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
Water is usually used to cool down automobile engines when they get hot, yes. Therefore, that means water has a high heat capacity.
That makes the answer letter D you provided above.
D) Water has a high heat capacity.
Another example would be trying to put out a fire with a bucket of water. Usually, you can put out the fire debating on the size!
Answer:
The correct answer is B
Explanation:
To calculate the acceleration we must use Newton's second law
F = m a
a = F / m
To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface
P = I / c absorbent surface
P = F / A
F / A = I / c
F = I A / c
The area of area of a circle is
A = π r²
We replace
F = I π r² / c
Let's calculate
F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸
F = 8.375 10⁻²³ N
Density is
ρ = m / V
m = ρ V
m = ρ (4/3 π r³)
m = 4500 (4/3 π (1 10⁻⁶)³)
m = 1,885 10⁻¹⁴ kg
Let's calculate the acceleration
a = 8.375 10⁻²³ / 1.885 10⁻¹⁴
a = 4.44 10⁻⁹ m/s² absorbent surface
The correct answer is B
Answer:
A.
Explanation: both triple by 3
The momentum goes to the wall
