A Color class has a method getColorName that returns a string corresponding to the common name for the color, e.g., yellow, blue
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Answer:
Review the explanation
Explanation:
Response 1) Keeping all data in a single row presents challenges; if one record in the sheet is deleted, all related records may also vanish unexpectedly. For instance, should the entry for author "James Taylor" be removed, all the books associated with him would get eliminated as well.
Response 2) In the first normal form (1NF), each row is supposed to have just one tuple, but in the case of the author "James Taylor," the title field contains two books, necessitating separation into distinct rows.
Response 3) The table shown in the image does not exhibit deletion anomalies; for the author "May Norton," if you erase this record, it will result only in the deletion of her sole book.
Thus, it’s essential to transition this table into the second normal form (2NF).
Create distinct tables for Authors and Books
Authors(Author Name, Author Country)
Books(Author Name, Book Title, Publisher, Publisher Location, Year, Price)
This cannot undergo further normalization.
Answer:
1. The capacitive load for the Pentium 4 Prescott processor is 32 nF. For the Core i5 Ivy processor, it is 29.05 nF.
2. The static power makes up 10% of the total power dissipated for the Pentium 4 Prescott processor. The static to dynamic power ratio is 0.11.
For the Core i5 Ivy Bridge processor, the static power percentage is 42.86%. The ratio of static to dynamic power stands at 0.75.
3. The voltage reduction for the Pentium 4 Prescott processor equals a decrease of 5.9 %.
The Core i5 Ivy Bridge processor sees a 9.8 % reduction in voltage.
Explanation:
1. Recognizing dynamic power, P as approximately 1/2 CV²f, where C is the transistor's capacitive load, v denotes voltage, and f is frequency.
Thus, C is found using the formula C ≈ 2P/V²f.
For the Pentium 4 Prescott processor, with V₁ = 1.25 V, f₁ = 3.6 GHz, and P₁ = 90 W, we denote its capacitive load as C₁. Thus, we find C₁ ≈ 2P/V²f = 2 × 90 W/(1.25 V)² × 3.6 × 10⁹ Hz = 3.2 × 10⁻⁸ F = 32 × 10⁻⁹ F = 32 nF.
For the Core i5 Ivy Bridge processor, with V = 0.9 V, f = 3.4 GHz, and P = 40 W, we define C₂ as its load. Therefore, C₂ ≈ 2P/V²f = 2 × 40 W/(0.9 V)² × 3.4 × 10⁹ Hz = 2.905 × 10⁻⁸ F = 29.05 × 10⁻⁹ F = 29.05 nF.
2. The summation of total power is derived from static plus dynamic power.
For Pentium 4 Prescott, static power adds to 10 W and dynamic power is 90 W. Hence, the overall power, P = 10 W + 90 W = 100 W.
The fraction of this total attributed to static power is calculated as static power over total power multiplied by 100, thus static power/total power × 100 = 10/100 × 100 = 10%.
The ratio of static to dynamic power equals static power over dynamic power = 10/90 = 0.11.
For the Core i5 Ivy Bridge, static power figures at 30 W, and dynamic power at 40 W, meaning the total power becomes P = 30 W + 40 W = 70 W.
The portion of the total power that is static is computed as static power/total power × 100 = 30/70 × 100 = 42.86%.
That ratio of static to dynamic stands at static power/dynamic power = 30/40 = 0.75.
3. The total power comprises static and dynamic contributions and resulting leakage current arises from static power. We understand that P = IV, hence leakage current, I = P/V.
With an intended total power reduction of 10%, we have P₂ = (1 - 0.1)P₁ = 0.9P₁, where P₁ is the initial dissipated power before the 10% decrement and P₂ represents the new dissipated power.
Hence, new total dissipated power P₂ = new static power I₂V₂ + new dynamic power 1/2C₂V₂²f₂ = 0.9P₁.
For the Pentium 4 Prescott with P₂ = I₂V₂ + 1/2C₂V₂²f₂ = 0.9P₁, given I₂ as leakage current which equals static power/voltage = 10 W/1.25 V = 8 A (since leakage remains constant), we determine
8 A × V₂ + 1/2 × 32 × 10⁻⁹ F × V₂² × 3.6 × 10⁹ Hz = 0.9 × 100.
This simplifies to 8V₂ + 57.6V₂² = 90, leading to the quadratic equation.
57.6V₂² + 8V₂ - 90 = 0, from which applying the quadratic formula yields
V₂ = .
Choosing the positive result, V₂ arrives at 1.18 V. The calculated reduction percentage is given by (new voltage - old voltage)/new voltage × 100% = (1.18 - 1.25)/1.18 × 100% = -0.07/1.18 × 100% = -5.9% with a 5.9% drop from 1.25V.
For the Core i5 Ivy Bridge processor, it follows that P₂ = I₂V₂ + 1/2C₂V₂²f₂ = 0.9P₁. With I₂ as leakage current equaling static power/voltage = 30 W/0.9 V = 33.33 A (again, leakage remains constant), we next evaluate
33.33 A × V₂ + 1/2 × 29.05 × 10⁻⁹ F × V₂² × 3.4 × 10⁹ Hz = 0.9 × 70.
This resolves to 33.33V₂ + 49.385V₂² = 63. Thus, it simplifies to the quadratic equation
49.385V₂² + 33.33V₂ - 63 = 0, whereby employing the quadratic formula lets us find
V₂ = .
Choosing the positive answer provides a new voltage of 0.82 V. The percentage reduction computes as (new voltage - old voltage)/new voltage × 100% = (0.82 - 0.9)/0.82 × 100% = -0.08/0.82 × 100% = -9.8% with a 9.8% decrease from 0.9V.