Answer:
Water leaves the sprinkler at a speed of 2.322 m/sec
Explanation:
We have given internal diameter of the garden ![d_1=1.1cm](https://tex.z-dn.net/?f=d_1%3D1.1cm)
Speed of water in the hose is ![v_1=0.95m/sec](https://tex.z-dn.net/?f=v_1%3D0.95m%2Fsec)
Number of holes n = 22
Diameter of each holes ![d_2=15cm](https://tex.z-dn.net/?f=d_2%3D15cm)
According to continuity equation ![A_1v_1=A_2v_2](https://tex.z-dn.net/?f=A_1v_1%3DA_2v_2)
![d_1^2\times v_1=22\times d_2^2v_2](https://tex.z-dn.net/?f=d_1%5E2%5Ctimes%20v_1%3D22%5Ctimes%20d_2%5E2v_2)
![1.1^2\times 0.95=22\times 0.15^2\times v_2](https://tex.z-dn.net/?f=1.1%5E2%5Ctimes%200.95%3D22%5Ctimes%200.15%5E2%5Ctimes%20v_2)
![v_2=2.322m/sec](https://tex.z-dn.net/?f=v_2%3D2.322m%2Fsec)
So water leaves the sprinkler at a speed of 2.322 m/sec
Answer:
C. Talk test
Explanation:
The talk test would be readily available to me at the the least cost.
The talk test is about the easiest way that one can monitor intensity as they exercise. Because the only thing needed here is the ability to talk and to breathe.
The intensity lies on if one can talk and breathe at the same time. The harder one exercises, the more breathless they become and they find it difficult to talk.
Acceleration= change in velocity/time=27.8-0/4.8=5.792 ms⁻²
Force= Mass*acceleration=66*5.792=382.25 N
Answer:
(a) 31.44 m/s (b) 164.74 m/s²
Explanation:
Given that,
The diameter of a disk, d = 12 cm
Radius, r = 6 cm
Angular speed = 5.24 rad/s
(a) Linear speed,
![v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s](https://tex.z-dn.net/?f=v%3Dr%5Comega%5C%5C%5C%5Cv%3D6%5Ctimes%205.24%5C%5C%5C%5Cv%3D31.44%5C%20m%2Fs)
(b) Centripetal acceleration,
![a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv%5E2%7D%7Br%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7B31.44%5E2%7D%7B6%7D%5C%5C%5C%5Ca%3D164.74%5C%20m%2Fs%5E2)
Answer:
Explanation:
The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.
Let Vector
is the tidal current velocity as shown in the diagram.
In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity,
must be in the north direction.
Let
is the speed of the kayaker having angle \theta measured north of east as shown in the figure.
For the resultant velocity in the north direction, the tail of the vector
and head of the vector
must lie on the north-south line.
Now, for this condition, from the triangle OAB
![|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|](https://tex.z-dn.net/?f=%7C%5Coverrightarrow%7BAB%7D%7C%5Csin%20%5Ctheta%3D%7C%5Coverrightarrow%7BOA%7D%7C)
![\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csin%5Ctheta%3D%5Cfrac%7B%7C%5Coverrightarrow%7BOA%7D%7C%7D%7B%7C%5Coverrightarrow%7BAB%7D%7C%7D%3D%5Cfrac%202%203)
![\Rightarrow \theta=\sin^{-1}\frac23](https://tex.z-dn.net/?f=%5CRightarrow%20%5Ctheta%3D%5Csin%5E%7B-1%7D%5Cfrac23)
![\Rightarrow \theta=41.81^{\circ}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Ctheta%3D41.81%5E%7B%5Ccirc%7D)
Hence, the kayaker must paddle in the direction of
in the north of east direction.