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 electromagnetic wa

Question

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 electromagnetic wave. During normal beating of the heart, the maximum value of the oscillating potential difference across 0.305 m of a person’s chest is 3.95 mV. This oscillating potential difference produces a 0.75-Hz electromagnetic wave.

What is the maximum electric field created?

in progress 0
Thanh Hà 4 years 2021-09-03T08:37:04+00:00 1 Answers 12 views 0

Answers ( )

  1. Thunguyet
    0
    2021-09-03T08:38:44+00:00
    Reply

    Answer:

    E = 0.0130 V/m.

    Explanation:

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

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

    Where:

    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!

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )