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Elza [17]
3 years ago
15

What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m?

Physics
1 answer:
polet [3.4K]3 years ago
4 0

Answer:

0.044 V

Explanation:

E = Electric field = 5.5\times 10^6\ V/m

d = Thickness of membrane = 8 nm

When the electric field strength is multiplied by the membrane thickness we get the voltage

Voltage across a gap is given by

V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V

The voltage across the membrane is 0.044 V

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A 720 g softball is traveling at 15.0 m/s when caught. If the force of the glove on the ball is 520 N, what is the time it takes
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The time it takes the ball to stop is 0.021 s.

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velocity of the ball, v = 15.0 m/s

applied force, F = 520 N

Apply Newton's second law of motion, to determine the time it takes the ball to stop;

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The reflective surface of a CD consists of spirals of equally spaced grooves. If you shine a laser pointer on a CD, each groove
Ipatiy [6.2K]

Answer:

d = 1.55 * 10⁻⁶ m

Explanation:

To calculate the distance between the adjacent grooves of the CD, use the formula, d = \frac{m \lambda}{sin(A_{m}) }..........(1)

The fringe number, m = 1 since it is a first order maximum

The wavelength of the green laser pointer, \lambda = 532 nm = 532 * 10⁻⁹ m

Distance between the central maximum and the first order maximum = 1.1 m

Distance between the screen and the CD = 3 m

A_{m} = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}

sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}

Putting all appropriate values into equation (1)

d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m

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3 years ago
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