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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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Viktor [21]

Answer: c Delta

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3 0
3 years ago
A student of mass 63.4 kg, starting at rest, slides down a slide 21.2 m long, tilted at an angle of 26.1° with respect to thehor
Vera_Pavlovna [14]

Answer:

fr = 65.46 N ,   a = 8.74 m / s²  and  vf = 19.25 m / s

Explanation:

We write a reference system with an axis parallel to the slide and gold perpendicular axis, in this system we decompose the weight

      sin 21.2 = Wx / W

      cos21.2 = Wy / W

      Wx = W sin21.2

      Wy = W cos 21.2

We form Newton's equations

X axis

      Wx -fr = m a

Y Axis

     N- Wy = 0

     N = Wy

     fr = μ N

     fr = μ (W cos 21.2)

     fr = 0.113 63.4 9.8 cos 21.2

     fr = 65.46 N

We replace and calculate the acceleration

     W sin 21.2 - μ W cos 21.2 = m a

     a = g (sin21.2 - μ cos 21.2)

     a = 9.8 (without 21.2 - 0.113 cos 21.2)

     a = 8.74 m / s²

This acceleration is along the slope of the slide, so we can calculate the distance

     d = 21.2 m

     vf² = v₀² + 2 a d

    vf² = 0 + 2 a d

    vf = √(2 8.74 21.2)

    vf = √ (370,576)

    vf = 19.25 m / s

3 0
3 years ago
How long would it take an object to stop if it is moving at 15m/s and has an acceleration -0.5m/s^2? Please show your work.
elena-14-01-66 [18.8K]

Time = 15/-0.5 = 14.5/-0.5 = 14.5s

The answer is 14.5s

Sorry no one would answer it sooner.

3 0
3 years ago
Gold has a specific heat of 0.130 J/g*C. If 195 joules of heat are added to 15 grams of gold how much does the temperature of th
Anna71 [15]

The correct answer to the question is :  100\ ^0C

EXPLANATION :

As per the question, the specific heat of gold is given as c = 0.130\ J/g^0C

The heat given to the gold dQ = 195 J

The mass of the gold is given as m = 15 gram.

We are asked to calculate the change in temperature.

Let the change in temperature is dT.

We know that dQ = mcdT

                      dT=\ \frac{dQ}{mc}

                             =\ \frac{195}{15\times 0.130}

                             =\ 100\ ^0C                   [ANS]

Hence, the change in temperature is 100 degree celsius.

5 0
4 years ago
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