Magnitude = √(vx²+vy²)
theta = tan-1(vy/vx)
You can post your answer for a check if you wish.
If Vx= 8.00 units and Vy= -6.40 unis, determine (a)the magnitude and (b)direction of ->V(vector)
b) theta= ______ degrees below the positive x-axis.
3 answers
mag= 10.24
theta= -38.66
theta= -38.66
I suppose you are the original poster, and not someone else. So I will give the answer on that basis. It would actually help if you use the same screen name all the time like most other students.
The magnitude of a vector is adding two orthogonal components (which are perpendicular to each other) to give the scalar value, much the same way as finding the hypotenuse of a right triangle. In fact, the formula is an application of Pythagoras theorem.
Magnitude
= √(vx²+vy²)
=√(8²+(-6.4)²)
=√(64+40.96)
=√(104.98)
=10.24 units
The angle can be visualized if you draw a graph and locate the point (8,-6.4) on the Cartesian plane. It will be below the x-axis, at 8 units to the right of the y-axis. The (negative) angle with the x-axis is the angle sought.
We look for tan(θ)=-6.4/8=0.8
We can find &theta using a calculator or tables to get -38.66°.
Hope this clarifies the answer.
The magnitude of a vector is adding two orthogonal components (which are perpendicular to each other) to give the scalar value, much the same way as finding the hypotenuse of a right triangle. In fact, the formula is an application of Pythagoras theorem.
Magnitude
= √(vx²+vy²)
=√(8²+(-6.4)²)
=√(64+40.96)
=√(104.98)
=10.24 units
The angle can be visualized if you draw a graph and locate the point (8,-6.4) on the Cartesian plane. It will be below the x-axis, at 8 units to the right of the y-axis. The (negative) angle with the x-axis is the angle sought.
We look for tan(θ)=-6.4/8=0.8
We can find &theta using a calculator or tables to get -38.66°.
Hope this clarifies the answer.
1 answer