the load which a crame can lift up is directly proportional to the mass and inversely proportional to the height through which the load is to be lifted.if a load of 40 and mass of 175gm is lifted through height of 286m,from what load will be carried through height 50m given the mass as 250gm?
"variation" is not your school subject. This looks like physics. For all I knew, you could have been asking about the variation of terrain in the Sierra foothills or the variables in a public opinion survey - anything.
Finally! Your variation description says that
L = km/h
so, Lh/m = k is constant. So, you want L such that
L*50/250 = 40*286/75
To find the load that can be carried through a height of 50m, given a mass of 250gm, we can use the proportional relationship stated in the question.
The load is directly proportional to the mass, and inversely proportional to the height. This means that we can set up the following equation:
Load₁ / Load₂ = (Mass₁ / Mass₂) * (Height₂ / Height₁)
Let's substitute the given values into the equation:
Load₁ / Load₂ = (175 / 250) * (50 / 286)
Simplifying the equation:
Load₁ / Load₂ = (7 / 10) * (5 / 286)
Load₁ / Load₂ = 35 / (10 * 286)
Load₁ / Load₂ = 35 / 2860
Now, we can solve for Load₂ by multiplying both sides of the equation by Load₂:
Load₁ = (35 / 2860) * Load₂
Next, substitute the known value of Load₁ (40) into the equation:
40 = (35 / 2860) * Load₂
To isolate Load₂, divide both sides by (35 / 2860):
Load₂ = (40 / (35 / 2860))
Simplifying the expression, we get:
Load₂ = 40 * (2860 / 35)
Load₂ ≈ 3251.42
Therefore, the load that can be carried through a height of 50m, given a mass of 250gm, is approximately 3251.42 units (the specific unit is not provided in the question).