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atroni [7]
2 years ago
8

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.

Physics
1 answer:
joja [24]2 years ago
7 0

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=Engine weight*2.50*sin32\\\\TW=14900*2.50*sin32

TW=19740N-m

b)

Generally the equation for the <em>Torq</em>ue due to thrust  is mathematically given as

TT=Engine thrust*2.50*cos32\\\\TT=61500*2.50*cos32\\\\

TT=130387.39N-m

For more information on this visit

brainly.com/question/23379286

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What is the mass of a bicycle that has 90 kg.m/s of momentum at a speed of 6 m/s?
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3 years ago
What is the distance that a car travels if it was brought to stop in 5 seconds and if it was traveling at 110 Km/h
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Answer:

Suppose that the acceleration is a constant, a.

a(t) = a.

To write the velocity equation, we must integrate over time, and the constant of integration will be equal to the initial velocity, in this case is 110km/h.

v(t) = a*t + 110km/h

And we know that at t = 5s, the car was brought to stop, so the velocity must be zero.

v(5s) = 0 = a*5s + 110km/h.

a = (110km/h)*(1/5s)

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Now the velocity equation is:

v(t) = -6.11m/s^2*t + 30.56m/s

To write the positon equation we must integrate over time again, we can get:

p(t) = (1/2)*(-6.11m/s^2)*t^2 + (30.56m/s)*t + p0

Where p0 is the initial position, here i will assume that is zero, because it does no really mater.

The total displacement of the car will be equal to p(5s)

p(5s) = (1/2)*(-6.11m/s^2)*(5s)^2 + (30.56m/s)*(5s) = 76.425 meters.

6 0
3 years ago
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