A 4 kg block is moving at 12 m/s on a horizontal frictionless surface. a constant force is applied such that the block slows wit
1 answer:
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Answer:
(a) Magnitude of Vector = 207.73 m
(b) Direction = 65.48°
Explanation:
(a)
The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:
<u>Magnitude of Vector = 207.73 m</u>
(b)
For the direction of the vector we have the formula:
<u>Direction = 65.48°</u>