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3 years ago

The amount of the lighted side of the moon you can see is the same during

Physics

1 answer:

attashe74 [19]3 years ago

5 0

Answer:

The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".

Explanation:

The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.

The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.

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The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ

Marizza181 [45]

The energy stored in the membrane is $6.44\cdot 10^{-14} J$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the material

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

$A=4.50\cdot 10^{-9} m^2$

$d=8.1\cdot 10^{-9} m$

Substituting, we find its capacitance:

$C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F$

Now we can find the energy stored: for a capacitor, it is given by

$U=\frac{1}{2}CV^2$

where

$C=2.26\cdot 10^{-11} F$ is the capacitance

$V=7.55\cdot 10^{-2} V$ is the potential difference

Substituting,

$U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J$

Learn more about capacitors:

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6 0

3 years ago

The formula for speed is v = s/t. a. True b. False

crimeas [40]

If you mean S is the distance then it is true
Velocity = Distance / time

3 0

3 years ago

At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw

allsm [11]

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

$f_i = f_0 \frac{c}{c+v}$

Where

$f_h$ = Approaching velocities

$f_i$ = Receding velocities

c = Speed of sound

v = Emitter speed

And

$f_h = f_0 \frac{c}{c+v}$

Therefore using the values given we can find the velocity through,

$\frac{f_h}{f_0}=\frac{c-v}{c+v}$

$v = c(\frac{f_h-f_i}{f_h+f_i})$

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

$v = 353(\frac{2.4-1}{2.4+1})$

$v = 145.35m/s$

Therefore the cars goes to 145.3m/s

7 0

2 years ago

Two point charges, 1.8 pC and −1.8 pC, are separated by 7 µm. What is the dipole moment of this pair of charges?

Salsk061 [2.6K]

Answer: 12,600,000Cm

Explanation:

From the data's;

Charges(q) = 1.8 PC equal to 1.8 x 10^¹²C

Distance = 7 micrometer, is equal to 0.0000070m

From the equation of electric dipole moment, p= q x d, where q= charge, d=distance and p is the dipole moment.

Then we have 1.8x10^¹² x 0.0000070= 12,600,000Cm

NB: The charges are identical.

3 0

2 years ago

What are the wavelength ranges for the following? (a) the AM radio band (540–1600 kHz) maximum wavelength m minimum wavelength m

Pie

Answer:

Explanation:

a ) AM radio band (540–1600 kHz)

frequency = 540 kHz = 540 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 540 x 10³

= 555.55 m

frequency = 1600 kHz = 1600 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 1600 x 10³

= 187.5 m

maximum wavelength = 555.55 m

minimum wavelength = 187.5 m

b )

AM radio band (88 - 108 MHz)

frequency = 88 MHz = 88 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 88 x 10⁶

= 3.41 m

frequency = 108 MHz = 108 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 108 x 10⁶

= 2.78 m

maximum wavelength = 3.41 m

minimum wavelength = 2.78 m

3 0

2 years ago