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Degger [83]
3 years ago
15

A 139 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 89 m below

the rim. A student (majoring in linguistics and physics) decides to perform a rescue/experiment using a nearby horse. After lowering a rope to her fallen hero and attaching the other end to the horse, the student measures how long it takes for the horse to pull the fallen physicist to the rim of the Grand Canyon. The acceleration of gravity is 9.8 m/s 2 . If the horse’s output power is truly 1 horsepower (746 W), and no energy is lost to friction, how long should the process take? Answer in units of s.
Physics
1 answer:
Illusion [34]3 years ago
4 0

In order to lift the fat (306 lb) physics professor 89 meters up to
the rim, he'll need more potential energy, equal to

      (mass) x (gravity) x (height) = (139 x 9.8 x 89) = 121,236 joules .

If the faithful horse delivers 1 constant horsepower = 746 watts,
AND if the cute-as-a-button student has instantly figured out a
way to keep the rope sliding around the edge without any friction,
then the soonest Prof. Tubby can arrive at the rim is

    (121,236 joules) / (746 joules/sec) = 162.5 seconds . 

Nowhere in this tense drama has the student needed her linguistics
skill yet, but I'll bet it comes in handy as she attempts gamely to
comprehend all of the various pleadings, prayers, and expletives
uttered by her heavy hero from the time he falls over the rim until
he's again lifted to it.

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Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

Initial speed of the electron, u=3\times 10^5\ m/s

Final speed of the electron, v=7\times 10^5\ m/s

Distance, d = 5 cm = 0.05 m

(a) The acceleration of the electron is calculated using the third equation of motion as :

a=\dfrac{v^2-u^2}{2d}

a=\dfrac{(7\times 10^5)^2-(3\times 10^5)^2}{2\times 0.05}

a=4\times 10^{12}\ m/s^2

Force exerted on the electron is given by :

F=m\times a

F=9.11\times 10^{-31}\times 4\times 10^{12}

F=3.64\times 10^{-18}\ N

(b) Let W is the weight of the electron. It can be calculated as :

W=mg

W=9.11\times 10^{-31}\times 9.8

W=8.92\times 10^{-30}\ N

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\dfrac{F}{W}=\dfrac{3.64\times 10^{-18}}{8.92\times 10^{-30}}

\dfrac{F}{W}=4.08\times 10^{11}

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8 0
4 years ago
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The weight of the bowling ball will balance the buouyant force

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