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:
Explanation:
= Cambio en la longitud de la cuerda = 0.25 cm
T = tensión en cuerda
A = Área de la cadena = 
d = Diámetro de la cuerda = 0.2 cm
L = Longitud original de la cuerda = 1.6 m
El cambio de longitud de una cuerda viene dado por
La tensión en la cuerda es 
.
4000 seconds
Explanation:
speed = distance / time
0.0004m/s = 1.6m / time
Subject time
time = 1.6 / 0.0004
time = 4000 seconds.
Hope this helps. Mark as brainliest if possible. tks
An Earth revolution is a trip around the sun in a closed path (relative to the sun).
The path is very nearly an ellipse with the sun at one focus, and a little less nearly a circle with the sun at the center.
One complete revolution takes roughly 365.24 days, and at that point, the Earth immediately begins another one.
We have a special word that we use to refer to that special period of time. In English, it's called a "year".
Answer:
Given: a projectile of initial launch velocity(V) and launch angle ∅ and no air resistance. At the maximum height, the projectile would have a zero contribution of speed from the vertical component(Vy) Therefore, if we say Vx=Vcos∅ is the only speed the projectile has at the instant of maximum height then we can replace Vx with 1/5V and write 1/5V=Vcos∅. Solving for the the launch angle ∅, gives Inverse Cos(1/5)=78.5 degrees.
