Answer:
According to Newton's third law, for every action force there is an equal (in size) and opposite (in direction) reaction force. Together, these two forces exerted upon two different objects form the action-reaction force pair.
Explanation:
Sana makatulong ^_^
Answer:
C
Explanation:
Generally, the speed of light slows down when passing through a medium that is not a vacuum. This is not always the case, but I will be ignoring the rare/exotic exceptions. Light has a harder time traveling through solids and liquids than it does with gases.
Both don’t have units beacuse they are ratios
When the substance are moved close together and they move more quickly they get compressed.
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
![Q=Av](https://tex.z-dn.net/?f=Q%3DAv)
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
![Q=1.20 m^3/s](https://tex.z-dn.net/?f=Q%3D1.20%20m%5E3%2Fs)
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
![A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%280.260%20m%29%5E2%3D0.212%20m%5E2)
So we can re-arrange the equation to find the speed of the water:
![v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7BQ%7D%7BA%7D%3D%5Cfrac%7B1.20%20m%5E3%2Fs%7D%7B0.212%20m%5E2%7D%3D5.66%20m%2Fs)
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
![Q_1 = Q_2\\A_1 v_1 = A_2 v_2](https://tex.z-dn.net/?f=Q_1%20%3D%20Q_2%5C%5CA_1%20v_1%20%3D%20A_2%20v_2)
where we have
![A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s](https://tex.z-dn.net/?f=A_1%20%3D%200.212%20m%5E2%5C%5Cv_1%20%3D%205.66%20m%2Fs%5C%5Cv_2%20%3D%203.60%20m%2Fs)
and where
is the cross-sectional area of the pipe at the second point.
Solving for A2,
![A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cfrac%7BA_1%20v_1%7D%7Bv_2%7D%3D%5Cfrac%7B%280.212%20m%5E2%29%285.66%20m%2Fs%29%7D%7B3.60%20m%2Fs%7D%3D0.333%20m%5E2)
And finally we can find the radius of the pipe at that point:
![A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cpi%20r_2%5E2%5C%5Cr_2%20%3D%20%5Csqrt%7B%5Cfrac%7BA_2%7D%7B%5Cpi%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.333%20m%5E2%7D%7B%5Cpi%7D%7D%3D0.326%20m)