In terms of the sequence of the scientific method, what is the immediate purpose of doing an experiment? A. asking the question.
2 answers:
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Answer:
a) t = 20 [s]
b) Can't land
Explanation:
To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = desacceleration = 5 [m/s^2]
t = time [s]
Note: the negative sign of the equation means that the aircraft slows down as it stops.
0 = 100 - 5*t
5*t = 100
t = 20 [s]
b)
Now we can find the distance using the following kinematics equation.
x - xo = distance [m]
x -xo = (0*20) + (0.5*5*20^2)
x - xo = 1000 [m]
1000 [m] = 1 [km]
And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land
Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²