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AysviL [449]
3 years ago
15

Drivers a and b travel west from boston. driver a leaves three hours earlier traveling at a constant speed of 68mph. driver b fo

llows at a constant speed of 85 mph. how many hours would it take for driver b to catch up to driver a?
Physics
1 answer:
Elza [17]3 years ago
5 0

We can answer this using one of the equations of linear motion:

v = d / t

where:

v = velocity

d = distance

t = time

<span>In the problem, we are asked to find for the time in which Driver B will catch up to Driver A. Therefore,  find the time when dA = dB. Rearranging the equation and equation dA and dB will result in:</span>

<span>vA * tA = vB * tB  ---> 1</span>

It was given that:

vA = 68 mph

tA = tB + 3 (since person A was travelling 3 hours earlier)

vB = 85 mph

tB = unknown

Substituting into equation 1:

68 * (tB + 3) = 85 * tB

68 tB + 204 = 85 tB

tB = 12 hrs

Therefore driver B would catch up to driver A after 12 hrs.

 

 

<span> </span>

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4 0
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Gaseous helium is in thermal equilibrium with liquid helium at 6.4 K. The mass of a helium atom is 6.65 × 10−27 kg and Boltzmann
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Answer:

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Explanation:

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v_{mp}=\sqrt{\frac{2K_bT}{m}}

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v_{mp}=\sqrt{\frac{2\times 1.38066\times 10^{-23} J/K\times 6.4 K}{6.65\times 10^{-27} kg}}

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I don’t think I understand the question but true?
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