Desired operation: A + B = C; {A,B,C) are vector quantities.
<span>Issue: {A,B} contain error (measurement or otherwise) </span>
<span>Objective: estimate the error in the vector sum. </span>
<span>Let A = u + du; where u is the nominal value of A and du is the error in A </span>
<span>Let B = v + dv; where v is the nominal value of B and dv is the error in B </span>
<span>Let C = w + dw; where w is the nominal value of C and dw is the error in C [the objective] </span>
<span>C = A + B </span>
<span>w + dw = (u + du) + (v + dv) </span>
<span>w + dw = (u + v) + (du + dv) </span>
<span>w = u+v; dw = du + dv </span>
<span>The error associated with w is the vector sum of the errors associated with the measured quantities (u,v)</span>
Answer:
Because of the knowledge of <u>relative size</u>, it will be assumed that the smaller jetliner is farther away.
Explanation:
According to the theory of relative size, the distance that an object has to the viewing individual affects the perception of the individual regarding the size of the object.
As stated in this case, one of the jetliners is farther away from the other. Therefore, even if the jets are of equal size, the one that is at a greater distance is perceived to be smaller as it is at a greater viewing range. The one that is nearer to the individual seems bigger in comparison to the one farther away due to a closer viewing range.
Therefore, the jet that is nearer appears larger.
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Work is obtained by multiplying the force and the object's displacement. The force and displacement and force should be in the same direction in order to have work.
W = F x d
d = W / F
Substituting the known values,
d = 352 J / 45 N = 7.82 m
Thus, the displacement of the student is 7.82 m.