What is electricity?

During normal beating, a heart creates a maximum 3.95-mV potential across 0.305 m of a person’s chest, creating a 0.75-Hz electr

WINSTONCH [101]

Answer:

E = 0.0130 V/m.

Explanation:

The electric field is related to the potential difference as follows:

$E = \frac{\Delta V}{d}$

<u>Where:</u>

E: is electric field

ΔV: is the potential difference = 3.95 mV

d: is the distance of a person's chest = 0.305 m

Then, the electric field is:

$E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m$

Therefore, the maximum electric field created is 0.0130 V/m.

I hope it helps you!

7 0

3 years ago

A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

$\lambda = \frac{2\pi}{k}$

$\lambda = (frac{2\pi}{\frac{\pi}{11})
$

$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0

4 years ago

Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in

Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

$L = \frac{N^2 \mu_0 A}{l}$

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

$A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2$

$L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH$

(b) The energy stored in the inductor when 21 A current ;

$E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J$

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

$emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s$

3 0

3 years ago

This picture represents the electric field diagram between two particles with static charges. Do the two particles have the same

dexar [7]

Answers:

No, They will attract each other, B, and neither direction

Explanation:

Since the two already presented particles in the diagram represent both opposing charges due to the direction of the arrows (the arrows facing away from the particle shows a positive charge and the particles facing towards the particle show a negative charge), not only because of this but as the arrows between the particles show an attracting magnetic field, then it can be concluded that the particles will attract to each other and if another particle was introduced into the diagram of a positive charge, then it would attract to the negatively charged particle. If you have any questions or need further explanation, please comment below. E2021, have a great day.

7 0

3 years ago

Please help on this one?

diamong [38]

Answer:

ITS C

Explanation:

7 0

3 years ago