A jet plane is traveling with a velocity of 720 kilometers/hour due north. There is a crosswind of 40 kilometers/hour from the east. Find the resultant velocity and the direction of the jet plane.
A.
720 kilometers/hour in the direction of 3.2° west of north
B.
720 kilometers/hour due north
C.
721 kilometers/hour in the direction of 3.2° west of north
D.
520 kilometers/hour in the direction of 45° east of north
Answers
Answer:
Option c is correct v = 721 km/h θ = 3.2°
Explanation:
In pic is further explanation: A triangle means division
The magnitude of the velocity of the plane can be found by calculating the resultant of the two velocities:
R= [tex]\sqrt{(720km ÷ h^)^2 + (40km ÷ h)^2) = 721.1 km ÷ h}[/tex]
The direction can be found as follows:
0 = tan^-1 (720km ÷ h/40km ÷h) = 86.8
where 720 km/h represents the velocity towards north, while 40 km/h represents the velocity towards west, so the angle is measured with respect to east. Therefore:
the direction is 86.8*north of west
(Hope this helps can I pls have brainlist (crown)1
Answer:
See image
Explanation:
Plato
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Answers
Answer:
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Answer:
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b).
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[tex]{ \rm{1 \: km \: \dashrightarrow \: 1000 \: m}} \\ { \rm{25 \: km \: \dashrightarrow \: (25 \times 1000)}} \\ \\ = { \underline{ \rm{ \: \: 25000 \: m \: \: }}}[/tex]