IMA = Ideal Mechanical Advantage
First class lever = > F1 * x2 = F2 * x1
Where F1 is the force applied to beat F2. The distance from F1 and the pivot is x1 and the distance from F2 and the pivot is x2
=> F1/F2 = x1 /x2
IMA = F1/F2 = x1/x2
Now you can see the effects of changing F1, F2, x1 and x2.
If you decrease the lengt X1 between the applied effort (F1) and the pivot, IMA decreases.
If you increase the length X1 between the applied effort (F1) and the pivot, IMA increases.
If you decrease the applied effort (F1) and increase the distance between it and the pivot (X1) the new IMA may incrase or decrase depending on the ratio of the changes.
If you decrease the applied effort (F1) and decrease the distance between it and the pivot (X1) IMA will decrease.
Answer: Increase the length between the applied effort and the pivot.
Answer:
gases because they have a straight line and if it was solid it would have a flat line to show it melt first before the temperature increase.
Hope it helps
Answer: sorry but I can’t help
Explanation:because I can’t see it
Answer:
1 / i + 1 / o = 1 / f thin lens equations
i = o f / (o - f) rearranging
Lens 1: object = 30 cm f = 15.2 cm
i1 = 30 * 15.2 / (30 - 15.2) = 30.8 cm
o2 = 40.2 - 30/8 = 9.4 cm distance of image 1 from lens 2
i2 = 9.4 * 15.2 / (9.4 - 15.2) = - 24.6 cm
The final image is 24.6 cm to the left of lens 2
The first image is inverted
The second image is erect (as seen from the first image)
So the final image is inverted
M = m1 * m2 = (-30.8 / 30) * (24.6 / 9.4) = -2.69
