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3 years ago

When a wave is acted upon by an external damping force, what happens to the energy of the wave

Physics

2 answers:

Rasek [7]3 years ago

8 0

The energy of the wave decreases gradually

bazaltina [42]3 years ago

3 0

Answer:

Explanation:

The energy of the wave its determined by the mass of the fuid of the wave and the speed. The amount of motion equation P= m.v

So according to the impulse and amount of motion theorem (ΔP= mvₙ- mv₀).

If the wave is acted upon by an external dumping force, the wave its going to decrease his energy because the reduction of the speed and the mass.

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À A car moves with an initial velocity of 18 m/s due north. Find the velocity of the car after 7 Os if

Nonamiya [84]

Answer:

(a) $v_f=28.5m/s$

(b) $v_f=7.5m/s$

Explanation:

Hello.

(a) In this case since the car is moving at an initial velocity of 18 m/s due north, the final velocity is computed considering the acceleration as positive since it is due north as well:

$v_f=v_0+at=18m/s+1.5m/s^2*7s\\\\v_f=28.5m/s$

(b) In this case, since the car is moving due north by the acceleration is due south it is undergoing a slowing down process, thereby the acceleration is negative therefore the final velocity turns out:

$v_f=v_0+at=18m/s-1.5m/s^2*7s\\\\v_f=7.5m/s$

Best regards.

3 0

3 years ago

Magnetic field lines are more concentrated at the poles of the magnet<br> True<br> False

yKpoI14uk [10]

Answer:

True! UwU

Explanation:

the magnetic field is strongest near to the poles of the magnet were the lines of flux are more closely spaced.

8 0

3 years ago

Chemical bonds form when atoms A.) gain protons. B.) combine nuclei C.) give up neutrons. D.) share or transfer electrons

frutty [35]

The correct answer is (D)

8 0

3 years ago

A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 5.9 m/s. if the coefficient of fricti

bagirrra123 [75]

Frictional force = vertical force x coeff of friction = 590N x 0.045 = 26.55N

Mass of boy and sled = 590N / g = 590N / 9.8m/s^2 = 60.20 kg

Deceleration due to friction = 26.55N / 60.20kg = 0.44 m/s^2.

For constant acceleration we have:

v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and ti is time. In this case v = 0, u = 5.9 m/s, a = -0.44 m/s^2. So we have
0 = 5.9 - 0.44t
which gives t = 5.9 / 0.44 = 13.409 s.

Distance traveled in this time d = ut + 0.5at^2 = 5.9 x 13.409 - 0.5 x 0.44 x 13.409^2 = 39.56 m

<span>Answer: </span>40 meters

8 0

3 years ago

Titanium metal requires a photon with a minimum energy of 6.94×10−19J to emit electrons. If titanium is irradiated with light of

butalik [34]

Answer:

a) $1.59(10)^{-19} J$

b) $2.34(10)^{12} electrons$

Explanation:

The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If the light is a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

<u>This is what Einstein proposed: </u>

Light behaves like a stream of particles called photons with an energy $E$ :

$E=\frac{hc}{\lambda}$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi=6.94(10)^{-19} J$ is the minimum amount of energy required to induce the photoemission of electrons from the surface of Titanium metal.

Knowing this, let's begin with the answers:

<h3 /><h3>a) Maximum possible kinetic energy of the emitted electrons ( $K$ )</h3>

From (1) we can know the energy of one photon of 233 nm light:

$E=\frac{hc}{\lambda}$

Where:

$h=6.63(10)^{-34}J.s$ is the Planck constant

$\lambda=233 (10)^{-9} m$ is the wavelength

$c=3 (10)^{8} m/s$ is the speed of light

$E=\frac{(6.63(10)^{-34}J.s)(3 (10)^{8} m/s)}{3 (10)^{8} m/s}$ (3)

$E=8.53(10)^{-19} J$ (4) This is the energy of one 233 nm photon

Substituting (4) in (2):

$8.53(10)^{-19} J=6.94(10)^{-19} J+K$ (5)

Finding $K$ :

$K=1.59(10)^{-19} J$ (5) This is the maximum possible kinetic energy of the emitted electrons

<h3>b) Maximum number of electrons that can be freed by a burst of light whose total energy is $2 \mu J=2(10)^{-6} J$ </h3>

Since one photon of 233 nm is able to free at most one electron from the Titanium metal, we can calculate the following relation:

$\frac{E_{burst}}{E}$

Where $E_{burst}=2(10)^{-6} J$ is the energy of the burst of light

Hence:

$\frac{E_{burst}}{E}=\frac{2(10)^{-6} J}{8.53(10)^{-19} J}=2.34(10)^{12} electrons$ This is the maximum number of electrons that can be freed by the burst of light.

4 0

3 years ago