What is a benefit of fuel cell cars over standard cars?
1 answer:
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Answer:
Explanation:
Radius of dee, r = 8 mm = 0.008 m
Electric field, e = 400 V/m
Magnetic field, B = 4.7 x 10^-4 T
mass of electron, m = 9.1 x 10^-31 kg
charge of electron, q = 1.6 x 10^-19 C
(a) Let v is the speed of electrons.
v = 661098.9 = 661099 m/s
(b)
e / m = 1.76 x 10^14 C / kg
(c) Let K be the kinetic energy
K = 0.5 x mv²
K = 0.5 x 9.1 x 10^-31 x 661099 x 661099
K = 1.99 x 10^-19 J
K = 1.24 eV
So, the potential difference is
V = 1.24 V
(d) if the acceleration voltage is doubled
V = 2 x 1.24 = 2.48 V
So, Kinetic energy
K = 2.48 eV
K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J
Let v is the speed
K = 0.5 x mv²
3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²
v = 933856.5 m/s
Let the new radius is r.
r = 0.0113 m = 1.13 cm
voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]
=1.33μf
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
Q=CV
given,V=8v
charge on 2.0μf capacitor is
=5.32v
learn more about series capacitance from here: brainly.com/question/28166078
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