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]
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
An object is accelerating if it is moving?
9514 1404 393
Answer:
Not Necessarily
Explanation:
If the object is changing speed or direction, then it is accelerating. If it is maintaining the same speed and direction, it is not accelerating.
What do mammoths and tigers need energy for
The smokestack of a stationary toy tra in consists
of a vertical spring gun chat shoots a steel ball a meter
or so straight into the air-----so straight that the ball
always falls back into the smokescack. Suppose the
train moves at constant speed along the straight track. Do you think the ball will still return to the smoke-
stack if shot from the moving train? What if the train
gains speed along the straight track? What if it moves at a constant speed on a circular track? Discuss why your answers differ,
Answer:
i)The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
ii)The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
iii) The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
Explanation:
The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
What is your role in motivating yourself?
i typically focus on motivating others.. i wanna be a motivational speaker someday. tbh if anyone needs anyone to talk to id be happy to listen and i would give the best advice i can. heres my discord: ionknow22#2868
An electron moving in the direction of the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the
Answer:
-z
Explanation:
The force on a moving charge due to a magnetic field follows the right hand rule, so a positive charge, experiencing a magnetic deflection in the -y direction, while it moves in the direction of the x-axis, will do it due to a magnetic field pointing in the +z direction.
As the electron has a negative charge, the magnetic field will point in the opposite direction, i.e., in the -z direction.
An object at rest starts accelerating.
If it travels 20 meters to end up going 10 m/s.
what was its acceleration?
Variables:
Equation and Solve:
Answer:
We are given:
initial velocity (u) = 0 m/s
final velocity (v) = 10 m/s
displacement (s) = 20 m
acceleration (a) = a m/s/s
Solving for 'a'
From the third equation of motion:
v² - u² = 2as
replacing the variables
(10)² - (0)² = 2(a)(20)
100 = 40a
a = 100 / 40
a = 2.5 m/s²
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground
This question is incomplete, the complete question is;
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.
How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J
Explanation:
Given that;
m = 9.9 kg
h = 4.9 m
d = 5 m
μ = 0.3
θ = 36.87°
Now from conservation of energy, the energy is;
Et = mgh
we substitute
Et = 9.9 × 9.8 × 4.9
= 475.398 J
Also the loss of energy i
E_loss = (umg cosθ) d
we substitute
E_loss = 0.3 × 9.9 × 9.8 × cos36.87° × 5
= 116.423 J
so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be
E = Et - E_loss
E = 475.398 J - 116.423 J
E = 358.975 J
Design a tension member and slip-critical splice to carry a factored load of 500 kips. Please use a wide-flange section for the tension member. Please use A572 Gr. 50 steel plates for the splice plates. Please use Group B, A490 bolts for the splice connection. The splice connection should be slip-critical, and have adequate strength after slip occurs as well. Please make any other assumptions you need in order to complete the problem. Provide detailed sketches and drawings for your design.
Answer:
Kindly check the explanation section.
Explanation:
For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.
From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.
Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.
Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).
If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.
The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500
The fracture strength = .75 × Ah × Fhb = 309 kips.
The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.
The light bulbs are identical. Initially both bulbs are glowing. What happens when the switch is closed
Answer:
They turn off
Explanation:
I need help please help ! For science
Answer:
1, 5, 4, 6
Explanation:
bc
2. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French class for the third time this week. She must get from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway, a distance of 35.0 m, at a speed of 3.50 m/s. The second hallway is filled with students, and she 4covers its 48.0 m length at a speed of 1.20 m/s. The final hallway is empty, and Suzette sprints its 60.0 m length at a speed of 5.00 m/s. How long does it take Suzette to make to class? Did Suzette beat the bell?
Answer:
62 secondsnoExplanation:
The total travel time Suzette experiences is the sum of the times in each hallway. Using
time = distance/speed
we can add the times.
(35.0 m)/(3.50 m/s) +(48.0 m)/(1.20 m/s) +(60 m)/(5.0 m/s)
= 10 s + 40 s + 12 s
= 62 s
It takes Suzette 62 seconds to get to class. She does not beat the bell.
Which of the following is the closest to the scientific fact
A. A hypothesis
B. A theory
C. An opinion
D. A prediction
A theory is the closest to a scientific fact (Option B).
A scientific theory is a well-sustained scientific idea that has been verified using the scientific method.A scientific theory can be refuted by the emergence of new lines of evidence against some aspect of this scientific statement.A hypothesis is a given explanation about a question that emerged by observing the natural world.In conclusion, a theory is the closest to a scientific fact (Option B).
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When you place leftover food in the refrigerator, what kind of energy do you
decrease in the food?
A. Nuclear energy
B. Electromagnetic energy
C. Thermal energy
D. Chemical energy
By cooling down the food, the thermal energy in the food molecules is reduced.
What is a refridgerator?A refrigerator is an appliance that is commonly used in the home for the purpose of cooling down a substnace especially water and drinks.
Due to the fact that the molecules that compose matter are is in a state of constant random motion, the food molecules contain thermal energy. Hence, by cooling down the food, the thermal energy in the food molecules is reduced.
Learn more about thermal energy: https://brainly.com/question/11278589
A race car accelerates from rest to a velocity of +90 m/s over a distance of 423m. Determine the acceleration of the race car.
Answer:
9.57m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Final velocity = 90m/s
Distance = 423m
Unknown:
Acceleration of the race car = ?
Solution:
To solve this problem, we should apply one of the appropriate motion equations;
V² = U² + 2as
Where V is the final velocity
U is the initial velocity
a is the acceleration
s is the distance
90² = 0² + 2 x a x 423
8100 = 846a
a = 9.57m/s²
The plate is a model for how sunlight hits Earth’s surface. Which parts of Earth are most similar to the plate with an axis angle of 0°?
Answer: The parts closer to the equator.
Explanation:
The parts of the earth closer to the equator are similar to the plate, because they receive more sunlights than other parts. Which makes them hotter than any other regions in the earth. Example of such countries are Gabon, Uganda, Kenya, Maldives, the democratic republic of Congo, sao tome and principle e.t.c
Hi, Please help.. I have assignments due tonight and I need someone to help me when a question I have generally..
Okay so if Density = Mass divided by Volume and I put that information into a calculator it comes out as
ex. 1.938773646 how do I make it look like something like this 1.4?
Answer:
did you tried putting it in standard form
Answer:
It may help to round all of the given numbers up to at least 1 or 2 decimal points or you could round up the number you get to 1 or 2 decimal places. For example, for this question round up your answer to 1.9 or 1.94
Explanation:
hope this helps!!
A block of mass m1 = 18.5 kg slides along a horizontal surface (with friction, μk = 0.22) a distance d = 2.3 m before striking a second block of mass m2 = 7.25 kg. The first block has an initial velocity of v = 8.25 m/s.
Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
How far does block two travel, d2 in meters, before coming to rest after the collision?
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
The velocity of block 2 and the distance traveled by it prior to being at rest post-collision are 19.5 m/s and 87.8 m. Check the calculations below:
FrictionGiven that,
[tex]m_{1}[/tex] = 18.5 kg
d = 2.3m
To find,
Acceleration of block 1:
∑[tex]F = ma[/tex]
⇒ -m₁gμ = m₁a
⇒ a = -gμ
⇒ a [tex]= -(9.8 m/s^2) (0.22)[/tex]
∵ a [tex]= -2.16 m/s^2[/tex]
Now,
To determine the velocity of block one prior to the collision:
We know,
The initial velocity of block 1 = 8.25 m/s
⇒ [tex]v^2 = v_{o}^2 + 2[/tex]aΔx
⇒ [tex]v^2 = (8.25 m/s)^2 + 2 (-2.16 m/s^2) (2.3 m)[/tex]
∵ [tex]v = 7.63 m/s[/tex]
We also know,
[tex]m_{2}[/tex] = 7.25 kg
Now,
The velocity of block 2 post collision:
⇒ [tex]m_{1} u_{1} + m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}[/tex]post-collision
Through this,
⇒ [tex](18.5 kg) (7.63 m/s) = (7.25 kg) v[/tex]
∵[tex]v = 19.5 m/s[/tex]
The distance can be found through:
⇒ [tex]v^2 = v_{o} ^{2} + 2[/tex][tex]a[/tex]Δ[tex]x[/tex]
⇒ [tex](0 m/s)^2 = (19.5 m/s)^2 + 2 (-2.16 m/s^2)[/tex]Δ[tex]x[/tex]
∵ Δ[tex]x = 87.8 m[/tex]
Thus, 19.5 m/s and 87.8 m are the correct answers.
Learn more about "Friction" here:
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Why is it better to use the metric system, rather than the English system, in scientific measurement?
A. The English system uses one unit for each category of measurement.
B. The metric system uses one unit for each category of measurement.
C. The English system uses consistent fractions that are multiples of 10.
D. The metric system utilizes a variety of number conversions.
A. The English system uses one unit for each category of measurement.
Answer:
A
Explanation:
kerosene is able to reach the oher end of a wick by
Answer:
Capillary Action
Explanation:
Narrow spacings or pores are present in the wick, due to which capillary action takes place, that makes the oil to reach the other end of wick. The ability of a liquid to flow in narrow spaces without any opposition or assistance of external force such as gravity is called as Capillary action.Which would increase the speed of a sound wave?
O A wave passes from a solid to a liquid while remaining the same temperature.
The medium increases in temperature while remaining in the same phase.
The medium decreases in temperature while remaining in the same phase.
O A wave passes from a liquid to a gas while remaining the same temperature.
Answer:
The medium increases in temperature while remaining in the same phase
Explanation:
The speed of a sound wave is increased when the medium increases in temperature while remaining in the same phase.
What is meant by the temperature coefficient of sound wave?The temperature coefficient of a sound wave is defined as the measure of increase in the velocity of the sound wave per unit rise in its the temperature.
Here,
The speed of a sound wave is affected by various factors in the medium through which it propagates. Among them, speed of the wave is mostly influenced by the temperature of the medium.
The speed of a sound wave in a medium is directly proportional to the square root of its absolute temperature. So,
v [tex]\alpha[/tex] √T
where v is thee speed of the sound wave and T is the absolute temperature.
Therefore,
As the temperature of the medium increases, the kinetic energy of the wave particles increases. Thus the speed of the sound wave is increased. As a result, the sound waves will move faster.
Hence,
The speed of a sound wave is increased when the medium increases in temperature while remaining in the same phase.
To learn more about temperature coefficient of sound, click:
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Please help me with this!
hope this helps good luck
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:
Engineers are using computer models to study train collisions to design safer
train cars. They start by modeling an elastic collision between two train cars
traveling toward each other. Car 1 is traveling north at 20 m/s and has a mass
of 12,745 kg. Car 2 is traveling south at 15 m/s and has a mass of 4,125 kg.
After the collision, car 1 has a final velocity of 3 m/s north. What is the final
velocity of car 2?
A. 56 m/s north
B. 56 m/s south
C. 38 m/s south
D. 38 m/s north
Answer:
did you get the answer???
Answer: 38 m/s north
Explanation:
The precision value of measuring tape is
1)0.1cm
2)0.1mm
3)1cm
4)0.01cm
C.1cm
Explanation:
precision is how close two or more measurements are to each other
4?
Explanation:
sorry im not sure but
You can always take a metre ruler as a starting point. Your metre ruler has the same precision as your 15.0cm or 30.0cm ruler, so bring a ruler during exams as they'll come in handy ;)
The order goes like this:
rulers: 0.1cm or 1mm
measuring tape: 0.01cm or 0.1mm
vernier calipers: 0.001cm or 0.01mm
micrometer screw gauge: 0.0001cm or 0.001mm
((if i'm not wrong))
The first ionization potential for calcium (Z = 20, A = 40) is 6.11 eV. Singly-ionized calcium (Ca+) produces two very strong absorption lines in the Sun’s spectrum discovered by Joseph Fraunhofer in 1814, who named them "H" and "K" (he didn’t know they were from calcium, as this was >100 years before the development of quantum mechanics). Both lines always appear together, with lambda subscript H equals 3968 end subscript Å and lambda subscript K equals 3933 Å; hence they are called a "doublet
A. What is the speed of an electron that has just barely enough kinetic energy to collisionally ionize a neutral calcium atom? What is the speed of a calcium ion with this same kinetic energy?
B. What is the temperature T of a gas in which the average particle energy is just barely sufficient to ionize a neutral calcium atom?
C. The lower energy level of both lines is the ground state of Cat. What is the difference in energy in eV) between the two states that correspond to the upper energy levels of the Hand Klines, respectively? How does this compare to the energy of a calcium K photon? Can these two lines can be formed by transitions to upper energy levels with different principal quantum numbers (different n), or do they represent transitions with the same n but some different higher-order quantum number? Explain your reasoning based on your understanding of the general behavior of atomic energy levels (En).
Answer:
A) v = 1.47 10⁶ m / s, v = 0.5426 10⁴ m / s , B) T = 4.7 10⁴ K, C) n₂ = 42
Explanation:
A) For this part, let's calculate the speed of an electron that has an energy of 6.11 eV.
Let's reduce the units to the SI system
E₀ = 6.11eV (1.6 10⁻¹⁹ J / 1eV) = 9.776 10⁻¹⁹ J
The kinetic energy of the electron is
K = ½ m v²
E₀ = K
v = √ 2E₀ / m
v = √ (2 9.776 10⁻¹⁹ / 9.1 10⁻³¹)
v = √ (2.14857 10¹²)
v = 1.47 10⁶ m / s
now the speed of a calcium ion is asked, let's find sum
m = 40 1.66 10⁻²⁷ = 66.4 10⁻²⁷ kg
v = √ (2E₀ / M)
v = √ (2 9.776 10⁻¹⁹ / 66.4 10⁻²⁷)
v = √ (0.2994457 10⁸)
v = 0.5426 10⁴ m / s
B) the terminal energy of an ideal gas is
E = 3/2 kT
T = ⅔ E / k
T = ⅔ (9,776 10-19 / 1,381 10-23)
T = 4.7 10⁴ K
C) To calculate the energy of these lines we use the Planck expression
E = h f
where wavelength and frequency are related
c =λ f
f = c /λ
let's substitute
E = h c /λ
let's look for the energies
λ = 396.8 nm
E₁ = 6.63 10⁻³⁴ 3 10⁸ / 396.8 10⁻⁹
E₁ = 5.0126 10⁻¹⁹ J
λ = 393.3 nm
E₂ = 6.63 10⁻³⁴ 3 10⁸ / 3.93.3 10⁻⁹
E₂ = 5.0572 10⁻¹⁹ J
The difference in energy between these two states is
ΔE = E₂ -E₁
ΔE = (5.0572 - 5.0126) 10⁻¹⁹ J
ΔE = 0.0446 10⁻¹⁹ J
let's reduce eV
ΔE = 0.0446 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
ΔE = 2.787 10⁻² eV
Now let's use Bohr's atomic model for atoms with one electron,
E = -13.606 Z² / n²
where 13,606 eV is the energy of the base state of the Hydrogen atom, Z is the atomic number of Calcium
n = √ (13.606 Z² / E)
λ = 396.8 nm
E₁ = 5.0126 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹J) = 3.132875 eV
n₁ = √ (13.606 20² / 3.132875)
n₁ = 41.7
since n must be an integer we take
n₁ = 42
λ = 393.3 nm
E₂ = 5.0572 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹ J) = 3.16075 eV
n₂ = √ (13.606 20² / 3.16075)
n₂ = 41.5
Again we take n as an integer
n₂ = 42
We can see that the two lines have the same principal quantum number, so for the difference of these energies there must be other quantum numbers, which are not in the Bohr model, because of the small difference they are possibly due to small numbers of the moment angular orbital or spin
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 is the answer to this ?
Is the answer clockwise (CW) or counter clockwise (CCW) ?
anyone to assist me on it ...especial page7 and 8
Answer:
i needed points it was an emergency sorry
Explanation: