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3 years ago

I have two questions

Physics

2 answers:

Anettt [7]3 years ago

8 0

Answer:

1)The energy transported by a wave is directly proportional to the square of the amplitude. So whatever change occurs in the amplitude, the square of that effect impacts the energy. This means that a doubling of the amplitude results in a quadrupling of the energy

2)Just as wavelength and frequency are related to light, they are also related to energy. The shorter the wavelengths and higher the frequency corresponds with greater energy. So the longer the wavelengths and lower the frequency results in lower energy.

Explanation:

Vesna [10]3 years ago

4 0

The correct answer to the question is : C) A pendulum swings back and forth.
EXPLANATION:
Before answering this question, first we have to understand the law of conservation of energy.
As per law of conservation of energy, energy can neither be created not be destroyed. It can only change from one form to another form, and the total energy of the universe is always constant.
As per the question, we have a pendulum which is moving back and forth.
Let us consider a pendulum which is taken to some height from its equilibrium or mean point. The energy possessed by the pendulum at this height is gravitational potential energy. When the pendulum is released, the potential energy is converted into kinetic energy. At mean point, whole of its potential energy is converted into kinetic energy.
Due to inertia, the pendulum reaches at the other extreme point. During its movement from mean point to extreme point, the kinetic energy is converted into potential energy. At extreme point, whole of its kinetic energy must have converted into potential energy. The same process will be repeated.
Hence, it obeys law of conservation of energy.
Hence, swinging of pendulum back and forth is the best example to demonstrate the law of conservation of energy.

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AveGali [126]

For no.4 is what seismologists use two main devices to measure an earthquake: a seismograph and aseismoscope. The seismographis an instrument that measures seismic waves caused by an earthquake. The seismographhas three main devices, theRichter Magnitude Scale, theModified Mercalli Intensity Scale, and the Moment-Magnitude Scale.

5 0

3 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

A piece of indium with a mass of 16.6 g is submerged in 46.3 cm3 of water in a graduated cylinder. The water level increases to

blagie [28]

Answer:

the density of indium is 7.2 g/cm^3

Explanation:

The computation of the density of indium is shown below:

Given that

Mass = 16.6 g

Volume = 48.6 c,^3 - 46.3cm^3 = 2.3 cm^3

Based on the above information

As we know that

Density = mass ÷ volume

So,

= 16.6g ÷ 2.3 cm^3

= 7.2 g/cm^3

hence, the density of indium is 7.2 g/cm^3

We simply applied the above formula so that the correct value could come

And, the same is to be considered

8 0

3 years ago

Consider a spherical planet of uniform density rho. The distance from the planet's center to its surface (i.e., the planet's rad

alisha [4.7K]

Answer:

a) $g(r) = 4\pi \cdot G \cdot \rho\cdot r$ , b) $g = 4\pi \cdot G \cdot \rho\cdot r_{P}$

Explanation:

a) The acceleration due to gravity inside the planet is:

$dg = G\cdot \frac{\rho \cdot dV}{r^{2}}$

$dg = G\cdot \frac{4\pi\cdot \rho \cdot r^{2}\,dr}{r^{2}}$

$dg = 4\pi\cdot G\cdot \rho \,dr$

$g(r) = 4\pi \cdot G \cdot \rho\cdot r$

b) The acceleration at the surface of the planet is:

$g = 4\pi \cdot G \cdot \rho\cdot r_{P}$

5 0

3 years ago

Most cancer-causing air pollutants are found outdoors. True False

natta225 [31]

This is most likely True. However, it could be also False. Let me explain.

There are many air pollutants that can be found outdoors and indoors. It does depend on what kind of indoor environment we're talking about. And also what kind of outdoor environment we're talking about. In both cases it depends on the context. If the indoor environment is quite toxic, then of course, there would be less cancer-causing air pollutants for a person outside. And the same is true for the other way around.

8 0

4 years ago

Read 2 more answers