Jsisisusuagahannananana this is worth 22 points g that’s crazy 25
Answer:
0.54m
Explanation:
Step one:
given data
length of seesaw= 3m
mass of man m1= 85kg
weight = mg
W1= 85*10= 850N
mass of daughter m2= 35kg
W2= 35*10= 350N
distance from the center= (1.5-0.2)= 1.3m
Step two:
we know that the sum of clockwise moment equals the anticlockwise moment
let the distance the must sit to balance the system be x
taking moment about the center of the system
350*1.3=850*x
455=850x
divide both sides by 850
x=455/850
x=0.54
Hence the man must sit 0.54m from the right to balance the system
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,



So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
Object should be placed at a distance, u = 7.8 cm
Given:
focal length of convex lens, F = 16.5 cm
magnification, m = 1.90
Solution:
Magnification of lens, m = -
where
u = object distance
v = image distance
Now,
1.90 = 
v = - 1.90u
To calculate the object distance, u by lens maker formula given by:
u = 7.8 cm
Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.
Trees. Every time the wind blows there is a wave of motion which is movement