A constant friction force of 21 N acts on a 63-kg skier for 25 s on level snow. What is the skier’s change in velocity?

1 answer:

erica [24]2 years ago

4 0

The skier’s change in velocity is 8.33.../8.4 m/s.
21/63= 0.33...
0.33...x 25 s = 8.33.../8.4 m/s

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bixtya [17]

Answer:42.43m/s

Explanation:According to vf=vi+at, we can calculate it since v0 equals to 0. vf=0+9.8m/s^2*4.33s= 42.434m/s

4 0

3 months ago

mr_godi [17]

They are related as $\bold{\underline {v}\,.\,\underline a }= \bold{0}$

In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
In mathematics, 2 vectors ( $\underline p$ , $\underline q$ ) that are perpendicular to each other have a quality that their dot product ( $\underline p\,.\, \underline q$ ) equal to zero vector ( $\bold 0$ ) which is written as $\undeline p\,.\, \underline q = \bold 0$ .
This quality can be considered when dealing with the velocity vector and the acceleration vector in a manner $\underline v\,.\, \underline a =\bold 0$ .