Asked by apoorva
a velocity vector of 40 degrees below the positive x-axis has a y-component of -10 m/s. What is the value of its x-component.
I know the answer is 11.9 m/s.
When I first tried it, I got 6.43 and then I just tried bunch of stuff and I got 11.9 but what I don't get is why or how I got there.
this is what I did:
I drew the picture.
Since a quadrant has 90 degrees, i assumed the other angle is 50 degrees.
Then for the y component i did:
10cos(50)=6.43
Since the problem told me the other side was 10, I just did square root of 6.43^2+10^2=11.9m/s
Is there another way to get this answer. Since it told me to find the x component, i thought it was just 10cos(50)
I know the answer is 11.9 m/s.
When I first tried it, I got 6.43 and then I just tried bunch of stuff and I got 11.9 but what I don't get is why or how I got there.
this is what I did:
I drew the picture.
Since a quadrant has 90 degrees, i assumed the other angle is 50 degrees.
Then for the y component i did:
10cos(50)=6.43
Since the problem told me the other side was 10, I just did square root of 6.43^2+10^2=11.9m/s
Is there another way to get this answer. Since it told me to find the x component, i thought it was just 10cos(50)
Answers
Answered by
bobpursley
The angle below the x axis is -40. The vertical is -10
-10/x=tan-40
x=-10/tan(-40)=11.8m/s
-10/x=tan-40
x=-10/tan(-40)=11.8m/s
Answered by
CIERRA
BECAUSE THERE FEW PEOPLE THAT KNOW PHYSICS
Answered by
pepe
hi
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