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Forward thrust has positive values and reverse thrust has negative values.
Thrust is a sudden push or pull in a certain direction.
a)
Flight speed u = 150 km/h
1 km/h = km/s
therefore, 150 km/h = 41.67 km / s
The thrust force represents the horizontal or x-component of momentum equation:
T = 50 * (150 - 41.67)
T = 5416.67 N
Therefore, the value of forward thrust is 5416.67 N.
b)
Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component,
thus thrust equation is:
T = 50 * (0 - 41.67)
T = -2083.5 N
Therefore, the thrust force T is -2083.5 N in the reverse direction.
c)
Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero because
T = 0
Therefore, there is no difference in two velocities in x direction.
The given question is incomplete, the complete question is,
"The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.
a. What is the forward thrust of this engine?
b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?
c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?"
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Answer:
a = 1.41 m/s²
Explanation:
Given that
mass ,m= 41 kg
F₁ = 65 N , θ = 59°
F₂ = 35 N ,θ = 32°
The component of Force F₁
F₁x= F₁cos59° i
F₁x= 65 x cos59° i = 33.47 i
F₁y= - F₁ sin 59° j
F₁y= - 65 x sin 59° j = - 55.71 j
The component of Force F₂
F₂x= F₂ sin 32° i
F₂x= 35 x sin 32° i = 18.54 i
F₂y= F₂ cos 32° j
F₂y= 35 x cos 32° j = 29.68 j
The total force F
F= 33.47 i + 18.54 i - 55.71 j + 29.68 j
F= 52.01 i - 26.03 j
The magnitude of the force F
F=58.16 N
We know that
F= m a
a= Acceleration
m=mass
58.16 = 41 x a
a = 1.41 m/s²
