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
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
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Answer:
V = 90.51 m/s
Explanation:
From the given information:
Initial speed (u) = 0
Distance (S) = 391 m
Acceleration (a) = 18.9 m/s²
Using the relation for the equation of motion:
v² - u² = 2as
v² - 0² = 2as
v² = 2as
v = 121.57 m/s
After the parachute opens:
The initial velocity = 121.57 m/ss
Distance S' = 332 m
Acceleration = -9.92 m/s²
How fast is the racer can be determined by using the relation:
V = 90.51 m/s
Answer:
t=0.42s
Explanation:
Here you have an inelastic collision. By the conservation of the momentum you have:
m1: mass of the bullet
m2: wooden block mass
v1: velocity of the bullet
v2: velocity of the wooden block
v: velocity of bullet and wooden block after the collision.
By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:
Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:
Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:
hence, the time is t=0.42 s