Answer:
h = 5.38 10¹⁶ m
Explanation:
Let's start this exercise by assuming that all the potential energy of the electron is converted into kinetic energy, let's use the conservation of energy
starting point. Just before ionization
Em₀ = U = qE
final point. Right after ionization
Em_f = K = ½ m v²
Energy is conserved
Em₀ = Em_f
q E = ½ m v²
v² = 2qE / m
Now we can use the relationship between net work and kinetic energy
W_net = ΔK
net work is the work done by the electron minus the binding energy with the atom, called the work function, Ф = 3 10-19 J
W - Ф = K_f - K₀
we assume that the electron converts all its initial initial kinetic energy to be zero
W -Ф = ½ m v² - 0
W = ½ m v² +Ф
we substitute
W = 1/2 m 2qE/m + E
W = qE +Ф
W = 1.6 10⁻¹⁹ 3 10⁶ + 3 10⁻¹⁹
W = 4.8 10⁻¹³ + 3 10⁻¹⁹
W = 4.8 10⁻¹³ J
When the electron is in air, its kinetic energy can be transformed into gravitational potential energy
As the electron is in the air, all work is transformed into scientific energy
W = K
starting point Em₀ = K = W
end point Em_F = U = m g h
energy conservation Em₀ = Em_f
W = m g h
h = [tex]\frac{W}{mg}[/tex]
let's calculate
h = [tex]\frac{4.8 \ 10^{-13} x}{9.1 \ 10^{-31} \ 9.8 }[/tex]
h = 5.38 10¹⁶ m
Electron involved in making in spark travel through the air before it collides with an atom will be at the distance of 5.38 10¹⁶ m.
What is an electric field?An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.
Let's begin this exercise by assuming that all of the electron's potential energy is turned into kinetic energy, and then we'll apply the law of conservation of energy.
Energy before ionization;
[tex]\rm Em_0 = U = qE[/tex]
Energy after ionization;
[tex]Em_f = K = \frac{1}{2} mv^2[/tex]
From the law of conservation of energy principle;
[tex]Em_0 = Em_f \\\\ q E =\frac{1}{2} m v^2\\\\ v^2 = \frac{2qE }{m}[/tex]
The relationship between net work and kinetic energy;
[tex]W_{net} = \triangle K[/tex]
The work function is defined as net work, which is the work done by the electron minus the binding energy with the atom.
[tex]W - \phi = K_f - K_0[/tex]
[tex]W = K_f+ \phi[/tex]
[tex]W = \frac{1}{2} m \times \frac{2qE}{m} + E\\ \\W = qE + \phi \\\\ \rm W = 1.6 \times 10^{-19}\times 3 \tims 10^6 3 10⁶ +3 \times 10^{-19} \\\\ W = 4.8 \times 10^{-13}+ 3 \times 10^{-19}\\\\ W = 4.8 \times 10^{-13} J[/tex]
EMF at starting point;
[tex]\rm Em_0 = K = W[/tex]
EMF at the endpoint;
[tex]\rm Em_F = U = m g h[/tex]
From the law of conservation of energy principle;
[tex]Em_0 = Em_f \\\\ W = m g \\\\ h = \frac{W}{mg}\\\\\ h = \frac{4.8 \timjes 10^{-13}}{9.1 \times 10^{-31} \times 9.81 }\\\\ \rm h= 5.38 \times 10^{16}[/tex]
Hence electron involved in making in spark travel through the air before it collides with an atom will be at a distance of 5.38 10¹⁶ m.
To learn more about the electric field refer to the link;
https://brainly.com/question/26690770
In many places on Earth, humans are responsible for the removal of grasses, shrubs, trees, and other plants with roots that hold soil in place. This activity is best described by which of the following? *
A) deforestation
B) urbanization
C) air pollution
D) rise in sea level
does altitude has an effect on weight? HELP
Answer: lose weight at high altitudes.
Explanation:
Answer:
Just a week at high altitudes can cause sustained weight loss, suggesting that a mountain retreat could be a viable strategy for slimming down. Overweight, sedentary people who spent a week at an elevation of 8,700 feet lost weight while eating as much as they wanted and doing no exercise
Coherent light with wavelength of 580 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is at 4.81 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?
Answer:
the required wavelength is 1.15815 μm
Explanation:
Given the data in the question;
The position of bright fringes [tex]y_m[/tex] on screen in double slit experiment is expressed as follows;
[tex]y_m[/tex] = mλD / d
solving for d, we substitute 1 for m
y₁ = (1)λD / d
d = λD / y₁
given that λ = 580 nm = 5.8 × 10⁻⁷ m, D = 3.00 m and y₁= y₀ = 4.81 mm = 0.00481 m
so we substitute
d = λD / y₁
d = ( 5.8 × 10⁻⁷ m × 3.00 m ) / 0.00481 m
d = 0.00000174 m² / 0.00481 m
d = 3.6117 × 10⁻⁴ m
Now, position of dark fringe [tex]y_m[/tex] on screen in double slit experiment is expressed as;
[tex]y_m[/tex] = ( m + 1/2 )λD / d
we substitute 0 for m
y₀ = ( 0 + 1/2 )λD / d
y₀ = λD / 2d
2y₀d = λD
λ = 2y₀d / D
we substitute
λ = ( 2(0.00481 m) ( 3.6117 × 10⁻⁴ m) ) / 3.0 m
λ = 1.15815 × 10⁻⁶ m
λ = 1.15815 μm
Therefore, the required wavelength is 1.15815 μm
does altitude has an effect on weight, yes or no
Answer: yes
Explanation:
Weight is the gravitational force experienced on a body. If you move up to higher altitudes, the distance between you and earth increases. ... Yes, weight drops as you go up in altitude (because of diminishing gravity), though your mass remains the same. However, the effect is not huge.
Consider a pulley of mass mp and radius R that has a moment of inertia 1/2mpR2. The pulley is free to rotate about a frictionless pivot at its center. A massless string is wound around the pulley and the other end of the rope is attached to a block of mass m that is initially held at rest on frictionless inclined plane that is inclined at an angle β with respect to the horizontal. The downward acceleration of gravity is g. The block is released from rest .
How long does it take the block to move a distance d down the inclined plane?
Write your answer using some or all of the following: R, m, g, d, mp,
Answer:
a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex] , t = [tex]\sqrt{ \frac{2d}{a} }[/tex]
Explanation:
To solve this exercise we must use Newton's second law
For the block
let's set a reference system with the x axis parallel to the plane
X axis
Wₓ - T = m a
Y axis
N- W_y = 0
N = W_y
for pulley
∑τ = I α
T R = (½ m_p R²) α
let's use trigonometry for the weight components
sin β = Wₓ / W
cos β = W_y / W
Wx = W sin β
angular and linear variables are related
a = α R
α = a / R
we substitute and group our equations
W sin β - T = m a
T R = ½ m_p R² (a / R)
W sin β - T = m a
T = ½ m_p a
we solve the system of equations
W sin β = (m + ½ m_p) a
a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]
let's find the time to travel the distance (d) through the block
x = v₀ t + ½ a t²
d = 0 + ½ a t²
t = [tex]\sqrt{ \frac{2d}{a} }[/tex]