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2 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

1 answer:

grandymaker [24]2 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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If the intensity level by 15 identical engines in a garage is 100 dB, what is the intensity level generated by each one of these

goblinko [34]

To develop this problem we will apply the concepts related to Sound Intensity. This can be defined as the power carried by sound waves per unit area in a direction perpendicular to that area. Mathematically it can be written as,

$\Delta L = 10log(n)$

$\Delta L$ = Level Difference

n = Number of equal loud sound sources

n = 15

Therefore replacing we have,

$\Delta L = 10 log (n)\\\Delta L = 10 log (15)\\\Delta L = 10*1.176\\\Delta L = 11.76dB$

Therefore over 100dB the total difference intensity level would be

$L_T = L_0 -\Delta L$

Here,

$L_0 = \text{ Total intensity}$

$\L_T = 100 -11.76$

$_T= 88.24dB$

The intensity level generated by one engine is

$L_T = 88.44dB$

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2 years ago

7. An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where the Earth’s horizontal magnetic field

VashaNatasha [74]

Answer:

0.00479 volts

Explanation:

From Faraday's law of electromagnetic induction, the induced emf equals the change in magnitude flux (magnetic field strength multiplied by the area = BA) divided by the time change

Therefore we have the equation

EMF = BA÷t

Since area A = 2πr

EMF = Bπr²÷t

B = 5.5×10^(-5)

Velocity = 100km/h = 27.7778m/s

r = 1m, t = r÷V = 0.036

EMF = Bπr²÷t = (5.5×10^(-5) x π x (1)²)÷0.036 = 0.00479 Volts

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2 years ago

What does the slope of a position time graph determine?

kupik [55]

The slope of a position/time graph is the speed of the motion.
(The slope doesn't 'determine' the speed; it only displays it.)

The slope of a speed/time graph is the acceleration of the motion.

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2 years ago

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$

5 0

2 years ago

Read 2 more answers

A 25.0 kg object is held 8.50 m above the ground. Calculate its PE

pantera1 [17]

Answer:=mgh

Explanation:

6 0

2 years ago