Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.
From this question, the plane is still up in the air.
We have wind blowing in [W 60° N ]
To solve the problem we have to make use of the sine rule
We put the values in the equation, we have:
50/Sinθ = 200/sin60°
The next step is to cross multiply
50 x sin60° = 200Sinθ
50 x 0.8660 = 200sinθ
We make Sin θ the subject
Sine θ = 43.30/200
sine θ = 0.2165
we find the value of θ
θ = sine⁻¹(0.2165)
θ = 12.50
So Lindsay has to fly this plane towards this direction
[W 12.5° S]
Here is a similar question brainly.com/question/13338067?referrer=searchResults
To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).
The answer is 36 kg m/s
(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
Learn more about work done here: brainly.com/question/25573309
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I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?
