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:
-6.44 m/s²
Explanation:
Given:
Δx = 60 m
v₀ = 27.8 m/s
v = 0 m/s
Find: a
v² = v₀² + 2aΔx
(0 m/s)² = (27.8 m/s)² + 2a (60 m)
a = -6.44 m/s²
Answer:
The magnification is
Explanation:
From the question we are told that
The power of the lens is
Generally
The object distance is the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )
Generally the focal length is mathematically represented as
=>
=>
converting to cm
=>
Generally from lens equation we have that
=>
=>
Generally the magnification is mathematically represented as
=>
=>
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