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:
Explanation:
the directions may change
Or they will repel and become opposite sides
Answer:
a) 1.73*10^5 J
b) 3645 N
Explanation:
106 km/h = 106 * 1000/3600 = 29.4 m/s
If KE = PE, then
mgh = 1/2mv²
gh = 1/2v²
h = v²/2g
h = 29.4² / 2 * 9.81
h = 864.36 / 19.62
h = 44.06 m
Loss of energy = mgΔh
E = 780 * 9.81 * (44.06 - 21.5)
E = 7651.8 * 22.56
E = 172624.6 J
Thus, the amount if energy lost is 1.73*10^5 J
Work done = Force * distance
Force = work done / distance
Force = 172624.6 / (21.5/sin27°)
Force = 172624.6 / 47.36
Force = 3645 N
Answer:
17.2 seconds
Explanation:
Given:
v₀ = 0 m/s
a₁ = 10.0 m/s²
t₁ = 3.0 s
a₂ = 16 m/s²
t₂ = 5.0 s
a₃ = -12 m/s²
v₃ = 0 m/s
Find: t
First, find v₁:
v₁ = a₁t₁ + v₀
v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)
v₁ = 30 m/s
Next, find v₂:
v₂ = a₂t₂ + v₁
v₂ = (16 m/s²) (5.0 s) + (30 m/s)
v₂ = 110 m/s
Finally, find t₃:
v₃ = a₃t₃ + v₂
(0 m/s) = (-12 m/s²) t₃ + (110 m/s)
t₃ = 9.2 s
The total time is:
t = t₁ + t₂ + t₃
t = 3.0 s + 5.0 s + 9.2 s
t = 17.2 s
Round as needed.
Answer:
no seasons
Explanation:
if the earth weren't tilted, it would rotate like that as it revolved around the sun, and we wouldn't have seasons—only areas that were colder (near the poles) and warmer (near the Equator).