The ball is moving 60 mph from the observers POV because the ball already has a speed of 50mph and the man is able to throw it at a speed of 10mph
and when you add 50mph and 10mph = 60mph.
Answer:
![A_2 = 2.5\ cm^2](https://tex.z-dn.net/?f=A_2%20%3D%202.5%5C%20cm%5E2)
Explanation:
given,
velocity factor = 4
Cross-sectional area of venules(A₁) = 10 cm²
cross sectional area of capillaries(A_2) = ?
continuity equation = Q = AV
now,
![\dfrac{A_1}{A_2} = \dfrac{V_2}{V_1}](https://tex.z-dn.net/?f=%5Cdfrac%7BA_1%7D%7BA_2%7D%20%3D%20%5Cdfrac%7BV_2%7D%7BV_1%7D)
![\dfrac{10}{A_2} =4](https://tex.z-dn.net/?f=%5Cdfrac%7B10%7D%7BA_2%7D%20%3D4)
![\dfrac{10}{4} =A_2](https://tex.z-dn.net/?f=%5Cdfrac%7B10%7D%7B4%7D%20%3DA_2)
![A_2 = 2.5\ cm^2](https://tex.z-dn.net/?f=A_2%20%3D%202.5%5C%20cm%5E2)
hence, the area of capillaries is equal to 2.5 cm₂
If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick
![T_R=T](https://tex.z-dn.net/?f=T_R%3DT)
whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that