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jolli1 [7]
3 years ago
6

A skier starting from rest skis straight down a slope 50 meters long in 5.0 seconds. What is the magnitude of the acceleration.

Physics
1 answer:
3241004551 [841]3 years ago
6 0
His average speed all the way is

                   (50 meters) / (5 sec)  =  10 m/s .

But, if the acceleration is uniform all the way down, then

            Average speed  =  (1/2) (start speed + end speed)

Start speed  =  0.
So
             Average speed  =  (1/2) (0 + end speed)

                                     =  (1/2) of end speed .

                   10 m/s        = (1/2) of end speed

               End speed (at the bottom)  =  20 m/s .

Magnitude of acceleration = (change in speed) / (time for the change)

                                      =     (20 m/s)            /    (5 sec)

                                       =                4 m/s²  .
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<h3><u>Answer;</u></h3>

= 20.436 seconds

<h3><u>Explanation;</u></h3>

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Therefore;

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An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

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Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

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