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myrzilka [38]
3 years ago
5

Large electric fields in cell membranes cause ions to move through the cell wall. The field strength in a typical membrane is 1.

0 times 107 N/C. What is the magnitude of the force on a calcium ion with charge +e?
Physics
1 answer:
serg [7]3 years ago
6 0

Explanation:

The given data is as follows.

          Electric field (E) = 1 \times 10^{7} N/C

          Charge (e) = 1.6 \times 10^{-19} C

Formula to calculate the magnitude of force is as follows.

                F = qE

                   = 1.6 \times 10^{-19} \times 1 \times 10^{7} N/C

                   = 1.6 \times 10^{-12} N

Therefore, we can conclude that magnitude of the force on a calcium ion with charge +e is  1.6 \times 10^{-12} N.

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Answer:

(from top to bottom)

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3 years ago
A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity
nadezda [96]

Answer:

u(t)=\frac{1}{5} sin\ (25t)

Explanation:

Given:

  • mass of the body stretching the spring, m=150\ g
  • extension in spring, \Delta x=1.568\ cm
  • velocity of oscillation, u'(0)=20\ cm.s^{-1}
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<u>According to given:</u>

m.g=k.\Delta x

150\times 980=k\times 1.568

k=93750\ dyne.cm^{-1}

<u>we know frequency:</u>

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{93750}{150} }

\omega=25

Now, for position of mass in oscillation:

u= A.sin\ (\omega.t)+B.cos\ (\omega.t)

u= A.sin\ (25.t)+B.cos\ (25.t)

at t=0;\ u(0)=0\ \Rightarrow A=0

∴u(t)=B.sin\ (25.t)

∵ at t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}

u(t)=\frac{1}{5} sin\ (25t)

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