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3 years ago

What is in the blank?

Physics

1 answer:

Vika [28.1K]3 years ago

8 0

This is in the thermosphere which is at an altitude of 85-520km

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uniform electric field of magnitude 365 N/C pointing in the positive x-direction acts on an electron, which is initially at rest

vfiekz [6]

Answer:

a) W = - 1.752 10⁻¹⁸ J, b) U = + 1.752 10⁻¹⁸ J

Explanation:

a) work is defined by

W = F . x

the bold letters indicate vectors, in this case the force is electric

F = q E

we substitute

F = q E x

the charge of the electron is

q = - e

F = - e E x

let's calculate

W = - 1.6 10⁻¹⁹ 365 3 10⁻²

W = - 1.752 10⁻¹⁸ J

b) the change in potential energy is

U = q ΔV

the potential difference is

ΔV = - E. Δs

we substitute

U = - q E Δs

the charge of the electron is

q = - e

U = e E Δs

we calculate

U = 1.6 10⁻¹⁹ 365 3 10⁻²

U = + 1.752 10⁻¹⁸ J

3 0

3 years ago

Which diseases are most likely to be treated with antibiotics?

Vitek1552 [10]

Answer:

c) those caused by parasites

4 0

3 years ago

The maximum amount of pulling force a truck can apply when driving on

kupik [55]

C the correct but not sure?

5 0

3 years ago

ASAP ONLY CORRECT ANSWERS PLZZZZZZZZZZZZZ

ycow [4]

Answer b

Explanation:

thank me later

5 0

3 years ago

Read 2 more answers

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

7 0

4 years ago