The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown i
n the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.
1 answer:
We have that for the Question, it can be said that With respect to this axis, the magnitude of the torque due to the weight and ,the thrust is
- TW=19740N-m
- TT=130387.39N-m
From the question we are told
The figure shows a jet engine suspended beneath the wing of an <em>airplane</em>. The weight of the <em>engine </em>is 14900 N and acts as shown in the <em>figure</em>. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the <em>figure </em>is <em>perpendicular </em>to the plane of the paper. With respect to this axis, find the <em>magnitude </em>of the torque due to (a) the <em>weight </em>and (b) the <u>thrust</u>.
a)
Generally the equation for the <u>Torque </u>due to weight is mathematically given as
TW=19740N-m
b)
Generally the equation for the <em>Torq</em>ue due to thrust is mathematically given as
TT=130387.39N-m
For more information on this visit
You might be interested in
Answer:
2) a_y= -g 3) vₓ=constant v_y = v_{oy} - g t, 4) vₓ = v₀ₓ - ax t
5) changes the horizontal speed, should change range
7) changes the vertical speed change the maximum height
Explanation:
1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.
2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration
3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression
v_y = v_{oy} - gt
at the point of maximum height, vy = 0 is equal to the maximum height
4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed
In the graph it would be directed to the left, therefore the velocity would be
vₓ = v₀ₓ - ax t
5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.
the equations of motion are
x = v₀ₓ t
y = v_{oy} t - ½ g t²
7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,
the equations of motion are the same.
