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  1. ... heated to 10 deg.C, the new volume is, V( new ) = 5x[ 1 + 0 .53x10^-4(10-5)] mL = 5.001325 mL

  2. ...the frequency of the fundamental frequency is -0.015, i.e. -1.5% Hence, new fundamental frequency = 400 × ( 1 - 0 .015) = 394 Hz (c) Percentage change...

  3. ...balance reading to increase. The answer is option B. 2. New current I' = ( 1 - 0 .25)I = 0.75I Initial power ...

  4. ..., when the small charge is moved to a distance of 1 . 0 m, New potential energy at is: W = Qq/4πε0 So work done...

  5. ...the stress will be increased by 10 times, implying that the new ratio of the stress to the breaking stress = 0.875.

  6. ...'=(V^2)/R------------------(2) Sub (2) into ( 1 ) 0 .5P'=(V^2)/R=20 0.5P'=20 P'=20/0.5 P'=40W So,the new power is 40W.

  7. (1) Using the period formula: T = 2π√(m/k) Then, New T/Old T = √(1.25/0.75) New T = 2&radic...

  8. ... connected in series. Capacitance of each of the new capacitor = e.A/[(d-l)/2] = 2eA/(d-l) where e is the permittivity of free space Hence, equivalent capacitance C' of two new capacitos in series = [2eA/(d-l)]/2 = eA/(d-l) (b) C' = eA/(d-0.4d) = eA/0.6d Hence, ratio = (eA/0.6d)/(eA/d) = 1 / 0 .6 = 5:3

  9. ...gas equation: PV/T=constant, (10)(0.26A)/285 = P(0.25A)/294 where P is the new pressure exerted by the trapped air solve for P gives P = 10.7 mmHg ...

  10. ...2(V2) v2 =0.353 ms^-1 <--- v2 = 0.112 m/s since now the new KE is (0.5)(1+ 1 )( 0 .353 )^2 =0.124 J <--- 0...

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