the stretch of a spring in millimeters is proportional to the weight in grams attached to the end of the spring. Find the unit rate.

Question

length of stretch (mm)
50-
45-
40-
35
30-
25
20-
15-
10-
5
0+
0
1
2
3
4
5
6
7 8 9
10
weight (g)

Answer 1

To find the unit rate, we need to divide the length of the stretch by the weight attached to the spring.

For the given data points, we can pair them up as follows:

(50, 0)
(45, 1)
(40, 2)
(35, 3)
(30, 4)
(25, 5)
(20, 6)
(15, 7)
(10, 8)
(5, 9)
(0, 10)

Now we can calculate the unit rate for each pair:

0/50 = 0
1/45 = 0.0222...
2/40 = 0.05
3/35 = 0.0857...
4/30 = 0.1333...
5/25 = 0.2
6/20 = 0.3
7/15 = 0.4666...
8/10 = 0.8
9/5 = 1.8
10/0 = undefined

Therefore, the unit rate for the stretch of the spring in millimeters to the weight in grams is as follows:

0, 0.0222..., 0.05, 0.0857..., 0.1333..., 0.2, 0.3, 0.4666..., 0.8, 1.8.

Answer 2

Huh ?
I think they are asking for the spring constant k = d F/ dx
for a change of mass of 1 gram the x changes by 5 millimeters
so if you are not a physicist you might say
k = 0.20grams / mm

of course grams is mass, not force and the units are nutty
if you are taking physics then
k = delta m g / delta x
delta m g = 0.001 kg * 9.81 newtons/kg
delta x = 0.005 meters
so
k = about 2 Newtons/meter

Answer 3

Apologies for the confusion. You are correct, the question is asking for the spring constant (k), which represents the relationship between the change in length of the spring (delta x) and the force applied (delta F).

In this case, the change in mass is 1 gram (0.001 kg), the acceleration due to gravity is approximately 9.81 m/s^2, and the change in length is 5 mm (0.005 meters). Therefore, we can calculate the spring constant (k) as follows:

k = (0.001 kg * 9.81 m/s^2) / (0.005 meters)
k ≈ 2 N/m

So, the unit rate or spring constant for this spring is approximately 2 Newtons per meter (N/m).

Answer 4

wtf does this sht mean