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Kisachek [45]
3 years ago
14

A kayaker paddles at 4.0 m/s in a direction 30° south of west. He then turns and paddles at 3.7 m/s in a direction 20° west of s

outh.
What is the magnitude of the kayaker’s resultant velocity? Round your answer to the nearest tenth.

___m/s
What is the direction of the kayaker’s resultant velocity?

___ ° south of west
please help
Physics
2 answers:
user100 [1]3 years ago
7 0
7.2 m/s 

49 degrees south of west

Gre4nikov [31]3 years ago
5 0

Answer: The magnitude of the kayaker's resultant velocity = 7.23 m/s

and the direction is 49.26° south of west.

Explanation:

Initial velocity, u = 4.0 m/s 30° south of west.

Final velocity, v = 3.7 m/s 20° west of south °≈ 70° south of west.

Writing in terms of component:

u = -4.0 cos 30 i - 4.0 sin 30 j = -3.46 i -2.0 j

v = -3.7 cos 70 i - 3.7 sin 70 j = -1.26 i - 3.48 j

Resultant velocity = addition of the two velocity vectors

R = (-3.46 i -2.0 j)+(-1.26 i - 3.48 j) = -4.72 i - 5.48 j

Magnitude of the vector = √(-4.72)²+( - 5.48)² = 7.23 m/s

Direction of the resultant velocity vector, θ = tan⁻¹ (-5.48÷ (-4.72)) = 49.26° south of west.

Thus, the magnitude of the kayaker's resultant velocity = 7.23 m/s

and the direction is 49.26° south of west.

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The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

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