A 300N box on a 43 degree angle.

A set of four capacitors are attached to a 12V battery in the circuit shown below. All capacitances are measured in milli-Farads

Bond [772]

The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is calculated as follows;

Capacitors in series;

1/Ct = 1/8 + 1/7.5

1/Ct = 0.25833

Ct = 3.87 mF

Capacitors is parallel;

Ct = 3.87 mF + 12 mF + 15 mF

Ct = 30.87 mF

Ct = 0.03087 F

<h3>Charge in each capacitor</h3>

Q = CV

Q = 0.03087 x 12

Q = 0.37 C

Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

Learn more about capacitors here: brainly.com/question/13578522

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3 0

1 year ago

Vector vector b has x, y, and z components of 4.00, 4.00, and 2.00 units, respectively. calculate the magnitude of vector

Sav [38]

Good morning.

We see that $\mathsf{\overset{\to}{b}} = \mathsf{(4.00, \ 4.00, \ 2.00)}$

The magnitude(norm, to be precise) can be calculated the following way:

$\star \ \boxed{\mathsf{\overset{\to}{a}=(x, y,z)\Rightarrow ||\overset{\to}{a}|| = \sqrt{x^2+y^2+z^2}}}$

Now the calculus is trivial:

$\mathsf{\|\overset{\to}{b} \| =\sqrt{4^2+4^2+2^2} =\sqrt{16+16+4}}\\ \\ \mathsf{\|\overset{\to}{b}\|=\sqrt{36}}\\ \\ \boxed{\mathsf{\|\overset{\to}{b}\| = 6.00 \ u}}$

7 0

3 years ago

A ray of light is incident on a body x what is the reftactive index

evablogger [386]

Answer:

1.63

Explanation:

If you have the following options:

B. 1.50

C. 1.49

D. 1.33

E. 1.02

6 0

2 years ago

A meter stick A hurtles through space at a speed v = 0.25c relative to you, with its length aligned with the direction of motion

yaroslaw [1]

Answer:

$L_0\approx1.0328\ m$

Explanation:

Given:

relativistic length of stick A, $L=1\ m$

relativistic velocity of stick A with respect to observer, $v=0.25c=7.5\times 10^{7}\ m.s^{-1}$

<em>Since the object is moving with a velocity comparable to the velocity of light with respect to the observer therefore the length will appear shorter according to the theory of relativity.</em>

<u> Mathematical expression of the theory of relativity for length contraction:</u>

$L=\frac{L_0}{\gamma}$

where:

L = relativistic length

$L_0=$ original length at rest

$\gamma =$ Lorentz factor $=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$

$\Rightarrow 1=\frac{L_0}{\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }}$

$L_0=\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }$

$L_0\approx1.0328\ m$

4 0

2 years ago

A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly ch

IceJOKER [234]

Answer:

Explanation:

Given that,

First Capacitor is 10 µF

C_1 = 10 µF

Potential difference is

V_1 = 10 V.

The charge on the plate is

q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC

q_1 = 100 µC

A second capacitor is 5 µF

C_2 = 5 µF

Potential difference is

V_2 = 5V.

Then, the charge on the capacitor 2 is.

q_2 = C_2 × V_2

q_2 = 5µF × 5 = 25 µC

Then, the average capacitance is

q = (q_1 + q_2) / 2

q = (25 + 100) / 2

q = 62.5µC

B. The two capacitor are connected together, then the equivalent capacitance is

Ceq = C_1 + C_2.

Ceq = 10 µF + 5 µF.

Ceq = 15 µF.

The average voltage is

V = (V_1 + V_2) / 2

V = (10 + 5)/2

V = 15 / 2 = 7.5V

Energy dissipated is

U = ½Ceq•V²

U = ½ × 15 × 10^-6 × 7.5²

U = 4.22 × 10^-4 J

U = 422 × 10^-6

U = 422 µJ

6 0

3 years ago