<u>Answer:</u>
Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s
Direction of resultant velocity of kayaker = 49.32⁰ South of west.
<u>Explanation:</u>
Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.
First kayaker paddles at 4.0 m/s in a direction 30° south of west, kayaker paddles at 4.0 m/s in a direction 210° anticlockwise from positive horizontal axis.
So velocity of kayaker = 4 cos 210 i + 4 sin 210 j = -3.46 i - 2 j
He then turns and paddles at 3.7 m/s in a direction 20° west of south, kayaker paddles at 3.7 m/s in a direction 250° anticlockwise from positive horizontal axis.
So that velocity = -1.27 i - 3.48 j
So resultant velocity of kayaker = -3.46 i - 2 j +(-1.27 i - 3.48 j) = -4.71 i - 5.48 j
Magnitude of resultant velocity of kayaker = 
Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s
Direction of resultant positive horizontal axis, θ = tan⁻¹(-5.48/-4.71) = 229.32⁰ = 49.32⁰ South of west.
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<span>
I think the the answer would be no. </span> If the instantaneous speed of an object remains constant, then its instantaneous velocity would not change. S<span>ince v = c for c is some constant, then |v| = |c|. This is regardless of the sign of c. Hope this answers the question.</span>
Answer:
<em>We need to (at least) apply a force of 9.8 N to move the block</em>
Explanation:
<u>Second Newton's Law</u>
If a net force
different from zero is applied to an object of mass m, then it will move at an acceleration a, given by

If we apply a force F to an object placed on a rough surface, the only way to make it move is to beat the friction force which is given by

Where
is the static friction coefficient and
is the normal force exerted by the table to the object. Since there is no motion in the vertical direction the normal force equals the weight of the object:

The friction force is

Thus, we need to (at least) apply a force of 9.8 N to move the block