Which best describes the motion of air particles when a transverse wave passes through them?

Which of the following are examples of some of the environmental costs of technological development? (Choose all that apply)

kirza4 [7]

Environmental costs of technological development include:

depletion of non-renewable resources
pollution of the atmosphere
increased recycling of heavy metals

<h3>What are environmental costs of technological development?</h3>

The environmental costs of technological development refers to the effects on the environment of technological development.

These effects of technological development on the environment are usually negative and detrimental to the environment and the organisms found in the environment.

Some environmental costs of technological development include:

depletion of non-renewable resources
pollution of the atmosphere
pollution of the land and water
increased recycling of heavy metals

In conclusion, the negative environmental costs of technological development damage the environment.

Learn more about environmental costs of technology and development at: brainly.com/question/9839688

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8 0

1 year ago

Q2/Deceleration of a particle is based on relation a=-3 v² m/s² where v in m/s. If it moves along a straight line and has veloci

kozerog [31]

Explanation:

Given: a = -3v^2

By definition, the acceleration is the time derivative of velocity v:

$a = \frac{dv}{dt} = - 3 {v}^{2}$

Re-arranging the expression above, we get

$\frac{dv}{ {v}^{2} } = - 3dt$

Integrating this expression, we get

$\int \frac{dv}{ {v}^{2} } = \int {v}^{ - 2}dv = - 3\int dt$

$- \frac{1}{v} = - 3t + k$

Since v = 10 when t = 0, that gives us k = -1/10. The expression for v can then be written as

$- \frac{1}{v} = - 3t - \frac{1}{10} = - ( \frac{30 + 1}{10} )$

or

$v = \frac{10}{30t +1 }$

We also know that

$v = \frac{ds}{dt}$

or

$ds = vdt = \frac{10 \: dt}{30t + 1}$

We can integrate this to get s:

$s = \int v \: dt = \int ( \frac{10}{30t + 1}) \: dt = 10 \int \frac{dt}{30t + 1}$

Let u = 30t +1

du = 30dt

so

$\int \frac{dt}{30t + 1} = \frac{1}{30} \int \frac{du}{u} = \frac{1}{30}\ln |u| + k$

$= \frac{1}{30}\ln |30t + 1| + k$

So we can now write s as

$s = \frac{1}{3}\ln |30t + 1| + k$

We know that when t = 0, s = 8 m, therefore k = 8 m.

$s = \frac{1}{3}\ln |30t + 1| + 8$

Next, we need to find the position and velocity at t = 3 s. At t = 3 s,

$v = \frac{10}{30(3) +1 } = \frac{10}{91}\frac{m}{s} = 0.11 \: \frac{m}{s}$

$s = \frac{1}{3}\ln |30(3) + 1| + 8 = 9.5 \: m$

Note: velocity approaches zero as t --> $\infty$

5 0

2 years ago

How is the energy of a photon related to its wavelength?

Delicious77 [7]

Answer:

Photons have no mass, but they have energy E = hf = hc/λ. ... The energy of each photon is inversely proportional to the wavelength of the associated EM wave. The shorter the wavelength, the more energetic is the photon, the longer the wavelength, the less energetic is the photon.

Explanation:

4 0

2 years ago

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4