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3 years ago

A 4V source is connected across a 2μF capacitor. Find the charge on this capacitor.

Physics

1 answer:

Feliz [49]3 years ago

6 0

Answer: $8$ μc

have: 2μF = $2.10^{-6}$ F

the charge on this capacitor is $2.10^{-6}.4=8.10^{-6}(c)=8$ (μc)

Explanation:

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The surface temperature of a planet depends on both the distance of the planet from the Sun and on how the panet's atmosphere di

BaLLatris [955]

Answer:

Aphelion: 6404 W/m2

Perihelion: 14978 W/m2

Explanation:

The solar energy flux depends on the solar power output divided by the surface of a sphere with a radius equal to the distance to the Sun.

$\Phi sol = \frac{Psol}{4 * \pi * d^2}$

The distances we need are the aphelion and perihelion of Mercury.

Planetary orbits are ellipses. In an ellipse the eccentricity is related to linear eccentricity and the length of the semi major axis:

$e = \frac{c}{a}$

Where

e: eccentricity

c: linear eccentricity

a: semi major axis

The linear eccentricity is equal to the distance of the focus of the center of the ellipse.

$c = a * c =$

a = 0.39 AU = 5.83e10 m

$c = 5.83e10e * 0.21 = 1.22e10 m$

In planetary orbits the Sun is in one of the fucuses. With this we can calculate the prihelion and aphelion as:

Ap = a + c = 5.83e10 + 1.22e10 = 7.05e10 m

Pe = a - c = 5.83e10 - 1.22e10 = 4.61e10 m

And the solar energy fluxes will be:

$\Phi Ap = \frac{4e26}{4 * \pi * 7.05e10^2} = 6404 W/m2$

$\Phi Pe = \frac{4e26}{4 * \pi * 4.61e10^2} = 14978 W/m2$

4 0

3 years ago

What is the maximum speed with which a 1200-kg car can round a turn of radius 90.0 m on a flat road if the coefficient of fricti

RideAnS [48]

Answer:

Maximum speed of the car is 26.56 m/s.

Explanation:

Given that,

Mass of the car, m = 1200 kg

Radius of the curve, r = 90 m

The coefficient of friction between tires and the road is 0.8.

We need to find the maximum speed of the car. On the circular curve, the centripetal force is balanced by the force of friction. So,

$\dfrac{v^2}{r}=\mu g\\\\v=\sqrt{\mu gr} \\\\v=\sqrt{0.8\times 9.8\times 90} \\\\v=26.56\ m/s$

So, the maximum speed of the car is 26.56 m/s.

5 0

4 years ago

A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it

mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

$radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm$

$r = 0.145\ cm = 0.145\times 10^{-2}\ m$

Charge ,

$Q = 30.0\ pC=30\times 10^{-12}$

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

$V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}$

Where,

V= Electric potential,

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge

r = Radius

Substituting the values we get

$V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}$

$V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt$

Therefore,

The potential (in V) near its surface is 186.13 Volt.

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