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.
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)
To find the unit rate, we will calculate the change in the length of stretch (mm) per change in weight (g).
Change in length of stretch = 50 - 0 = 50 mm
Change in weight = 10 - 0 = 10 g
Unit rate = Change in length of stretch / Change in weight = 50 mm / 10 g = 5 mm/g
Therefore, the unit rate is 5 mm/g.
wtf does this sht mean
To find the unit rate, we need to determine how many millimeters of stretch there are for every gram of weight attached to the spring.
To do this, we can calculate the ratio of the length of stretch (mm) to the weight (g) for each data point given. By dividing the length of stretch by the weight, we can find the unit rate.
Here's how we can calculate the unit rate for each data point:
1. For the data point 50-: Since there is no weight mentioned, we cannot calculate the unit rate.
2. For the data point 45-: Again, there is no weight mentioned, so we cannot determine the unit rate.
3. For the data point 40-: No weight mentioned, so we cannot determine the unit rate.
4. For the data point 35: The length of stretch is 35mm, and there is no weight mentioned, so we cannot determine the unit rate.
5. For the data point 30-: No weight mentioned, so we cannot determine the unit rate.
6. For the data point 25: The length of stretch is 25mm, and there is no weight mentioned, so we cannot determine the unit rate.
7. For the data point 20-: No weight mentioned, so we cannot determine the unit rate.
8. For the data point 15-: No weight mentioned, so we cannot determine the unit rate.
9. For the data point 10-: No weight mentioned, so we cannot determine the unit rate.
10. For the data point 5: The length of stretch is 5mm, and the weight is 10g, so the unit rate is 5mm/10g or 0.5mm/g.
11. For the data point 0+: No weight mentioned, so we cannot determine the unit rate.
12. For the data point 0: The length of stretch is 0mm, and the weight is 0g, so the unit rate is 0mm/0g. However, we cannot divide by zero, so we are unable to determine the unit rate in this case.
13-21. For the data points 1-9 and 10: Unfortunately, without the lengths of stretch corresponding to these weights, we cannot calculate the unit rate.
Therefore, based on the information given, the only unit rate we can determine is 0.5mm/g for the data point 5g.
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.
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
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).