Calculations and Concepts Using Freebody Diagrams A type of elevator called a cage is used to raise and lower miners in a mine shaft. Suppose the
cage carries a group of miners down the shaft. If the unbalanced force on the cage is 60.0 N, and
the mass of the loaded cage is 1.5 x 102kg, what is the acceleration on the cage?
Miners do not like using the cage, so they walk down stairs to enter the mine
O 0.400 m/s^2
4.0 m/s^2
1.5 x 10^2 kg
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what is the image?

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step-by-step explanation:

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)