Physics, 06.07.2019 03:30 xavier0329

A15-uf capacitor is connected to a 50-v battery and becomes fullycharged. the battery is
removed and a slab of dielectric that completely fills the spacebetween the plates is inserted.
if the dielectric has a dielectric constant of 5.0:
a. what is the capacitance of the capacitor after the
slab is inserted?
b. what is the voltage across the capacitor's plates
after the slab is inserted?

Answers: 1

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Answer from: recon12759

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

A. The capacitance of the capacitor after the slab is inserted is 75-uF.B. The voltage across the capacitor's plates after the slab is inserted is 10 volts.What is capacitance of capacitor?

The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference.

The capacitance of the capacitor is 15-uF and it is connected to 50-V battery. The value of the dielectric constant is 5.0.

A. The capacitance of the capacitor after the slab is inserted

The dielectric constant of the slab is the ratio of capacitance of the capacitor after the slab and the capacitance of the before the slab inserted.

As the value of the dielectric constant is 5.0. Thus the capacitance of the capacitor after the slab is inserted given as,

$0.5=\dfrac{C}{15\times10^{-6}}\\C=75\rm \mu F$

Thus, the capacitance of the capacitor after the slab is inserted is 75-uF.

B. The voltage across the capacitor's plates after the slab is inserted

The dielectric constant of the slab is the ratio of voltage across the capacitor's of the capacitor before the slab inserted and the voltage across the capacitor after the slab inserted.

As the value of the dielectric constant is 5.0. Thus the voltage across the capacitor's plates after the slab is inserted can be given as,

$V=\dfrac{50}{5.0}\\C=10\rm V$

Thus, the voltage across the capacitor's plates after the slab is inserted is 10 volts.

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

A. The capacitance of the capacitor after the slab is inserted is 75-uF.B. The voltage across the capacitor's plates after the slab is inserted is 10 volts.

Learn more about the capacitance of capacitor here;

link

Answer from: andrealeon2000

A. C = 75μF and B. V = 10V.

We have to use the equation k = C/C₀ and k = V₀/V which both are the dielectric constant.

A. The capacitance after the slab is inserted.

With C₀ = 15μF and k = 5.0. Clear k for the equation k = C/C₀:

C = k*C₀

C = (5.0)(15x10⁻⁶F) = 0.000075F

C = 75μF

B. The voltage across the capacitor's plates after the slab is inserted.

With V₀ = 50V and k = 5.0. Clear V from the equation k = V₀/V:

V = V₀/k

V = 50V/5.0

V = 10V

Answer from: Quest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

Answer from: Quest

5.5 n====== $\sum f = ma$ assuming that $\sum f = f_{app}$ we have m = 0.145 kg a = 38 m/s²therefore $f_{app} = (0.145\text{ kg})(38\text{ m/s}^2) = 5.51\text{ n}$ 5.51 n, but two significant figures.final answer: 5.5 n

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A15-uf capacitor is connected to a 50-v battery and becomes fullycharged. the battery is
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