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3 years ago

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Low energy waves have Select one: A. short wavelengths and low frequencies. B. long wavelengths and low frequencies. C. short wa

velengths and high frequencies. D. long wavelengths and high frequencies.

1 answer:

Licemer1 [7]3 years ago

3 0

The answer would be : <span>B. long wavelengths and low frequencies.

Hope this helps !

Photon</span>

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What are one of the ways the atmosphere recycle gases through interactions with the biosphere?

tino4ka555 [31]

Answer:

Though the process photosynthesis

Explanation:

One of the important ways by which the atmosphere recycle gases through interactions with the the biosphere is by photosynthesis.

The biosphere is the living component on the earth. The atmosphere is the gaseous envelope round the earth.

Photosynthesis is the process whereby green plants manufacture their food using carbon dioxide and water in the presence of sunlight to produce glucose and oxygen gas.
This way, atmospheric carbon dioxide is exchanged for oxygen gas.

This way, there is a sustained interaction between the atmosphere and biosphere.

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3 years ago

Read 2 more answers

Suppose you read in the newspaper that a new planet has been found. Its average speed in its orbit is 33 kilometers per second (

Harlamova29_29 [7]

Answer:

E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.

Explanation:

We can answer this question by using Kepler's second law of planetary motion, which states that:

"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"

This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).

In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is

E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.

5 0

3 years ago

Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30

beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

$J=-D\frac{\Delta C}{\Delta x}$

$\Delta C$ is the rate in concentration

$\Delta x$ is the rate in thickness

D is the diffusion coefficient, where,

$D= D_0 exp(\frac{Q_d}{RT})$

Replacing D in the first law,

$J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}$

clearing T,

$T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}$

Replacing our values

$T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}$

$T=-\frac{80000}{-138.09}$

$T=575.16K$

4 0

4 years ago

A tuning fork generates sound waves with a frequency of 240 Hz. The waves travel in opposite directions along a hallway, are ref

Bingel [31]

Answer:

The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

Explanation:

Given that,

Frequency of sound wave = 240 Hz

Distance = 46.0 m

Distance of fork = 14 .0 m

We need to calculate the path difference

Using formula of path difference

$\Delta x=2(L_{2}-L_{1})$

Put the value into the formula

$\Delta x =2((46.0-14.0)-14.0)$

$\Delta x=36\ m$

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{v}{f}$

Put the value into the formula

$\lambda=\dfrac{343}{240}$

$\lambda=1.42\ m$

We need to calculate the phase difference

Using formula of the phase difference

$\phi=\dfrac{2\pi}{\lambda}\times \delta x$

Put the value into the formula

$\phi=\dfrac{2\pi}{1.42}\times36$

$\phi=159.29\ rad$

$\phi\approx 68.2^{\circ}$

Hence, The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

7 0

3 years ago

A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

3 years ago