"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
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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Mechanical energy is the sum of kinetic energy and potential energy
Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes
(6.4)^2 x 10^12
= 40.96 x 10^12 .
Now it's just a matter of mashing out the fraction.
The 'mantissa' (the number part) is
6/40.96 = 0.1465
and the order of magnitude is
10^24 / 10^12 = 10^12 .
Put it all together and you've got
1.465 x 10^11 .
156 is the answer. so 156.25 is almost the same thing, you just round. It's not hard. Thank you!!!