All categories

3 years ago

When a 12 V battery is connected to a 6 uF capacitor, how much energy is stored

Physics

1 answer:

adoni [48]3 years ago

4 0

The energy stored in a capacitor is

E = (1/2) · (capacitance) · (voltage)²

E = (1/2) · (6 x 10⁻⁶ F) · (12 V)²

E = (3 x 10⁻⁶ F) · (144 V²)

<em>E = 4.32 x 10⁻⁴ Joule</em>

(That's 0.000432 of a Joule)

You might be interested in

What's effect called?<br>

k0ka [10]

Answer:

what do u mean

Explanation:

8 0

3 years ago

Read 2 more answers

If a satellite travels 62,000,000 miles from earth, how can we write it in scientific notation

Fudgin [204]

Answer:

6.2 x 10^6

Explanation:

The 6 at the end is an exponent the carrot is the symbol

6 0

3 years ago

Read 2 more answers

Mass m, moving at speed 2v, approaches mass 4m, moving at speed v. The two collide elastically head-on. Part A Find the subseque

meriva

Thank you for posting your question here at brainly. A mass of m moves with 2V towards in the opposite direction of a mass, 4m moving at a speed of V, the speed of m was 2/5V and the mass of 4m was 7.5V. I hope it helps.

4 0

3 years ago

Read 2 more answers

Given: g = 9.8 m/s 2 . A small glass of water is placed on carousel inside a microwave oven, at a radius 6.5 cm from the center.

pashok25 [27]

Answer:

0.17 degrees

Explanation:

From the question we are given the following:

acceleration due to gravity = $9.8 \frac{m}{s^{2} }$

radius (r) = 6.5 cm = 0.065 m

time (s) = 9.3 s

angle = ?

we first have to calculate the distance the water moves, and this is the circumference of the microwave

circumference = 2 x π x r = 2 x π x 0.065 = 0.408 m

Now we find the speed with which it moves at

speed = $\frac{distance}{time}$

= $\frac{o.408}{9.3}$ = 0.044

The next step is to find the centripetal acceleration

a = $\frac{v^{2} }{r}$

= $\frac{0.044^{2} }{0.065}$ = 0.03

being a vector quantity, acceleration has direction, the centripetal acceleration is the x component while the y component would be the acceleration due to gravity (g)

A = 0.03 x + 9.8 y

to find the angle we can apply tan θ = $\frac{ 0.03 }{9.8$

θ = 0.17 degrees

7 0

3 years ago

What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g

aalyn [17]

Answer:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle $\theta$ , given by:

$f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)$

Where:

$Z_1 e$ represent the charge of the projectile (Z1=2)

$Z_2 e$ is the target charge (z2=79)

$K= 5x10^6 eV$ represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

$t= 10^{-8] m$ represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

$n =\frac{\rho N_A N_M}{M_g}$

Where:

$\rho = 19.3 g/m^3= 19300 Kg/m^3$

$N_A = 6.022 x10^{23} molecules/mol$ the Avogadro's number

$N_M = 1$ represent the atoms per molecule

$M_g = 197 g/mol = 0.197 Kg/mol$ represent the molecular weigth

If we replace we got:

$n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3$

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness $t=10^{-8}m$

And using the first formula we got:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

4 0

3 years ago