nothing seriously‚ all he likes to do is eat and train. The only time to take things seriously is when he is battling. When it comes to battling‚ he is very strategic. In the Dragon Ball Z universe people are able to use Ki. Ki is the force that’s forms of any living thing. Goku uses Ki to perform destructive‚ and powerful attacks. He has an arsenal of
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divisor is 10 times the quotient and 5 times the remainder. If the remainder is 46‚ the dividend is: (1) 4236 (2) 4306 (4) 5336 m sT 2. If 1.5 x= 0.04 y‚ then the value of (y-x) (y+x) is: od ay (3) 4336 (1) 730/77 (2) 73/77 kE xa (3) 7.3/77 (4) 703/77 .B an 3. An employee may claim Rs. 7.00 for each km when he travels by taxi and Rs. 6.00 for each km if he drives his own car. If in one week he claimed Rs. 595 for traveling km. How many kms did he travel by taxi
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produced. This iodine will be converted into I3- which will react stoichiometrically with the thiosulfate. The thiosulfate to be used for titration was first standardized using KIO3 as the primary standard. KI crystals and sulfuric acid was added to the KIO3 solution one after the other. KI crystals should be added first or the solution may produce more O2 and add to the analyzed dissolved oxygen of the sample: The resulting solution was titrated right away with Na2S2O3 as the titrant and starch
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Sukses Berwirausaha: Bangkit dari Kegagalan Ada pepatah kuno yang mengajarkan pada kita bahwa sesuatu yang kecil‚ kalau kita tekun membesarkannya‚ lama-lama akan menjadi besar juga. Sedikit-sedikit lama-lama menjadi bukit. Semakin besar bisnis yang ingin kita capai‚ maka kita perlu menyederhanakannya dengan mulai melangkah‚ sekecil apapun langkah itu. Banyak pengusaha yang memulai dari bisnis kecil-kecilan namun perlahan-lahan kemudian menjadi pengusaha besar. Lalu kita hanya membutuhkan kesabaran
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Motor parameters used for simulation are given in Table 1. Tables 2 and 3 shows the GA-PID and ACO-PID controller’s parameters. the response of Kp‚ Ki and Kd parameters are summarized on table 4 which shows that the proportional controller “Kp” will have the effect of reducing rise time and will reduce the steady state error while the integral controller “Ki” will have the effect of eliminating the steady state error but will have the worst transient response on a system while the derivative
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reaction CH3 NC(g) −→ CH3 CN(g) are 4.0 × 1013 s−1 and 272 kJ/mole‚ respectively. Calculate the values of ∆S ◦‡ ‚ ∆H ◦‡ ‚ and ∆G◦‡ at 300 K. 7. [20 points] An enzyme-catalyzed reaction (KM = 2.7 × 10−3 M) is inhibited by a competitive inhibitor I (KI = 3.1 × 10−5 M). Suppose that the substrate concentration is 3.6 × 10−4 M. How much of an inhibitor is needed for 65% inhibition? How much does the substrate concentration have to be increased to reduce the inhibition to 25%? Constants R = 8
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Qualitative Observations of Double Displacement Reactions Lab Table 1.0 Qualitative Observation of Products Formed |Balanced Chemical Equations |Qualitative Observations | |BaCl2 (aq) + 2NaOH (aq)( BaOH2(aq) + 2NaCl(s) |An aqueous solution formed | | |Precipitate
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Nike and Flying Fish Nike and Flying Fish is a 1998 artwork fathered by Ke Francis ;an art teacher‚ owner of Hoop snake Press‚ and manager of the university of central Florida’ publishing company‚ Flying Horse Press. A layered style supports the arrangement of items painted. “A rabbit trap layered under a funnel with a flying fish that’s face to face with a drawn image of Nike‚ the Winged Victory of Samothrace‚ a Greek pagan goddess. Francis’s work is a seventy-two and a forth by seventy-eight
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Project 1: 1. Results for a: ht=500/ spd=0mph —too slow! Satellite crashed into Earth! Results for b: ht=500/ spd=100mph —too slow! Satellite crashed into Earth! Results for c: ht=500/ spd=25000mph —too fast! Satellite left viewing area and took off in space somewhere! I was determined to get a 100%‚ so I played around with the numbers until I found that at 500 miles above the Earth‚ the satellite would have to be launched at 16‚700mph in order to reach a perfect orbit. Below is the explanation
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000 lb Altitude = 10‚000 ft Airspeed = 300 KTAS PE = mgh PE = (50000)(10000) PE = 5*108 KE = ½mv2 KE = ½(50000/32.17)(300*1.6878)2 KE = 1.99*108 TE = 6.99*108 7. Given conditions of #6‚ find new altitude if the aircraft raises the nose to capture 250 KTAS but keeps the thrust at the same setting and‚ for simplicity‚ assume the drag stays the same. Think of this as a Constant Energy Problem. TE = KE + PE
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