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Verdich [7]
3 years ago
11

Two climbers are on a mountain. Simon, of mass m, is sitting on a snow covered slope that makes an angle θ with the horizontal.

The coefficient of static friction between his body and the snow is μ. He is tied into one end of a massless rope that runs over a frictionless pulley. Joe, of mass M, is at the other end of the rope. He has fallen and is hanging motionless below an overhang. Derive an expression for the maximum value of Joe’s mass M so that Simon is not pulled down the slope, in terms of relevant system parameters.
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0

Answer:

Explanation:

It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.

Let in equilibrium , tension in rope be T

For balancing Joe

T = M g

For balancing Simon

friction + T = mgsinθ

μmgcosθ+T = mgsinθ

μmgcosθ+Mg = mgsinθ

M = (msinθ - μmcosθ)

M = m(sinθ - μcosθ)

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In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro
blondinia [14]
In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass. 
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were. 

Ok, Rameses:

Elementary charge . . . . .  1.6 x 10⁻¹⁹  coulomb
                                        negative on the electron
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Electron rest-mass . . . . .  9.11 x 10⁻³¹  kg


a).  The force between two charges is

      F  =  (9 x 10⁹) Q₁ Q₂ / R²

          =  (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

          =     ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

          =          8.05 x 10⁻⁸  Newton .


b).  Centripetal acceleration  = 

                                               v² / r  .

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                     =      7.7 x 10²²  m/s² .

That's an enormous acceleration ... about  7.85 x 10²¹  G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use  "  F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether  F = M A .

I'm going to leave that step for you.   Good luck !
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Soloha48 [4]
I think this is the answer. I hope you can understand.

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