Position of paul with respect to john is given as
14 m due west of john

position of George with respect to Paul is given as 36 m in direction 37 degree south of east

now we need to find the position of George with respect to John
![r_{GJ} = r_G - r_j[\tex]now for the above equation we can add the two equations[tex]r_{Gj} = -14\hat i + 36 cos37\hat i - 36sin37\hat j](https://tex.z-dn.net/?f=r_%7BGJ%7D%20%3D%20r_G%20-%20r_j%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Enow%20for%20the%20above%20equation%20we%20can%20add%20the%20two%20equations%3C%2Fp%3E%3Cp%3E%5Btex%5Dr_%7BGj%7D%20%3D%20-14%5Chat%20i%20%2B%2036%20cos37%5Chat%20i%20-%2036sin37%5Chat%20j)

so the magnitude is given as

and direction is given as

<em>so it is 26.2 m at an angle 55.75 degree South of east</em>
Answer: colder and drier
Explanation:
Trees don’t grow well in wet climates, nor when they are cold. The wider the rings, the healthier the tree is. Since it was narrow and then wide, was bad climate and then good.
Answer:
A. 7.1m
B. 3.55m/s
C. 1.775m/s^2
Explanation:
First step is to identify given parameters;
Ball 1: m₁ = 0.5kg, u (initial velocity) =0, t = 2seconds
Ball 2: m₂ = 0.25kg, u = 15m/s, t = 2seconds
<u>Second step:</u> we determine the y-coordinate of ball 1 after 2 seconds, using the equation of motion under gravity as shown below;



Recall, that the ball was thrown from a height of 25m, total y-coordinate of ball 1 after 2 seconds becomes 25m +(-19.6m)
[tex]y_{1} = 5.4m[/tex]
<u>Third step</u>: we determine the y-coordinate of ball 2 after 2 seconds


<u>Fourth step: </u>we determine the y-component of the center mass of the two balls


y = 7.1m
<u>Fifth step:</u> we solve B part of the question; velocity of the center mass of the two balls


velocity = 3.55m/s
<u>Sixth step:</u> we solve C part of the question; acceleration of the center mass of the two balls


acceleration = 1.775 m/s^2
Answer:
Coefficient of friction is
Ū = 0.31
Explanation:
T2 = T1* e^(ūơ)
Where T2 = 7500n = tension in the belt, T1 = 150n = reaction force,
ū = coefficient of friction
Ơ = 2pai * N
Where N = number of turns = 2
Ơ = 4pai
7500 = 150e^(ū*4pai)
50 = e^(ū *4pai)
lin 50 = 4pai * ū
Ū = 3.91/4pai
Ū = 0.31
the formula for (momentum) P = (force) F * (Mass) M
P = FM
if mass = 0 momentum = 0
Hope this helps...