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
Answer:
a) h / h₀ = 1,682
, b) h / h₀ = 1.11
, c) h / h₀ = 2.07
Explanation:
For this exercise let's look for the growth equation, as they indicate that it is exponential
h = h₀ 
The initial height is 9 ”, so the constant
h₀ = 9”
Let's use the given values
h = 15.1655”
t = 5 days
h / h₀ =
α = 1 / t ln h / h₀
α = 1/5 ln (15.1655 / 9)
α = 0.104
h = 9 e^{0.104 t}
a) the growth factor is the relationship between the initial value and the current value
h / h₀ = 
h / h₀ = 1,682
b) for t = 1 day
h / h₀ = 
h / h₀ = 1.11
c) for t = 7 days
h / h₀ = e^{0.104 7}
h / h₀ = 2.07
vf² = vi² + 2ad
vf = vi +at
vf = final velocity
vi = initial velocity
decelerated = a = - (negative)
a. stop ⇒ vf = 0
vf = vi+at
0 = 20 - 1.5t
1.5t=20
t = 13.3 s
b. vi = 20 m/s
0 = 20² - 1.5d
1.5d = 20²
1.5d = 400
d =266.67 m
c. d = vi.t -1/2at²
d = 20.3 - 1/2.1.5.3²
d = 53.25 m
Answer:
magnification is - 159
objective distance is 3.85 cm
Explanation:
Given data
focal length f1 = 1.40 cm
focal length f2 = 2.20 cm
separated d = 19.6 cm
to find out
angular magnification and How far from the objective
solution
we know magnification formula that is
magnification = ( - L / f1 ) (D/f2)
here D = 25 cm put all value
magnification = ( - 19.6 / 1.40 ) (25/2.20)
magnification = - 159
and
now we apply lens formula
i/f = 1/q + 1/p
p = f2 = 2.20
so
q = f2 p / p -f2
q = 1.4(2.20) / ( 2.2 - 1.4 )
q = 3.85 cm
so objective distance is 3.85 cm
Answer:
If sound waves of the same energy were passed through a block of wood and a block of steel, which is more dense than the wood, the molecules of the steel would vibrate at a slower rate. Sound moves faster through denser air because the molecules are closer together in dense air and sound can be more easily passed on.
Explanation:
