Answer:
a) A1 = [tex]\frac{\pi (d1)^{2} }{4}[/tex]
b) A1 = 2.688 cm[tex]^{2}[/tex]
c) Q1 = A1 x v1
d) v1 = 3.1994 m/s
e) A2 = [tex]\frac{A1 X v1}{v2}[/tex]
f) A2 = 0.7963cm[tex]^{2}[/tex]
Explanation:
a) Area = [tex]\pi r^{2}[/tex]
r = [tex]\frac{d}{2}[/tex]
thus,
area = [tex]\pi (\frac{d}{2})^{2}[/tex]
A1 = [tex]\frac{\pi (d1)^{2} }{4}[/tex]
b) d1 = 1.85 cm
substituting in the above equation,
A1 = [tex]\frac{\pi (d1)^{2} }{4}[/tex]
A1 = [tex]\frac{\pi (1.85)^{2} }{4}[/tex]
A1 = 2.688 cm[tex]^{2}[/tex]
c) Flow rate = Area x velocity ( refer https://brainly.com/question/13997998)
Q1 = A1 x v1
d) From the above equation,
v1 = [tex]\frac{Q1}{A1}[/tex] = [tex]\frac{860}{2.688}[/tex] = 319.94 cm/s = 3.1994 m/s
e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.
i.e. A1 x v1 = A2 x v2
thus,
A2 = [tex]\frac{A1 X v1}{v2}[/tex]
f) v2 = 10.8 m/s.
substituting the values in the above equation,
A2 = [tex]\frac{2.688 X 3.1994}{10.8}[/tex] = 0.7963cm[tex]^{2}[/tex]
This is defined as the area of an object if you view it as a 2D object.
We can calculate A2 through the formula
A2 = A1V1 / V2
where v2 = 10.8 m/s we have:
(2.688×3.1994)/ 10.8 = 0.7963cm².
Read more about Cross-sectional area here https://brainly.com/question/17143004
