An airplane is flying at a constant velocity through the air. What is the relationship between the magnitudes of the four forces
shown in the diagram?
A. L = W; F < D
B. L = W; F = D
C. L = W = F = D
D. L = W; F > D
A. L = W; F < D
B. L = W; F = D
C. L = W = F = D
D. L = W; F > D
1 answer:
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Answer:
Explanation:
information we know:
Total force:
Weight:
distance:
vertical component of the force:
-------------
In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).
horizontal component:
vertical component:
but from the given information we know that
so, equation these two and
and we know the force , thus:
now we clear for
the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:
whith this horizontal component we calculate the work to move the crate a distance of 4 m:
the work done is W=173.48J
Complete question is;
Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.
Block 1 is removed without disturbing block 2.
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.
Answer:
a = (m2)g/(m3 + m2)
Explanation:
Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;
T = (m3)a - - - (eq 1)
where;
T is the tension in the string
a is acceleration
m3 is mass of block 3
Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a
Making T the subject gives us;
T = (m2)g - (m2)a - - - (eq 2)
where;
g is acceleration due to gravity
T is the tension in the string
a is acceleration
m2 is mass of block 2
To solve for the acceleration, we will just substitute (m3)a for T in eq 2.
Thus;
(m3)a = (m2)g - (m2)a
(m3)a + (m2)a = (m2)g
a(m3 + m2) = (m2)g
a = (m2)g/(m3 + m2)
