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4 years ago

Helppp me please !!

Physics

1 answer:

vagabundo [1.1K]4 years ago

3 0

Answer:

the frequency of this light is $1.5\,\,10^{16}$ Hz (which coincides with the fourth option listed among possible answers)

Explanation:

Recall the relationship between wavelength and frequency of radiation based upon the speed of light in vacuum:

$frequency\,*\,wavelength\,=\, speed\,\,of\,light\\frequency\,*\,wavelength\,=\,3\,\,10^8\,\frac{m}{s} \\frequency\,*\,20 \,\,10^{-9}\,m\,=\,3\,\,10^8\,\frac{m}{s}\\frequency\, =\,\frac{3\,\,10^8}{2\,\,10^{-9}} \,\frac{1}{s} \\frequency\,=1.5\,\,10^{16}\,\,Hz$

In the process shown above, we have written the speed of light in meters per second, and also converted the 20 nano-meters into meters, in order to reduce SI units properly and have the result shown in units of 1/second (that is hertz's - abbreviate "Hz")

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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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2 years ago

What is a likely consequence of preventing prescribed burns to forest ecosystems?

sattari [20]

<span>Your answer is A. Natural wildfires will burn longer and hotter when they occur because there is more underbrush and fuel available.</span>

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3 years ago

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The impulse given to a ball with mass of 2 kg is 16 N*s. If the ball starts from rest, what is its final velocity?

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The impulse is equal to the variation of momentum of the object:
$I=\Delta p = m \Delta v$
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The ball starts from rest so its initial velocity is zero: $v_i=0$ . So we can rewrite the formula as
$I=m v_f$
or
$v_f = \frac{I}{m}$

and since we know the impulse given to the ball (I=16 Ns) and its mass (m=2 kg), we can find the final velocity of the ball:
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3 years ago

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A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$