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### A 7561.0

**pF**capacitor holds 0.173 microCoulombs of charge. What is the voltage across the capacitor?Capacitance C = Q/V => V = Q/C C = 7561.0×10^-12 F Q = 0.173×10^-6 C => V = 0.173×10^-6 / 7561.0×10^-12 = 22.88 V

1 Answers · Science & Mathematics · 27/02/2015

Volume V = 20*30*thck = 600*thck dm^3 Volume V = mass/density = 324/2.7 = 120.0 dm^3 thickness = 120/600 = 0.200 dm = 2.00 cm

1 Answers · Science & Mathematics · 16/12/2016

The formula to be used is the parallel plate capacitor formula: C = ε0 S / d where C = capacitance, S plate area, d plate distance. Then: d = ε0 S / C = 8.85 x 10^-12 x 3.14 x (0.40 x 10^-3)² / (100.0 x 10^-12) = 0.044 x 10^-6 m = 0.044 µm

1 Answers · Science & Mathematics · 15/12/2010

The basic formula is q =V *C where q is the charge and V is the emf across the capacitor, C is the capacitance. Therefore V= q/c C= 7.731 * 10 ^ -9 F q=1.83 * 10 ^ -7 C Dividing out I get 23.7...

1 Answers · Science & Mathematics · 13/11/2012

E = (1/2) CV^2 = (0.5)(2400x10-12)(650)^2 = 5.1x10^-4 J

2 Answers · Science & Mathematics · 03/02/2013

Q = CV V = Q/C = 18*10^(-8)/5.2*10^(-9) V = 34.6V

1 Answers · Science & Mathematics · 27/05/2009

I don't all the things you're allowed to assume, but from the Bolzano-Weierstrauss Theorem we know that K contains all of its limit points (accumulation points, or whatever naming convention you wish to use...

1 Answers · Science & Mathematics · 03/06/2012

V=Q/C=15.5E-8/5.8E-9=26.7 volts.

1 Answers · Science & Mathematics · 13/05/2011

Assuming you mean that the density is 19.3 g/cm³ (cm cubed) Volume you can calculate from sphere V = ⁴/₃πr³ multiply by the density to get mass in grams convert that mass into pounds. 1 US pound = 454 grams.

3 Answers · Science & Mathematics · 21/01/2013

V = Q / C = 27.2v

1 Answers · Science & Mathematics · 10/12/2012

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