Given values:
Mass of the steel ball, m = 100 g = 0.1 kg
Height of the steel ball, h1 = 1.8 m
Rebound height, h2 = 1.25 m
a. PE= mgh
0.1 x 9.8 x 1.8 =
1.764 Joules
b. KE = PE ->
1.764 Joules
c. KE= 1/2 mv square
so v = square root 2ke/m
square root 2 x 1.764/ 0.1
= 5.93 m/s
d. KE=PE=mgh square
0.1 x 9.8 x 1.21 =
1.186 joules
velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s
Each increase in the prefix is a division by 1000.
Answer:
1. Urea, (d)Fertilizer
2. Combines, (c) Harvesting and threshing
3. Silos, (b) Storage of grains
4. Seed drills, (a) Sowing seeds
5. Irrigation, (f) Sprinklers
6. Tilling, (e) Preparation of soil
Explanation:
1) Urea is a widely used and important fertilizer in the agricultural industry
2) The combine harvester combines three categories of harvesting grain drops such as threshing reaping and winnowing
3) Silos are used to store grains
4) A seed drill is a seed planting mechanism for burying seeds to a particular depth during seed planting
5) Irrigation is the application of required volume of water to plants
6) Tilling is the digging, turning, and staring operations meant to prepare the soil.
C. sound waves and water waves
Answer:
(D) 4
Explanation:
The percentage error in each of the contributors to the calculation is 1%. The maximum error in the calculation is approximately the sum of the errors of each contributor, multiplied by the number of times it is a factor in the calculation.
density = mass/volume
density = mass/(π(radius^2)(length))
So, mass and length are each a factor once, and radius is a factor twice. Then the total percentage error is approximately 1% +1% +2×1% = 4%.
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If you look at the maximum and minimum density, you find they are ...
{0.0611718, 0.0662668} g/(mm²·cm)
The ratio of the maximum value to the mean of these values is about 1.03998. So, the maximum is 3.998% higher than the "nominal" density.
The error is about 4%.
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<em>Additional comment</em>
If you work through the details of the math, you will see that the above-described sum of error percentages is <em>just an approximation</em>. If you need a more exact error estimate, it is best to work with the ranges of the numbers involved, and/or their distributions.
Using numbers with uniformly distributed errors will give different results than with normally distributed errors. When such distributions are involved, you need to carefully define what you mean by a maximum error. (By definition, normal distributions extend to infinity in both directions.) While the central limit theorem tends to apply, the actual shape of the error distribution may not be precisely normal.
