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stepladder [879]
4 years ago
9

In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn

er B starts from rest and accelerates at 2.0 m/s^2 for the first 30 m and then runs at constant speed. Runner A befins running as soon as the race begins but B first takes a nap to rest up. What is the longest nap that B can take and still not lose the race?
Physics
1 answer:
kifflom [539]4 years ago
4 0

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

s=ut+\dfrac{1}{2}at_{A}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times1.6\times t_{A}^2

t_{A}=\sqrt{\dfrac{30\times2}{1.6}}

t_{A}=6.12\ sec

We need to calculate the time for B

Using equation of motion

s=ut+\dfrac{1}{2}at_{B}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times2.0\times t_{B}^2

t_{B}=\sqrt{\dfrac{30\times2}{2.0}}

t_{B}=5.48\ sec

We need to calculate the time for longest nap

Using formula for difference of time

t'=t_{A}-t_{B}

t'=6.12-5.48

t'=0.64\ s

Hence, B can take 0.64 sec for the longest nap .

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Two parallel conducting plates are separated by 9.2 cm, and one of them is taken to be at a potential of zero volts.What is the
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Answer:

E=54V/cm

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The equation for change of voltage between two points separated a distance d inside parallel conducting plates (<em>which have between them constant electric field</em>) is:

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So to calculate our electric field strength we use the fact that the potential 8.8 cm from the zero volt plate is 475 V:

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