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Zielflug [23.3K]

3 years ago

5

What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radiu

s r1 = 0.775 m ?

1 answer:

alexira [117]3 years ago

7 0

The expression for the flux through a sphere is
flux = charge/Enot (i.e. epsillon not: a constant)

this shows that flux doesn't vary with radius but with the charge of the points

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A horizontal force in used to pull a 5 kilogram cart at a constant speed of 5 meters per second across the floor as shown in the

ddd [48]

A cart is pulled by horizontal force such that it moves with constant velocity

So here since velocity is constant we can say that its acceleration will be ZERO

now here for zero acceleration we can say

$F_{net} = 0$

So here we will have

$F_{net} = F_{ap} - F_f = 0$

here we know that

$F_f = 10 N$

so we will have

$F_{ap} - 10 = 0$

$F_{ap} = 10 N$

so here applied force on handle will be

<em>b. 10 N</em>

5 0

3 years ago

What is most likely to happen during deposition?

Iteru [2.4K]

Answer:

a person appears at a specified time and place and gives sworn testimony

Explanation:

Hope this helps

6 0

3 years ago

In an amusement-park ride, riders stand with their backs against the wall of a spinning vertical cylinder. The floor falls away

seraphim [82]

Answer:

90.78 rev/min

Explanation:

In first place, we have to do the force balance to determine the minimum angular speed required to avoid slipping. The forces acting here are friction and the force due to circular movement, that is centripetal force. Then, we have:

$f=ma_{c}$

μmg=mRω^2

ω= $\sqrt{\frac{μg}{R} } \\$

Then, replacing the given values in the expression we have the following result:

ω=1.51 rev/s*60s=90.78 rev/min

4 0

3 years ago

What is the value of the boltzmann constant of 1.7 moles of gas at 290 K that has an average kinetic energy of 1.4 MJ? A. 7x10^-

fredd [130]

Answer:

The value of of the Boltzmann constant is $3.14\times 10^{-23} J/K$ .

Explanation:

Generally the formula for average kinetic energy of a molecule :

$K.E=\frac{3}{2}kT$

where,

k = Boltzmann’s constant

T = temperature = 290 K

Given:

Number of molecules in 1.7 moles:

$1.7\times 6.022\times 10^{23} molecules=1.0237\times 10^{24} molecules$

Average kinetic energy of 1.7 moles = 1.4 MJ = $1.4\times 10^6 J$

Average kinetic energy of 1 molecule = $\frac{1.4\times 10^6 J}{1.023\times 10^{24}} =1.3675\times 10^{-18} J$

$1.3675\times 10^{-18} J=\frac{3}{2}\times k\times (290K)$

$k=3.14\times 10^{-23} J/K$

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