The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
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Explanation:
it dosent depend on the weights of the items. I'll reach the ground at same time taking as no air friction or restrictions.
i.e
v = u + gt
whte v is final velocity of the object
u is initial velocity of the object
g is acceleration due to gravity and
t is time. thanks
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Answer:
what was the answer
Explanation:
im taking the quiz and could use some help
Answer:
g'(10) = 
Explanation:
Since g is the inverse of f ,
We can write
g(f(x)) = x <em> </em><em>(Identity)</em>
Differentiating both sides of the equation we get,
g'(f(x)).f'(x) = 1
g'(10) = 
--equation[1] Where f(x) = 10
Now, we have to find x when f(x) = 10
Thus 10 = 
+ 2
= 8
x = 
Since f(x) = + 2
f'(x) = -
f'() = -4 × 4 = -16
Putting it in equation 1, we get:
We get g'(10) = -
See from periodic table the proton or atomic number of elements u want to know about then as u know first electron shell can hold 2 electrons 2nd can hold eight as well as all others........protons are equal to electrons so divide proton number into shells but remember to use amounts which it can hold
CHEERS !!
