A force of 182.6 lb should be applied to lift 700 lb object.
<h3>How to calculate the force?</h3>
Let the force required to lift be F
Total length = 29 in.
Distance of short end from pivot = in.
Distance from long end = 29-6 = 23 in.
Now Equating moments
F × 23 = 700x6
23F = 4200
F = 4200/23
= 182.6 lb
A force of 182.6 lb should be applied to lift 700 lb object.
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Hello,
f(x) has as inverse f^(-1)(x)
and (fof^(-1))(x)=(f^(-1)of)(x)= x (the neutral function)
so h(f(x))=x
We know that
If the scalar product of two vectors<span> is zero, both vectors are </span><span>orthogonal
</span><span>A. (-2,5)
</span>(-2,5)*(1,5)-------> -2*1+5*5=23-----------> <span>are not orthogonal
</span><span>B. (10,-2)
</span>(10,-2)*(1,5)-------> 10*1-2*5=0-----------> are orthogonal
<span>C. (-1,-5)
</span>(-1,-5)*(1,5)-------> -1*1-5*5=-26-----------> are not orthogonal
<span>D. (-5,1)
</span>(-5,1)*(1,5)-------> -5*1+1*5=0-----------> are orthogonal
the answer is
B. (10,-2) and D. (-5,1) are orthogonal to (1,5)
C because the equation goes across on its own without any blockage.
