Answer:
Temperature of the air in the balloon = 272°C
Explanation:
Given:
Volume of balloon = 500 m³
Air temperature = 15° C = 273 + 15 = 288 K
Total weight = 290 kg
Density of air = 1.23 kg/m³
Find:
Temperature of the air in the balloon
Computation:
Density of hot air = Density of air - [Total weight / Volume of balloon]
Density of hot air = 1.23 - [290 - 500]
Density of hot air = 0.65 kg/m³
[Density of hot air][Temperature of the air in the balloon] = [Density of air][Air temperature ]
Temperature of the air in the balloon = [(1.23)(288)]/(0.65)
Temperature of the air in the balloon = 544.98
Temperature of the air in the balloon = 545 K
Temperature of the air in the balloon = 545 - 273 = 272°C
Answer:
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The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Explanation:
Since block moves with constant speed
So, frictional force
f = FCosq
Work done by friction
W = - fd
W = - fd Cos q
The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.
Yo don’t click the link that guy says dude I think it’s like tracking or something
Answer: Part 1: Propellant Fraction (MR) = 8.76
Part 2: Propellant Fraction (MR) = 1.63
Explanation: The Ideal Rocket Equation is given by:
Δv = 
Where:
is relationship between exhaust velocity and specific impulse
is the porpellant fraction, also written as MR.
The relationship
is: 
To determine the fraction:
Δv = 

Knowing that change in velocity is Δv = 9.6km/s and
= 9.81m/s²
<u>Note:</u> Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.
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<u>Part 1</u>: Isp = 450s

ln(MR) = 
ln (MR) = 2.17
MR = 
MR = 8.76
<u>Part 2:</u> Isp = 2000s

ln (MR) = 
ln (MR) = 0.49
MR = 
MR = 1.63
k e is energy of movement. When thing solidifies, atoms stop moving ke is less.