Physics, 10.12.2019 03:31 deepunalli300p3ur3i

10. a girl coasts down a hill on a sled, reaching the bottom with a speed of 7.0 m/s. the coefficient of friction between the sled's runners
and the snow is 0.05, and the girl and sled together weigh 645 n. how far does the sled travel on the flat surface before coming to rest?

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Answer from: 2000070489

distance traveled before coming to rest is: 50 meters

Explanation:

As the girl+sled reach the bottom of the hill and start the horizontal travel with a speed of 7 m/s, there are three forces acting on them (please see attached free body diagram):

1) The weight "w" pointing down and known to be 645 N,

2) the normal force "n" of the snow on the sled+girl system which equals and counteracts the weight since there is no vertical displacement any more,

and 3) the force of friction with the snow, which we can calculate from the formula: $F_f=\mu\,*\,n=0.05*645\,N=32.25\,N$

Therefore, the net resultant force for this system is that of the force of friction: 32.25 N acting to stop the sled. Since this is the net force, we can estimate what (negative) acceleration is being imparted on the system via Newton's second law (F=m*a) if we know the girl+sled mass.

We can determine such knowing that the weight of girl+sled is 645 N:

mass of girl + sled = $\frac{645}{g} \,kg$

which results in kilograms given that all other quantities are in SI units.

Now we use it to estimate the acceleration that the friction is imparting on the sled+girl system to stop it: $a=32.25 / \frac{645}{g} =\frac{32.25\ *\ g}{645} \,\frac{m}{s^2} = 0.49\,\frac{m}{s^2}$

Now that we know the acceleration acting (remember that it is negative - acting against the initial travelling velocity), we can use kinematic equations for an accelerated movement:

$v_f^2-v_i^2=2\,a\,(x_f-x_i)$

We know that the initial velocity is 7 m/s and the final (when the sled stops) is zero, we also have now the negative acceleration exerted by the friction force: "-0.49 $\frac{m}{s^2}$ ", and (x_f-x_i) is the distance traveled which is what we want to find. So, we solve for this distance:

$v_f^2-v_i^2=2\,a\,(x_f-x_i)\\\frac{v_f^2-v_i^2}{2\,a} = (x_f-x_i)\\\frac{0-7^2}{2(-0.49)} =(x_f-x_i)\\50\,m=(x_f-x_i)$

And we know that the answer comes in meters since all quantities used in the formula were given in SI units.

10. a girl coasts down a hill on a sled, reaching the bottom with a speed of 7.0 m/s. the coefficien

Answer from: Quest

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Answer from: Quest

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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