We drop a 0.1 kg apple 1 m. What is its potential energy?

392 N/m

Step-by-step explanation:

We know that the total mechancial enery will be conserved from when the apple drops to when it hit the spring.  Therefore, we can set the potential energy of the apple equal to the elastic potential energy of the spring.

PE = mgh and SE = 1/2kx² — PE is the initial potential energy of the apple and SE is the elastic potential energy of the spring.

mgh = 1/2kx².  (.1)(9.8)(2) = (1/2)(k)(.1²), solve for k. k = 392

sample answer: you can determine whether each element of the domain is paired with exactly one element of the range. for example, if given a graph, you could use the vertical line test; if a vertical line intersects the graph more than once, then the relation that the graph represents is not a function.

an = 22 + (n-1) *6

110 seats

step-by-step explanation:

an=a1+(n-1)(d)

a1 = 22   this is the first item in the sequence

d =6   this is how the sequence goes up

a)   an = 22 + (n-1) *6

b) a12 = 22 + (12-1) *6

    a12 = 22+11*6

  a12 = 22 +88

  a12 = 110