On a coordinate plane, line segment A prime B has points (negative 2, negative 4) and (2, 1). Find the pre-image given the image
2 answers:
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Answer:
Effort and distance = Load x distance
7.30 x 4.45x10^2N = 3.2485 X 10^3N
We then know we can move 3 points to the right and show in regular notion.
= 3248.5
Divide by 2 = 3248.5/2 = 1624.25 force
Step-by-step explanation:
In the case of a Second Class Lever as attached diagram shows proof to formula below.
Load x distance d1 = Effort x distance (d1 + d2)
The the load in the wheelbarrow shown is trying to push the wheelbarrow down in an anti-clockwise direction whilst the effort is being used to keep it up by pulling in a clockwise direction.
If the wheelbarrow is held steady (i.e. in Equilibrium) then the moment of the effort must be equal to the moment of the load :
Effort x its distance from wheel centre = Load x its distance from the wheel centre.
This general rule is expressed as clockwise moments = anti-clockwise moments (or CM = ACM)
This gives a way of calculating how much force a bridge support (or Reaction) has to provide if the bridge is to stay up - very useful since bridges are usually too big to just try it and see!
The moment of the load on the beam (F) must be balanced by the moment of the Reaction at the support (R2) :
Therefore F x d = R2 x D
It can be seen that this is so if we imagine taking away the Reaction R2.
The missing support must be supplying an anti-clockwise moment of a force for the beam to stay up.
The idea of clockwise moments being balanced by anti-clockwise moments is easily illustrated using a see-saw as an example attached.
We know from our experience that a lighter person will have to sit closer to the end of the see-saw to balance a heavier person - or two people.
So if CM = ACM then F x d = R2 x D
from our kitchen scales example above 2kg x 0.5m = R2 x 1m
so R2 = 1m divided by 2kg x 0.5m
therefore R2 = 1kg - which is what the scales told us (note the units 'm' cancel out to leave 'kg')
But we can't put a real bridge on kitchen scales and sometimes the loading is a bit more complicated.
Being able to calculate the forces acting on a beam by using moments helps us work out reactions at supports when beams (or bridges) have several loads acting upon them.
In this example imagine a beam 12m long with a 60kg load 6m from one end and a 40kg load 9m away from the same end n- i.e. F1=60kg, F2=40kg, d1=6m and d2=9m
CM = ACM
(F1 x d1) + (F2 x d2) = R2 x Length of beam
(60kg x 6m) + (40Kg x 9m) = R2 x 12m
(60kg x 6m) + (40Kg x 9m) / 12m = R2
360kgm + 360km / 12m = R2
720kgm / 12m = R2
60kg = R2 (note the unit 'm' for metres is cancelled out)
So if R2 = 60kg and the total load is 100kg (60kg + 40kg) then R1 = 40kg
From the diagram above,
XZ = 10 in and OX = 10 in
we are to find length of OY
XZ is a chord and line OY divides the chord into equal length
Hence, ZY=YX= 5 in
Now we solve the traingle OXY
To find OY we solve using pythagoras theorem
applying values from the triangle above
Therefore,
Length of OY =
