Question

few physics questions

Answer 1

for the 2nd attachement i only ned help on 20

Answer 2

Are you familiar with breaking down vectors into its components given its magnitudes?

Answer 3

From the first attachment

Answer 4

i believe here we are suppose to get ax, ay, and bx, and by. the resultant vector would be the square root of the squares of ax+ay and bx+by

Answer 5

i still keep getting it wrong

Answer 6

Try the root of (ax+bx)^2+(ay+by)^2
In this case ay is 0

Answer 7

i did

Answer 8

should a calculator be in degrees or radians for this?

Answer 9

Degrees. Speaking of which, what angle are you using?

Answer 10

95?

Answer 11

I think you should be using 85!
You can't use obtuse angles for breaking down vectors. This is because we can only apply the property that \(\vec{A}=<|A| \cos(\theta), |A| \sin(\theta)>\) for right triangles

Answer 12

Besides, there are no obtuse angles in a right triangle. 95 is obtuse!

Answer 13

i should use 85 for ax as well?

Answer 14

and also, would any of these values be negative?

Answer 15

Bx will be negative. You can't do that for Ax because the only direction is Ax.

Think of it this way, if you break a bar of candy into 2 pieces, piece 1 and piece 2, but then you state that piece 2 has 0 candy, that must mean that all of the candy is in piece 1! Similarly, we have Ay=0, so the magnitude is the same as Ax!

Proof:
\(\large \vec{A}=<|A| \cos(\theta), |A| \sin(\theta)>\)
We already said that the vector A is completely in the x direction! Therefore the angle is 0!
\(\large \vec{A}=<|A| \cos(0), |A| \sin(0)>=<|A|, 0>\)
\[\large|\vec{A}|=\sqrt{|A|^2+0^2} \rightarrow |\vec{A}|=|A|\]

Answer 16

so bx would be -.4619?

Answer 17

and ax would stay the same at 19.9?

Answer 18

Yes!

Answer 19

So now what's By?

Answer 20

5.279

Answer 21

@IrishBoy123 Does my proof contain enough substance?

@dtan5457 Yep!
Now you need the vector addition A+B. What do you do when you add vectors components?

Answer 22

is the resultant vector 20.14?

Answer 23

i did the rx^2+ry^2 and the sqrt

Answer 24

In magnitude, yes, that's what I got

Answer 25

unfortunetely i got the question wrong so many times i cant check the answer but it seems right

Answer 26

is bx negative cause its technically in quadrant 2?

Answer 27

Yes, you already asked that! But the x-component vector of the vector sum A+B (you should've gotten 19.43) contains a positive value because as you can see in your diagram A+B still points rightward.

Answer 28

alright i get that one

Answer 29

next question?

Answer 30

You want to know that angle that I drew there, ya?

Answer 31

absolutely

Answer 32

oh wait now that i know the resultant vector and by can i just use cos

arccos=19.9/20.14

Answer 33

You want to know that angle that I drew there, ya? We can see that if we break B properly, we can get a right triangle and apply the rule that
\[\huge \tan(\theta)=\frac{ y }{ x }=\frac{ B_y }{ |A| }=\frac{ B_y }{ A_x }\]

Answer 34

Give me a sec here to draw it out