The mass of fuel the engine burn each second to produce a thrust of 7.66×10⁵ N is 2.5×10² kg/s.
<h3 /><h3>What is mass?</h3>
Mass can be defined as the quantity of matter contained in a body. The S.I unit of mass is kilogram(kg)
To calculate the mass the engine burns each seconds, we use the formula below.
Formual:
- M = T/v............. Equation
Where:
- M = Mass per seconds of the rocket
- T = Thrust
- v = Velocity
From the question,
Given:
- T = 7.66×10⁵ N
- v = 3.05×10³ m/s
Substitute these values into equation 1
- M = (7.66×10⁵)/(3.05×10³)
- M = 2.5×10² kg/s
Hence, the mass of fuel burned in each second is 2.5×10² kg/s.
Learn more about mass here: brainly.com/question/25121535
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Answer:
The only work done is when the person lifts the sack over a distance, W = 78.48 [N]
Explanation:
We have to remember the definition of work, which tells us that work is the result of a force by a distance, we must apply this concept in each of the movements of the person in the problem described.
W = F * d
where:
F = force [N]
d = distance [m]
The force is given by the producto of the mass by the gravity.
F = 5 * 9.81 = 49.05 [N]
W = 49.05 * 1.6 = 78.48 [N]
Answer:
they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta
Explanation:
use in your own words teachers know when your not trust me.
I think frequency it sounds like the correct answer but I am not completely sure if I am correct
Answer:
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Explanation:
Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.
Substituting the values of the variables into the equation, we have
ΔV = V₂ - V₁.
ΔV = 175 V - 33 V.
ΔV = 142 V
The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.
So, substituting the values of the variables into the equation, we have
ΔU = eΔV
ΔU = eΔV
ΔU = -1.602 × 10⁻¹⁹ C × 142 V
ΔU = -227.484 × 10⁻¹⁹ J
ΔU = -2.27484 × 10⁻²¹ J
ΔU ≅ -2.275 × 10⁻²¹ J
So, the required equation for the electric potential energy change is
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
