Answer
given,
mass of jogger = 67 kg
speed in east direction = 2.3 m/s
mass of jogger 2 = 70 Kg
speed = 1.3 m/s in 61 ° north of east.
jogger one
![P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}](https://tex.z-dn.net/?f=P_2%20%3D%2070%5Ctimes%20v%20cos%20%5Ctheta%20%5Chat%7Bi%7D%20%2B70%5Ctimes%20v%20sin%20%5Ctheta%20%5Chat%7Bj%7D%20)
now
P = P₁ + P₂
magnitude
![P = \sqrt{198.22^2 + 79.59^2}](https://tex.z-dn.net/?f=P%20%3D%20%5Csqrt%7B198.22%5E2%20%2B%2079.59%5E2%7D)
![P =213.60 kg.m/s](https://tex.z-dn.net/?f=P%20%3D213.60%20kg.m%2Fs)
![\theta = tan^{-1}\dfrac{79.59}{198.22}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%5Cdfrac%7B79.59%7D%7B198.22%7D)
![\theta = 21.87](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2021.87)
the angle is
north of east
energy extracted out of liquids an atoms are left to come closer arrange themselves shorter distance and then they solidify
Answer:
τ=0.060 N.m
Explanation:
By kinematics:
![\omega f = \omega o-\alpha*t](https://tex.z-dn.net/?f=%5Comega%20f%20%3D%20%5Comega%20o-%5Calpha%2At)
Solving for α:
![\alpha=\frac{\omega o-\omega f}{t}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7B%5Comega%20o-%5Comega%20f%7D%7Bt%7D)
where ωo = 600*2*π/60; ωf = 0; t=10s
![\alpha=6.283rad/s^2](https://tex.z-dn.net/?f=%5Calpha%3D6.283rad%2Fs%5E2)
The sum of torque is:
![\tau=I*\alpha](https://tex.z-dn.net/?f=%5Ctau%3DI%2A%5Calpha)
![\tau=M*R^2/2*\alpha](https://tex.z-dn.net/?f=%5Ctau%3DM%2AR%5E2%2F2%2A%5Calpha)
![\tau=0.060 N.m](https://tex.z-dn.net/?f=%5Ctau%3D0.060%20N.m)