Respuesta :
Answer:
The speed as it exits the shower-head openings is 9.6 m/s.
Explanation:
Given that,
Radius [tex]r_{1}= 1.0\ mm[/tex]
Number of circular = 20
Radius of pipe[tex]r_{2} = 0.80\ cm[/tex]
Speed = 3.0 m/s
We need to calculate the area of circle
Using formula of area of circle
[tex]A=\pi r^2[/tex]
We need to calculate the speed
Using equation of continuity
[tex]nA_{1}v_{1}=A_{2}v_{2}[/tex]
[tex]v_{1}=\dfrac{A_{2}v_{2}}{nA_{1}}[/tex]
Where, A =area of cross section
v = speed
Put the value in the equation
[tex]v_{1}=\dfrac{\pi r_{2}^2\times3.0}{\pi r_{1}^2\times20}[/tex]
[tex]v_{1}=\dfrac{\pi\times(0.80\times10^{-2})^2\times3.0}{\pi\times(1.0\times10^{-3})^2\times20}[/tex]
[tex]v_{1}=9.6\ m/s[/tex]
Hence, The speed as it exits the shower-head openings is 9.6 m/s.
Speed is defined as the rate of change of the distance or the height attained. It travels at a speed of 9.6 m/s when it exits the showerhead apertures.
What is speed?
Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while u for the final speed. its si unit is m/sec.
Given,
Number of circular opening (n) = 20
the radius of the opening of the circular head =[tex]\rm{r_1= 1.0 mm = 0.001 m}[/tex]
the radius of the pipe = [tex]\rm{r_2= 0.80 cm = 0.008 m[/tex]
speed of water in pipe = [tex]\rm{V_2= 30. m/sec}[/tex]
The area of a circle must be calculated.
Using the continuity equation
Where A denotes the cross-sectional area.
v stands for velocity.
[tex]\rm{nA_1v_1=A_2V_2}[/tex]
[tex]\rm{V_1=}\frac{A_2V_2}{nA_1}[/tex]
[tex]\rm{V_1=}\frac{\pi(r_2) ^{2} V_2}{n\pi(r_1)^2}[/tex]
[tex]\rm{V_1=}\frac{3.1\times (0.80) ^{2} \times 3.0}{20\pi(1.0)^2}[/tex]
[tex]\rm{V_1=9.6 m/sec}[/tex]
Hence, the speed at which it escapes the shower-head holes is 9.6 meters per second.
To know more about the speed refer to the link;
brainly.com/question/7359669
