1o4 Fahrenheit is equal to
2 answers:
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Answer:
v = 719.2 m / s and a = 83.33 m / s²
Explanation:
This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is
v - v₀ = ln (M₀ / M)
where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket
In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s
m_fuel = 75 10
m_fuel = 750 kg
As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is
M = 3000 -750 = 2250 kg
let's calculate
v - 0 = 2500 ln (3000/2250)
v = 719.2 m / s
To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is
Push = v_{e} dM / dt
let's calculate
Push = 2500 75
Push = 187500 N
If we use Newton's second law
F = m a
a = F / m
let's calculate
a = 187500/2250
a = 83.33 m / s²
Answer:
D₂ = 2.738 cm
Explanation:
Continuity equation
The continuity equation is nothing more than a particular case of the principle of conservation of mass. It is based on the flow rate (Q) of the fluid must remain constant throughout the entire pipeline.
Since the flow rate is the product of the surface of a section of the duct because of the speed with which the fluid flows, we will have to comply with two points of the same pipeline:
Q = v*A : Flow Equation
where:
Q = Flow in (m³/s)
A is the surface of the cross sections of points 1 and 2 of the duct.
v is the flow velocity at points 1 and 2 of the pipe.
It can be concluded that since the flow rate must be kept constant throughout the entire duct, when the section decreases, the flow rate increases in the same proportion and vice versa.
Data
D₁= 6.0 cm : faucet diameter
v₁ = 5 m/s : speed of fluid in the faucet
v₂ = 20 m/s : speed of fluid in the nozzle
Area calculation
A = (π*D²)/4
A₁ = (π*D₁²)/4
A₂ = (π*D₂²)/4
Continuity equation
Q₁ = Q₂
v₁A₁ = v₂A₂
v₁(π*D₁²)/4 = v₂(π*D₂²)/4 : We divide by (π/4) both sides of the equation
v₁ (D₁)² = v₂(D₂)²
We replace data
6 *(5)² = 20*(D₂)²
150 = 20*(D₂)²
(150 /20) = (D₂)²
7.5 = (D₂)²
D₂ = 2.738 cm : nozzle diameter