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Nesterboy [21]
3 years ago
8

A 40 kg skier starts at the top of a 12 meter high slope. At the bottom, she is traveling 10 miles. How much energy does she los

e to friction?
Physics
2 answers:
givi [52]3 years ago
8 0
Use conversion of energy. mgh+.5mv^2 initial = final. at the top there's no kinetic energy, at the bottom theres no potential. so mgh at the top=.5mv^2 at the bottom.
mgh at the top = 40 (9.8)(12)
not sure what you mean by 10 miles, do you mean 10 mph or 10 meters/second. if it isnt in SI units, do the proper conversions and put it in m/s
.5mv^2=.5 (40)(final velocity)
DaniilM [7]3 years ago
7 0
2,704 J i believe. Not sure
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Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

v = \frac{m + M}{m} \cdot \sqrt{2gh}

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.   </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}           (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}  

\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}

\frac{v_{2}}{v_{1}} = 1.41  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!

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