

t_1 = \frac{d_1}{v_1}=\frac{300}{3.0}=100 s

In the second part of the run, the athlete covers an additional distance of

d_2 = 300 m

with a speed

v_2 = 6.0 m/s

So, the time taken in this second part is

t_2 = \frac{d_2}{v_2}=\frac{300}{6.0}=50 s

So, the total distance covered is

d = 300 m + 300 m = 600 m

And the total time taken

t = 100 s + 50 s = 150 s

Therefore, the average speed for the entire trip is

v=\frac{d}{t}=\frac{600}{150}=4.0 m/s

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Explanation:

a)

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

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For the capacitor in this problem, before insering the dielectric, we have:

C=13.5 \mu F = 13.5\cdot 10^{-6}F is its capacitance

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b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

C'=kC

where

k = 3.55 is the dielectric constant of the material

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Therefore, the energy stored now in the capacitor is:

U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2

where:

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C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

U=3.27\cdot 10^{-3} J

The final energy stored in the capacitor, after the dielectric is inserted, is

U'=11.60\cdot 10^{-3} J

Therefore, the change in energy of the capacitor during the insertion is:

\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J

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Using the relation :

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Force (F) * 0.12 = (500,000 - 100,000)

0.12F = 400,000 J

Force = (400,000 J) / 0.12s

Force = 33333.333

Force = 33,333.33 N

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Part b)

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