Answer:
v₂ = 7/ (0.5)= 14 m/s
Explanation:
Flow rate of the fluid
Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.
The formula for calculated the flow rate is:
Q= v*A Formula (1)
Where :
Q is the Flow rate (m³/s)
A is the cross sectional area of a section of the pipe (m²)
v is the speed of the fluid in that section (m/s)
Equation of continuity
The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:
Q₁= Q₂
Data
A₁ = 2m² : cross sectional area 1
v₁ = 3.5 m/s : fluid speed through A₁
A₂ = 0.5 m² : cross sectional area 2
Calculation of the fluid speed through A₂
We aply the equation of continuity:
Q₁= Q₂
We aply the equation of Formula (1):
v₁*A₁= v₂*A₂
We replace data
(3.5)*(2)= v₂*(0.5)
7 = v₂*(0.5)
v₂ = 7/ (0.5)
v₂ = 14 m/s
Answer:
3.6 seconds
Explanation:
Given:
y₀ = y = 0 m
v₀ = 31 sin 35° m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 = 0 + (31 sin 35°) t + ½ (-9.8 m/s²) t²
0 = 17.78t − 4.9t²
0 = t (17.78 − 4.9t)
t = 0 or 3.63
Rounded to the nearest tenth, the ball lands after 3.6 seconds.
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If the soloist produces "x" decibels and the 10-person choir produces "y" decibels, combined they will produce "x+y" decibels.
The second choir has 90 additional singers, we base our description on the first choir. If a 10-person choir produces "x+y" decibels, then the 90 person choir produces 10 (x+y) decibels.