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2 years ago

6

What is superstring theory?

1 answer:

Stels [109]2 years ago

4 0

Superstring theory is an attempt to explain all of the particles and fundamental forces of nature in one theory by modeling them as vibrations of tiny supersymmetric strings.

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Malika is writing an essay about the Sun. Below is the first paragraph in the essay. The Sun is an average-size star that is loc

djverab [1.8K]

The 109 Earth's the sun can hold over billion Earth's

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2 years ago

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Let x define the position of an object such that x =

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See the attached picture for answers

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according to the theory developed by Charles darwin what process causes organisms to develop traita that help them adapt to chan

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If a photon has frequency = 2.00 x 1014s-1 and the speed of light = 3.00 x 108ms-1, then what is its wavelength?

butalik [34]

Answer:

The photon has a wavelength of $1.5x10^{-6}m$

Explanation:

The speed of a wave can be defined as:

$v = \nu \cdot \lambda$ (1)

Where v is the speed, $\nu$ is the frequency and $\lambda$ is the wavelength.

Equation 1 can be expressed in the following way for the case of an electromagnetic wave:

$c = \nu \cdot \lambda$ (2)

Where c is the speed of light.

Therefore, $\lamba$ $\lambda$ can be isolated from equation 2 to get the wavelength of the photon.

$\lambda = \frac{c}{\nu}$ (3)

$\lambda = \frac{3.00x10^{8}m/s}{2.00x10^{14}s^{-1}}$

$\lambda = 1.5x10^{-6}m$

Hence, the photon has a wavelength of $1.5x10^{-6}m$

<em>Summary: </em>

Photons are the particles that constitutes light.

3 0

3 years ago

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun

Dominik [7]

Answer:

a. $K_{Axis}=2.574x10^{29}J$

b. $K_{Orbit}=2.6577x10^{33}J$

Explanation:

$K_{Axis}=\frac{1}{2}I*w^2$

$I_{Sphere}=\frac{2}{5}*m*r^2$

$w=\frac{2\pi }{T}$ , $T=24hrs*\frac{3600s}{1hr} =86400s$

radius earth = 6371 km

mass earth = 5,972*10^24 kg

a.

$K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2$

$K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2$

$K_{Axis}=2.574x10^{29}J$

b.

$T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s$

$K_{Orbit}=\frac{1}{2}*I*w$

$I=m*r^2$

$K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2$

$K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2$

$K_{Orbit}=2.6577x10^{33}J$

4 0

3 years ago