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3 years ago

A 19.0-μF capacitor and a 44.0-μF capacitor are charged by being connected across separate 45.0-V batteries.

Physics

1 answer:

nikitadnepr [17]3 years ago

3 0

Answer:

Explanation:

C1 = 19 micro farad

C2 = 44 micro farad

V = 45 V

The relation between the charge and the potential is

q = C x V

where, c is the capacitance and V be the potential difference

q1 = C1 x V = 19 x 45 = 855 micro Coulomb

q2 = C2 x V = 44 x 45 = 1980 micro Coulomb

Now the charge is shared, the charge on both the capacitors = (q1 + q2)/ 2

q = ( 855 + 1980) / 2 = 1417.5 micro Coulomb

Potentia across 44 micro farad

V' = q/C2 = 1417.5 / 44 = 32.2 V

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Deffense [45]

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2 years ago

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

3 years ago

g Which of the following wavefunctions are: a) square-integrable on the interval provided (1 point); b) valid wavefunction satis

Goshia [24]

Answer:

The answer is given in the attachment

Explanation:

5 0

4 years ago

Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart

Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0

3 years ago

You're in the middle of moving and need to bring furniture and into your new house. You want to prop the door open, but your dog

Mamont248 [21]

Answer:

C) Put the brick as far from the hinges as possible

Explanation:

As torque is the product of the force around the rotation point and the distance to the pivot point, and the mass (force) of the brick stays constant, what we can do to maximize the torque is maximize the distance to the pivot point, aka the hinge. So we should put the brick as far from the hinges as possible.

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3 years ago