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β 12y agoRelative to the starting point, the coordinates of the end point are (10-7, 5) or (3, 5)
So the magnitude of the resultant vector is sqrt(32 + 52) = sqrt(9 + 25) = sqrt(34) = approx 5.83 km.
The bearing (measured from North) is arctan(3/5) = arctan(0.6) = 031 degrees.
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β 12y agoThe resultant velocity of a plane is 75 km/hr.
If the airspeed is maintained at 200 km/hour with a 50 km/hour tailwind, then the speed over ground will be 250 km/hour (resultant velocity).
If two vectors with equal magnitudes 'M' have perpendicular directions, then the resultant ismidway between them ... 45 degrees from each ... and the magnitude of the resultant isM sqrt(2).84 km/hr North + 84 km/hr East = 84 sqrt(2) = 118.794 km/hr Northeast (rounded).
zero, in any direction
4*sin(10) = 0.6945927107 or about 0.7 km
The resultant velocity of the plane relative to the ground can be calculated using vector addition. Given the plane's speed due north (100 km/h) and the crosswind speed (100 km/h westward), use the Pythagorean theorem to find the resultant velocity. The resultant velocity will be 141 km/h at an angle of 45 degrees west of north.
The plane's resultant velocity can be found by adding the vectors of the plane's velocity and the wind's velocity. The plane's velocity to the north is 2.5 x 10^2 km/h, and the wind's velocity toward the southeast is 75 km/h. Using vector addition, the resultant velocity will be a combination of these two velocities.
The resultant velocity can be found using vector addition. The component of velocity in the x-direction is 90 km/h, and in the y-direction is 50 km/h. Using the Pythagorean theorem, the resultant velocity magnitude is β(90^2 + 50^2) km/h = β(8100 + 2500) = β10600 β 103 km/h. The direction of the resultant velocity can be found using trigonometry: tanΞΈ = (50 / 90), so ΞΈ β 30.96 degrees. So, the resultant velocity of the airplane is approximately 103 km/h at an angle of 30.96 degrees from the original direction.
A = 230kmB = 340kmR = ?First, find the horizontal and vertical components of A& B.Ay = 0kmAx= 230kmNote: The angle, theta, is 45°, since NE is 45° from horizontal.By = 340km * sin(45°) = 240.42kmBx = 340km * cos(45°) = 240.42kmNow, add the respective components to get the Rcomponents.Ry = Ay + By = 0km + 240.42km = 240.42kmRx = Ax + Bx = 230km + 240.42km = 470.42kmFinally, use the Pythagorean theorem to get R.R = sqrt(Rx2 + Ry2) = 528.29kmTherefore, the resultant vector, R, is 528.29km.
100 km and 75 km are displacements, NOT velocities. The resultant displacement is 25 km north,
The resultant velocity of a plane is 75 km/hr.
The resultant velocity of a plane is 125 km/hr.
If the airspeed is maintained at 200 km/hour with a 50 km/hour tailwind, then the speed over ground will be 250 km/hour (resultant velocity).
No, km/s is not a vector quantity. It is a scalar quantity that represents speed, which describes how fast an object is moving without specifying its direction.
The resultant velocity of a boat is 17 km/hr and the direction of the boat is SW.
The resultant is 2 km South.
If two vectors with equal magnitudes 'M' have perpendicular directions, then the resultant ismidway between them ... 45 degrees from each ... and the magnitude of the resultant isM sqrt(2).84 km/hr North + 84 km/hr East = 84 sqrt(2) = 118.794 km/hr Northeast (rounded).