The professor gave us no background material or explanations for any of these questions.
1. A contemporary computer chip dissipates 54 watts of power on a area of 1 square cm. due to the transistors inside this area. Assuming that transistors in succeding computer generations require constant power independent of their size, estimate the power that will be burnt in a 1 cm square computer chip 3 years from now. Here we are going to assume “Moore's law”. This law claims that the transistor count (# transistors per unit area) doubles every 1.5 years.
A. 700 Watts
B. 108 Watts
C. 10 Kilowatts
D. 216 Watts
I put D
2. Due to continuous use a mascot’s sword has lost its luster. A sculptor decides to propose a replacement. She builds a small scale model and discovers she needs 200 grams of bronze. Her final sword will be 6 times as big as her model in all dimensions. How much bronze will she need for the final product?
A. 11 kgs
B. 22 kgs
C. 43.2 kgs
D. 6 kgs
I feel like I don't have enough information to answer this problem.
3. When the student is finished with her small scale model in the above problem, she finds that she can give it two coats of paint using exactly one small can. How many cans does she need for the full scale model for the same number of coats?
A. 72 cans
B. 36 cans
C. 216 cans
D. 6 cans
Again I feel like there isn't enough information to answer this.
4. Eighteen grams of water is known to contain 6.02 x 1024 molecules of H20.
What is the mass in kilograms of one molecule of H20?
A. 2.99 x 10-24 kg
B. 54 x 10-27 kg
C. 6.02 x 10-24 kg
D. 2.99 x 10-27 kg
My answer matches none of these options.
Thanks a lot!
I kept working on these and now I have C for 2 and B for 3 but I'm still unsure.
Also, I came out with D for 4 but again I'm not sure.
1 correct
2. the enlargement factor will be six cubed.
3. Surface area increases by a factor of six squared. Think on that.
4..018/(6.02E24) matches a number. In reality, Alvarado's number is listed erroneously, most accept it as 6.02E23
1. To estimate the power that will be burnt in a 1 cm square computer chip 3 years from now, we can use Moore's law. According to Moore's law, the transistor count doubles every 1.5 years. Since the power dissipated is assumed to be constant regardless of the size, we can assume that the power density (power per unit area) remains constant as well.
Given that the power dissipated on a 1 cm square computer chip is 54 watts, we can determine the power dissipated on a 1 cm square computer chip 3 years from now as follows:
- In 1.5 years, the transistor count doubles, which means the area needed for the same number of transistors halves. So, the area will be 0.5 cm square (half of 1 cm square).
- In another 1.5 years, the transistor count doubles again, reducing the area to 0.25 cm square (half of 0.5 cm square).
Therefore, the power dissipated on a 1 cm square computer chip 3 years from now would be distributed on an area of 0.25 cm square. To determine the power, we can use the conservation of power density:
Power density before = Power density later
Power before / Area before = Power later / Area later
54 watts / 1 cm^2 = Power later / 0.25 cm^2
Solving for Power later:
Power later = (Power before / Area before) * Area later
Power later = (54 watts / 1 cm^2) * 0.25 cm^2
Power later = 13.5 watts
Therefore, the estimated power that will be burnt in a 1 cm square computer chip 3 years from now is 13.5 watts. So, the correct answer is not listed in the options.
2. To determine how much bronze the sculptor will need for the final product, we can use the information provided. The scale model requires 200 grams of bronze, and the final sword will be 6 times as big as the model in all dimensions.
Since the dimensions are scaling up uniformly by a factor of 6, the volume of the final sword will increase by a factor of 6^3 = 216. The mass of the bronze will also increase by the same factor since it is assumed to have a uniform density.
So, the mass of the bronze needed for the final product would be:
Mass of bronze needed for final product = Mass of bronze needed for model * Scaling factor
Mass of bronze needed for final product = 200 grams * 216
Mass of bronze needed for final product = 43,200 grams
Converting grams to kilograms:
Mass of bronze needed for final product = 43,200 grams / 1000 = 43.2 kilograms
Therefore, the sculptor will need 43.2 kilograms of bronze for the final product. So, the correct answer is option C.
3. Unfortunately, based on the information provided, we cannot determine the number of cans needed for the full-scale model's coats of paint. It is unspecified how the scale or proportions of the small can and full-scale model compare. Therefore, there is not enough information to answer this question.
4. To determine the mass of one molecule of H2O, we can use the provided information that 18 grams of water contains 6.02 x 10^24 molecules of H2O.
We know that 18 grams of water is equal to 6.02 x 10^24 molecules of H2O. Therefore, we can set up a proportion:
Mass of 1 molecule of H2O / Number of molecules of H2O = 18 grams / (6.02 x 10^24)
Solving for the mass of 1 molecule of H2O:
Mass of 1 molecule of H2O = (18 grams / (6.02 x 10^24)) * 1 molecule of H2O
Mass of 1 molecule of H2O = 2.99 x 10^-24 grams
Converting grams to kilograms:
Mass of 1 molecule of H2O = 2.99 x 10^-24 grams / 1000 = 2.99 x 10^-27 kilograms
Therefore, the mass of one molecule of H2O is approximately 2.99 x 10^-27 kilograms. So, the correct answer is option D.