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fgiga [73]
3 years ago
7

An object falls freely from the rest of three seconds. The acceleration of the object is

Physics
2 answers:
IrinaK [193]3 years ago
8 0
Whenever an object is under free fall motion, assuming that there is no air present in its surrounding, acceleration of object will always be 9.8 m/s²
morpeh [17]3 years ago
6 0
<span>If we drop an object near the surface of the Earth, it falls towards the surface with constant acceleration, g. This means that, as it falls, it speeds up (specifically, its speed increases by 9.8 m/s for each second it is falling). If we measure the distance fallen, we find that the object has dropped 4.9 meters after 1 second, 18.6 m after 2 sec, 44.1 m after 3 sec, etc. The equation for the distance fallen after t seconds is  </span>
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Calculate the acceleration of a skier heading down a 10.0º slope, assuming the coefficient of friction for waxed wood on wet sno
lutik1710 [3]

(a) 0.74 m/s^2

Explanation:

There are two forces acting on the skier: the component of the weight parallel to the slope, which acts downward, and the frictional force, which acts upward along the incline.

The component of the weight parallel to the inclined plane is:

W= m g sin \theta

where m is the mass of the skier, g=9.81 m/s^2 and \theta=10^{\circ}.

The frictional force is instead

F_f = -\mu m g cos \theta

\mu=0.1 is the coefficient of friction for waxed wood on wet snow.

If we apply Newton's second law, we can write that the net force must be equal to the product of mass per acceleration:

mgsin \theta -\mu mg cos \theta =ma

And symplifying m, we can find the acceleration:

a=g sin \theta-\mu g cos \theta=

=(9.81 m/s^2)(sin 10^{\circ})-(0.1)(9.81 m/s^2)(cos 10^{\circ})=0.74 m/s^2


(b) 5.7^{\circ}

Explanation:

This time, the skier is moving at constant velocity. Therefore, the acceleration is zero (a=0) and Newton's second law becomes:

mg sin \theta - \mu m g cos \theta=0

By simplifying, we get

tan \theta = \mu

From which we can find the angle at which the skier could coast at a constant velocity:

\theta= tan^{-1} (\mu) = tan^{-1} (0.1)=5.7^{\circ}



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