- #1
yyttr2
- 46
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I am doing physics homework and I came upon a problem that I had forgotten how to do, and i do not believe my answer is correct. I have tried to Google this but failed, can you please tell me if i have done this correct?
the question:
"4. Give the horizontal and vertical components of:
a.70m/s West
b. 42m North"
At first I was wondering how do i solve the horizontal and vertical components of a single variable? So i combined them and hoped they were the same problem.
were the 42m is at the right and 70m/s is the base.
I then found the unknown angle showed in the picture (ignore all the symbols up there)
[tex]\phi[/tex]=arctan[tex]\frac{42}{70}[/tex]
[tex]\phi[/tex]=80.54[tex]\circ[/tex]
and I found the length of the hypotenuses.
hy=[tex]\sqrt{(l1)^2+(l2)^2}[/tex]
hy=[tex]\sqrt{(42)^2+(70)^2}[/tex]
hy=[tex]\sqrt{6664}[/tex]
hy=81.6
I then graphed the hypotenuse on a Cartesian Coordinate graph were x= east -x= west
y=north and -y=south
and drew my line 81m at 80.54[tex]\circ[/tex] east of north
and then found my vertical component.
[tex]\frac{sin\phi}{1}[/tex]=[tex]\frac{op}{hy}[/tex]
op(1)=sin[tex]\phi[/tex]hy
op=sin80.54[tex]\circ[/tex]81
op=79.898
vertical component= 79.898
and next my horizontal component.
hy=[tex]\sqrt{(l1)^2+(l2)^2}[/tex]
[tex]hy^{2}[/tex]=[tex]l1^{2}[/tex]+[tex]l2^{2}[/tex]
[tex]hy^{2}[/tex]-[tex]Vc^{2}[/tex]=[tex]Hc^{2}[/tex]
[tex]81^{2}[/tex]-[tex]79.898^{2}[/tex]=[tex]Hc^{2}[/tex]
Hc=[tex]\sqrt{81^2-79.898^2}[/tex]
Hc=13.4
If i have done this wrong please explain to me how, If i have done them right please explain to me which units I should use for the Hc and Vc and why.
the question:
"4. Give the horizontal and vertical components of:
a.70m/s West
b. 42m North"
At first I was wondering how do i solve the horizontal and vertical components of a single variable? So i combined them and hoped they were the same problem.
were the 42m is at the right and 70m/s is the base.
I then found the unknown angle showed in the picture (ignore all the symbols up there)
[tex]\phi[/tex]=arctan[tex]\frac{42}{70}[/tex]
[tex]\phi[/tex]=80.54[tex]\circ[/tex]
and I found the length of the hypotenuses.
hy=[tex]\sqrt{(l1)^2+(l2)^2}[/tex]
hy=[tex]\sqrt{(42)^2+(70)^2}[/tex]
hy=[tex]\sqrt{6664}[/tex]
hy=81.6
I then graphed the hypotenuse on a Cartesian Coordinate graph were x= east -x= west
y=north and -y=south
and drew my line 81m at 80.54[tex]\circ[/tex] east of north
and then found my vertical component.
[tex]\frac{sin\phi}{1}[/tex]=[tex]\frac{op}{hy}[/tex]
op(1)=sin[tex]\phi[/tex]hy
op=sin80.54[tex]\circ[/tex]81
op=79.898
vertical component= 79.898
and next my horizontal component.
hy=[tex]\sqrt{(l1)^2+(l2)^2}[/tex]
[tex]hy^{2}[/tex]=[tex]l1^{2}[/tex]+[tex]l2^{2}[/tex]
[tex]hy^{2}[/tex]-[tex]Vc^{2}[/tex]=[tex]Hc^{2}[/tex]
[tex]81^{2}[/tex]-[tex]79.898^{2}[/tex]=[tex]Hc^{2}[/tex]
Hc=[tex]\sqrt{81^2-79.898^2}[/tex]
Hc=13.4
If i have done this wrong please explain to me how, If i have done them right please explain to me which units I should use for the Hc and Vc and why.