Students drop a 2 kg object from rest out a window that has a height of 10 m during a lab experiment. The students time the

object to hit the ground and measure 1.2 s. Calculate the

difference between the PE at the top and the KE at the bottom.

a. greater than 75 J

b. less than 25 J

c. between 25 J & 50 J

d. between 50 J & 75 J

a. greater than 75 J

Step-by-step explanation:

Potential Energy at the top = mgh

Mass m = 2kg

g = 9.8m/s^2

height h  = 10m

Potential Energy at the top = 2 * 10 * 9.8

Potential Energy at the top = 196Joules

The object is not moving at the bottom, hence the KE of the body at the bottom is 0Joules

difference between the PE at the top and the KE at the bottom = 196 - 0 = 196Joules

Hence the difference between the PE at the top and the KE at the bottom is greater than 75Joules

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.

add a pic of the graph, then ill edit this "answer" and solve it for you : 3

step-by-step explanation: