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torisob [31]
3 years ago
15

During normal beating, the heart creates a maximum 3.60 mV potential across 0.250 m of a person's chest, creating a 1.00 Hz elec

tromagnetic wave. (a) What is the maximum electric field strength (in V/m) created
Physics
1 answer:
grandymaker [24]3 years ago
5 0

Answer:

The maximum electric field strength is 0.0144 V/m.

Explanation:

Given that,

Electric potential created in the heart, V = 3.6 mV

Distance, d= 0.25 m

Frequency of the the electromagnetic wave, f = 1 Hz

We need to find the maximum electric field strength created. We know that the electric potential is given by :

V=Ed

E is the maximum electric field strength

E=\dfrac{V}{d}\\\\E=\dfrac{3.6\times 10^{-3}\ V}{0.25\ m}\\\\E=0.0144\ V/m

So, the maximum electric field strength is 0.0144 V/m. Hence, this is the required solution.

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For your senior project, you are designing a Geiger tube for detecting radiation in the nuclear physics laboratory. This instrum
Paha777 [63]

Answer:

Maximum linear charge density = 84.14 nC/m

Explanation:

Looking at this question, The electric field of a line charge of infinite length is given by : Er = (1/(2πεo)) x (λ/r)

r = the distance from the center of the line of charge

λ = the linear charge density of the wire.

Now looking at the equatiom, due to the fact that Er varies inveresely with r, its maximum value will occur at the surface of the wire where r = R, the radius of the wire:

And so, Emax = (1/(2πεo)) x (λ/R)

Let's make λ the subject of the equation and we get;

λ = 2πεo(REmax)

From the question, R = 0.55/2 = 0.275cm

Also, Emax = 5.50 × 10^(6) N/C

Let's take the value of the electric constant to be εo = 8.854 x 10^(-9) C^(2) / Nm^2

R = 0.275mm = 0.000275m

Plugging these values into the equation, we get;

λ = 2π x 8.854 x 10^(-12) x 0.000275 x 5.50 × 10^(6) = 84.14 nC/m

4 0
3 years ago
Select all the correct answers.
amid [387]

Answer:

A

Explanation:

because in the moving object there's a certain energy applied

8 0
3 years ago
Einstein calculated that ripples of gravity travel at exactly the speed of _____
damaskus [11]

Answer:

299,792,458 m/s = speed of light

Explanation:

6 0
2 years ago
A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6
Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6  = V/0.0001

V = 2.0 × 10^6  * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q =  1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0
2 years ago
With certain exceptions, Class E airspace extends upward from either 700 feet or 1,200 feet AGL to, but does not include,A) 14,5
DedPeter [7]

Answer:

B) 18,000 feet MSL

Explanation:

There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.

6 0
3 years ago
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