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2 years ago

7

A race car accelerates uniformly from 18.5 m/s in 2.47 seconds m. Determine the acceleration of the car in m/sec^2 and the dista

nce

1 answer:

Diano4ka-milaya [45]2 years ago

6 0

The change of velocity is from 18.5 to 46.1 m/s or 46.1 - 18.5 = 27.6 m/s. The time of change is 2.47 s. So the acceleration is 27.6 m/s / 2.47 s = 11.17 m/s^2. The distance traveled is average velocity times time of travel.

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Read 2 more answers

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1. False

Explanation: it's exactly the opposite. In fact, Lenz's law states that the induced current and the induced emf in a conductor are generated in such a direction that opposes to the change in magnetic flux through the coil. This is summarized by the negative sign in the Faraday's Newmann Lenz law:

$\epsilon= -\frac{d\Phi }{dt}$

where $\epsilon$ is the emf induced and $\frac{d\Phi}{dt}$ is the rate of change of magnetic flux through the coil.

Lenz's law is a consequence of the law of conservation of energy. In fact, we have:

- If the magnetic flux through a coil is increasing, then the induced current has a direction such that the magnetic field generated by the coil is opposite to the direction of the external magnetic field, in order to decrease the total flux

- If the magnetic flux through a coil is decreasing, then the induced current has a direction such that the magnetic field generated by the coil is in the same direction as the external field, in order to increase the total flux

2. False

Explanation: the magnetic flux through a surface is given by

$\Phi = BA cos \theta$

where

B is the magnitude of the magnetic field

A is the surface area

$\theta$ is the angle between the direction of B and the normal vector to the surface

As we see, the magnetic flux depends not only on B and A, but also on the orientation of the coil with respect to the magnetic field.

3. False

Explanation:

- Sound waves are mechanical waves: mechanical waves are waves consisting of oscillations of the particles in a medium. Due to their nature, therefore, mechanical waves can propagate only in the presence of a medium

- Electromagnetic waves are NOT mechanical waves: they consist of oscillations of electric and magnetic fields, in a direction perpendicular to the direction of propagation of the wave. Since em waves are not mechanical waves, they do NOT need a medium to propagate, since they can also travel through a vacuum; therefore the original statement is false.

5 0

3 years ago

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

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

If we consider the light as 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 <u>kinetic energy. </u>

This is what Einstein proposed:

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

$E=h.f$ (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 $\Phi$ is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal. </u>

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the <u>frequency has an inverse relation with the wavelength </u>

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

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

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago