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 from what load will be carried through height 50m given the mass as 250gm? Solve using variation
L = km/h^2
So, Lh^2/m = k is constant
You want L such that
L*50^2 = 40*h^2/175
If you will supply the original height, then you can solve for L.
a crame lift a load at 100kg very sloly through a distance of 150cm calculate the work done agaihst gravity of (g=10m/2s)
Next answer please
To solve this problem using variation, we need to understand the direct and inverse variations involved.
The problem states that the load (L) that a crane can lift up is directly proportional to the mass (M) and inversely proportional to the height (H) through which the load is to be lifted. Mathematically, this can be represented as:
L ∝ M
L ∝ 1/H
To solve the problem, we can write the equation of variation as follows:
L = k * (M / H)
Where k is the constant of variation.
To find the value of k, we can use the given values for the load (L = 40+) and mass (M = 175g):
40+ = k * (175 / H)
Next, we can use the new values for the height (H) and mass (M) to find the load (L) through the new height (50m) and mass (250g):
L = k * (250 / 50)
Now, let's solve these equations to find the load:
To find k:
40+ = k * (175 / H)
k = (40+ * H) / 175
To find L:
L = ((40+ * H) / 175) * (250 / 50)
Simplifying the expression:
L = (40+ * 250 * H) / (175 * 50)
Finally, substitute the value of H (50) into the equation to get the load (L):
L = (40+ * 250 * 50) / (175 * 50)
L = (1000 * H) / 7
Therefore, the load carried through a height of 50m with a mass of 250g would be (1000 * 50) / 7.
Please note that the "+" symbol after the number 40 indicates that the load is at least 40 units, but the exact value is not specified.