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postnew [5]
3 years ago
7

If you start skating down this hill, your potential energy will be converted to kinetic energy. At the bottom of the hill, your

kinetic energy will be equal to your potential energy at the top. What will be your speed at the bottom of the hill?
Physics
1 answer:
baherus [9]3 years ago
6 0
Your potential energy at the top of the hill was (mass) x (gravity) x (height) .

Your kinetic energy at the bottom of the hill is (1/2) x (mass) x (speed)² .

If there was no loss of energy on the way down, then your kinetic energy
at the bottom will be equal to your potential energy at the top.

(1/2) x (mass) x (speed)² = (mass) x (gravity) x (height)

Divide each side by 'mass' :

(1/2) x (speed)² = (gravity) x (height) . . . The answer we get
will be the same for every skater, fat or skinny, heavy or light.
The skater's mass doesn't appear in the equation any more.

Multiply each side by 2 :

(speed)² = 2 x (gravity) x (height)

Take the square root of each side:

<u>Speed at the bottom = square root of(2 x gravity x height of the hill)</u>

We could go one step further, since we know the acceleration of gravity on Earth:

Speed at the bottom = 4.43 x square root of (height of the hill)

This is interesting, because it says that a hill twice as high won't give you
twice the speed at the bottom.  The final speed is only proportional to the
<em>square root </em>of the height, so in order to double your speed, you need to
find a hill that's <em>4 times</em> as high.






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Spiral fracture of bone: Spiral fracture of bone occurs due to twisting of the limb, and is a very common skiing accident. The f
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Answer:

principal stresses :б1 = 32.62mPa  б2 = 31.38mPa

Max Shear stress : 16.31 mPa

Orientation of max principle plane = 44.43°

Orientation of minimum principal plane = 134.43°

Explanation:

Given data:

Torque = 50 N-m

weight = 80 kgs

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radius of bone = 10 mm = 0.010 m

<u>a) Determine the principal stresses and shear stress</u>

first calculate the max shear stress ( this will occur in the outermost element

= 16T / π*d^3   where : T = 50 , d = 0.020 m

hence max shear stress = 32 mPa

next determine compressive stress

= ( 40*g)  / π/4*d^2 . where : d = 0.020 m , g = 9.81

hence compressive stress = 1.24 mPa

draw and calculate the radius of Mohr's circle

radius of Mohr's circle = 32.0060

Hence principal stresses = 32.0060 ± 0.62

б1 = 32.62mPa  

б2 = 31.38mPa

attached below is the remaining part of the solution

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Answer:

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It is given that,

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d_x=d\ cos\theta

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