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labwork [276]
2 years ago
15

You are skiing in ... preparation for a competition. being a dedicated physics student, you happen to have a scale with you. you

notice that the mountainside provides you with 346 n of support force. the snow under your wooden skis is fairly wet. how much friction force acts on your skis as you move down the mountain?
Physics
1 answer:
kolezko [41]2 years ago
4 0

Answer: 34.6N


Explanation:


1) The friction force that acts on the skis as you move down the mountain follows the rule:


Friction force = friction kinetic coefficient × Normal force


2) The friction kinetic coeffcient for waxed wood skis on snow is reported in some literature as 0.10.


3) The normal force is given in the statement: 346 N


4) With that you calculate the friction force as:


Friction force = 0.1 × 346N = 34.6N

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3 years ago
A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the
yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

       x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

       x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

        x = ½ a₁ t²

        x = ½ a₂ (t-4)²

Let's solve

       a₁  t² = a₂ (t-4)²

      a₁/a₂ t² = t² -2 4 t + 16

      t² (1- 2.0 / 4.0) - 8 t +16

      t² 0.5 - 8 t +16 = 0

      t² -16 t + 32 = 0

Let's solve the second degree equation

     t = [16 ±√( 16² - 4 32)] / 2  

     t = ½ (16 ± 11,3)

Solutions

     t1 = 13.66 s

     t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

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2 years ago
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There is not enough information to draw a conclusion about
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2 years ago
In an incline projection the horizontal motion is uniform motion (vx)= constant why?​
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2 years ago
The Hubble Space Telescope was released from the Space Shuttle, which was in a circular orbit at 400km earth altitude. The relat
snow_lady [41]

Answer:

d = (75 i ^ + 93 j ^ + 27 k ^) m ,  d2 = (900 i ^ + 1116 j ^ + 324 k ^) m

Explanation:

The two objects are in circular orbit together, therefore with the same angular velocity, after the launch they move with the relative velocity, so we can use the kinematic relation

       

         v = d / t

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Y axis  

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Z axis

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        z = 0.09 300

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For the time of 1 h

       t2 = 1 h (3600s / 1 h) = 3600

       x2 = 900 m

       y2 = 1116 m

       z2 = 324 m

      d2 = (900 i ^ + 1116 j ^ + 324 k ^) m

5 0
3 years ago
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