Answer:
The value is [tex]H = 18*10^{2} \ Atom / sec [/tex]
Explanation:
From the question we are told that
The atom fraction of metal A at point G is [tex] A = 0.30 \ m[/tex]
The atom fraction of metal A at a distance 5000nm from G is [tex]A_2 = 0.35[/tex]
The number of atoms per [tex]m^3[/tex] is [tex]N_h = 9 * 10^{28}[/tex]
The diffusion coefficient is [tex]D = 2* 10^{-14 } m^2/s[/tex]
Generally of the concentration of atoms of metal A at G is
[tex] N_A = A * N_h [/tex]
=> [tex] N_A = 0.3 * 9 * 10^{28}[/tex]
=> [tex] N_A = 2.7 * 10^{28} 2.7 atoms/m^3[/tex]
Generally of the concentration of atoms of metal A at a distance 5000nm from G is
[tex]D = 0.35 *9 * 10^{28}[/tex]
=> [tex]D = 3.15 * 10^{28} \ atoms / m^3[/tex]
The concentration gradient is mathematically represented as
[tex]\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }[/tex]
=> [tex]\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }[/tex]
=> [tex]\frac{dN_A}{dx} = 9 * 10^{20} / m^4[/tex]
Generally the flux of the atoms per unit area according to Fick's Law is mathematically represented as
[tex]J = -D* \frac{d N_A}{dx}[/tex]
=> [tex]J = -2* 10^{-14 * 9 * 10^{20} [/tex]
=> [tex] J = 18*10^{6}\ atoms\ crossing\ /m^2 s [/tex]
Generally if the cross-section area is [tex] a = 1 cm^2 = 10^{-4} \ m^2[/tex]
Generally the number of atom crossing the above area per second is mathematically is
[tex]H = 18*10^{6} * 10^{-4} [/tex]
=> [tex]H = 18*10^{2} \ Atom / sec [/tex]
Weight of a person's muscles, bones, tendons, and ligaments.
A. flexibility
B. lean mass
C. aerobic
Consider a particle of mass m which can move freely along the x axis from -a/2 to a/2, but which is strictly prohibited from being found outside this region. The wave function of the particle within the allowed region is
Answer:
φ = B sin (2π n/a x)
Explanation:
In quantum mechanics when a particle moves freely it implies that the potential is zero (V = 0), so its wave function is
φ = A cos kx + B sin kx
we must place the boundary conditions to determine the value of the constants A and B.
In our case we are told that the particle cannot be outside the boundary given by x = ± a / 2
therefore we must make the cosine part zero, for this the constant A = 0, the wave function remains
φ = B sin kx
the wave vector is
k = 2π /λ
now let's adjust the period, in the border fi = 0 therefore the sine function must be zero
φ (a /2) = 0
0 = A sin (2π/λ a/2)
therefore the sine argument is
2π /λ a/2 = n π
λ= a / n
we substitute
φ = B sin (2π n/a x)
Pls help pls pls pls pls
describe the energy conversion that occurs in a diesel engine
One student runs with a velocity of +10 m/s while a second student runs with a velocity of –10 m/s. Which student has the faster velocity? Why?
Answer:
The one with the faster velocity is the one with a velocity of -10m/s
Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
Answer:
The two charged objects will exert equal and opposite forces on each other.
Explanation:
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of charges on the objects and inversely proportional to the square of the distance between the two objects.
This force of attraction or repulsion between the two charged objects is always equal and opposite.
Therefore, the two charged objects will exert equal and opposite forces on each other.
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just bef
Answer:
(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the bowling ball is 113.272 joules.
Explanation:
The statement is incomplete. The complete question is:
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(a) What is the kinetic energy of the ball just before it hits the mattress?
(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?
(c) How much work does the gravitational force do on the ball while it is compressing the mattress?
(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）
(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:
[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.
Now we expand the expression by definition of gravitational potential energy:
[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]
[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the bowling ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:
[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]
[tex]K_{2} = 102.974\,J[/tex]
The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]
[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]
[tex]\Delta W = 102.974\,J[/tex]
The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]
Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.
[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]
Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]
[tex]\Delta W = 10.298\,J[/tex]
Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:
[tex]\Delta W' = K_{2}+\Delta W[/tex]
([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])
[tex]\Delta W' = 113.272\,J[/tex]
The work done by the mattress on the bowling ball is 113.272 joules.
3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage dropped across each phase?
Answer:
[tex]E_s = 277.13V[/tex]
Explanation:
Given
[tex]Load\ Voltage = 480V[/tex]
Required
Determine the voltage dropped in each stage.
The relation between the load voltage and the voltage dropped in each stage is
[tex]E_l = E_s * \sqrt3[/tex]
Where
[tex]E_l = 480[/tex]
So, we have:
[tex]480 = E_s * \sqrt3[/tex]
Solve for [tex]E_s[/tex]
[tex]E_s = \frac{480}{\sqrt3}[/tex]
[tex]E_s = \frac{480}{1.73205080757}[/tex]
[tex]E_s = 277.128129211[/tex]
[tex]E_s = 277.13V[/tex]
Hence;
The voltage dropped at each phase is approximately 277.13V
How long would it take you to walk 3,962 km from New York to Los
Angeles?
Answer:
913 hours ur welcome :)
Luck walked to a store that is 250m away and it took him 50secs while Layne walked to the mall that is 1000m away and took her 200s to do. What do they have in common?
A. Average speed
B. Acceleration
C. Displacement
D.mass
Answer:
Average speed
Explanation:
250/50=5
1000/20=5
Define reflection of sound?
Explanation:
When sound travels in a given "medium", it would touch the surface of another "medium" and will bounce back in some other direction, this occurrence is called the reflection of sound.
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration?
Answer:68.15m/s
Explanation:
Given:
v₁=15m/s
a=6.5m/s²
v₁=?
x=340m
Formula:
v₁²=v₁²+2a (x)
Set up:
=[tex]\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)[/tex]
Solution:68.15m/s
what happens to the temperature of water as time elapses? IF YOU ANSWER IT I WILL MARK YOU A BRAINLEST ANSWER
Answer:
I think it will get colder
Explanation:
Answer:
The water molecules go faster as it gets colder they go slower
Explanation:
trust me thats the answer
A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m
Answer:
35.35 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =.?
The range (i.e how far away) of the ball can be obtained as follow:
R = u² Sine 2θ /g
R = 20² Sine (2×30) / 9.8
R = 400 Sine 60 / 9.8
R = (400 × 0866) / 9.8
R = 346.4 / 9.8
R = 35.35 m
Therefore, the range (i.e how far away) of the ball is 35.35 m
Which statement accurately describes impulse?
State corrrect ans
Answer:
2nd option on edge2021
Explanation:
What amount of work is done on a cart that is pushed 4.0 meters across a floor by a horizontal 40-N net force?
Answer:
The answer is 160 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 40 × 4
We have the final answer as
160 JHope this helps you
What is the correct answer?
Answer:
2156 N
Explanation:
Data obtained from the question include:
Mass of satellite (m) = 220 Kg
Force (F) of gravity =?
The force of gravity exerted on the satellite on the surface of the earth can be obtained by using the following formula:
Force (F) of gravity = mass (m) × acceleration due to gravity (g)
F = mg
Mass of satellite (m) = 220 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) of gravity =?
F = mg
F = 220 × 9.8
F = 2156 N
Thus, the force of gravity exerted on the satellite on the surface of the earth is 2156 N
What do mammoths and tigers need energy for
A sleigh is being pulled horizontally by a train of horses at a constant speed of 6.38 m/s. The magnitude of the normal force exerted by the snow-covered ground on the sleigh is 7.50 ✕ 103 N.
(a) If the coefficient of kinetic friction between the sleigh and the ground is 0.26, what is the magnitude of the kinetic friction force experienced by the sleigh?
N
(b) If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh, what must be the magnitude of this force?
N
Answer:
(a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force
Explanation:
Given that,
Constant speed = 6.38 m/s
Force [tex]F=7.50\times10^{3}\ N[/tex]
Kinetic friction = 0.26
(a). We need to calculate the friction force
Using formula of friction force
[tex]f_{k}=\mu F_{N}[/tex]
Put the value into the formula
[tex]f_{k}=0.26\times7.50\times10^{3}[/tex]
[tex]f_{k}=1950\ N[/tex]
(b). If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh,
We need to calculate the magnitude of this force
According to given data,
The same force will be applied to keep constant velocity.
Hence, (a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force.
(a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force
The calculation is as follows;a. The magnitude of the kinetic friction force experienced by the sleigh is
[tex]= 0.76 \times 7.50 \times 10^3[/tex]
= 1950 N
b. It should be equivalent to the friction force.
Learn more: https://brainly.com/question/24908711?referrer=searchResults
1. What quantity of heat is required to raise?
the temperature of 450 grams of water
from 35°C to 85°C?
capacity of water is 4.18 J/g °C.
Answer: Calculate the energy required in joules to raise the temperature of 450 grams of water from 15°C to 85°C? (The specific heat capacity of water is 4.18 J/g/°C)
Explanation:
The quantity of heat is required to raise the temperature of of water is 94050 joule.
What is law of conservation of energy?Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.
The law of energy conservation is adhered to by all energy forms. The law of conservation of energy essentially says that the total energy of the system is conserved in a closed system, also known as an isolated system.
Mass of water: m = 450 grams
Initial temperature of water = 35° C
Final temperature of water = 85° C
Capacity of water is 4.18 J/g °C.
Hence, the quantity of heat is required to raise = mass × Capacity × raise in temperature
= 450 × 4.18 × (85 - 35) joule
= 94050 joule.
Learn more about energy here:
https://brainly.com/question/1932868
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Work Done by a Varying Force 06 Work and Energy
w of Energy
100%
2.) The force required to stretch a spring by 1 m from its unstretched length is
150 N. What is the force required to stretch the spring by 3 m?
A. 600 N
B. 450 N
C. 300 N
D. 200 N
Answer:
B. 450 N
Explanation:
Use Hooke's law:
F = kx
150 N = k (1 m)
k = 150 N/m
F = kx
F = (150 N/m) (3 m)
F = 450 N
A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1.3 dB. What is the power output P
Answer:
The value of the power is [tex]P_c = 38.55 \ W [/tex]
Explanation:
From the question we are told that
The power rating [tex]P_{1000} =P_b= 52 \ W[/tex]
The frequency is [tex]f = 1000 \ Hz[/tex]
The frequency at which the sound intensity decreases [tex]f_k = 20 \ Hz[/tex]
The decrease in intensity is by [tex]\beta = 1.3 dB[/tex]
Generally the initial intensity of the speaker is mathematically represented as
[tex]\beta_1 = 10 log_{10} [\frac{P_b}{P_a} ][/tex]
Generally the intensity of the speaker after it has been decreased is
[tex]\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ][/tex]
So
[tex]\beta_1-\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ][/tex]
=> [tex]\beta = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3[/tex]
=> [tex]\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3[/tex]
=> [tex]\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3[/tex]
=> [tex]10log_{10} [\frac{P_b}{P_c} ] = 1.3[/tex]
=> [tex]log_{10} [\frac{P_b}{P_c} ] = 0.13[/tex]
taking atilog of both sides
[tex][\frac{P_b}{P_c} ] = 10^{0.13}[/tex]
=>[tex][\frac{52}{P_c} ] = 10^{0.13}[/tex]
=> [tex]P_c = \frac{52}{1.34896}[/tex]
=> [tex]P_c = 38.55 \ W [/tex]
A boat initially moving at 10 m/s accelerates at 2 m/s for 10 s. What is the velocity of the boat after 10 seconds?
Answer:
30 m/s
Explanation:
v = u + at
given that,
u = 10 m/s (initial speed)a = 2 m/s^2 t = 10sv =?(final speed)v = 10 + ( 2 × 10)
v = 10 + 20
v = 30 m/s
the radius of the earth social
What type of research based on approach that used to describe variables rather than to test a predicted relationship between variables?
Answer:
Correlational research can be used to see if two variables are related and to make predictions based on this relationship.
what phase changes take place when you are adding energy to the substance
Answer:
During a phase change, a substance undergoes transition to a higher energy state when heat is added, or to a lower energy state when heat is removed. Heat is added to a substance during melting and vaporization. Latent heat is released by a substance during condensation and freezing. Explanation:
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration? (68.15 m/s)
v² - u² = 2 a ∆x
where u = initial velocity, v = final velocity, a = acceleration, and ∆x = distance traveled.
So
v² - (15 m/s)² = 2 (6.5 m/s²) (340 m)
v² = 4645 m²/s²
v ≈ 68.15 m/s
Write a function to accept a vector of masses (m) from the user and gives the corresponding energy to them. Energy vector is the output of the program. Constant c is the speed of light which is 2.9979 x 108 m/s inside the function.
Answer:
Written in Python
def energyvector(mass):
c = 2.9979 * 10**8
energy = mass * c ** 2
print(round(energy,2))
Explanation:
This line defines the function
def energyvector(mass):
This line initializes the speed of light
c = 2.9979 * 10**8
This line calculates the corresponding energy
energy = mass * c ** 2
This line prints the calculated energy
print(round(energy,2))
What's the difference between an open cluster and a globular cluster
An open cluster is a group of up to a few thousand stars that were formed from the same giant molecular cloud, and are still loosely gravitationally bound to each other. In contrast, globular clusters are very tightly bound by gravity. ... Open clusters are very important objects in the study of stellar evolution.
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