Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The temperature change is [tex]\frac{dT}{dt} = 1.016 ^oC/m[/tex]
Explanation:
From the question we are told that
The velocity field with which the bird is flying is [tex]\vec V = (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s[/tex]
The temperature of the room is [tex]T(x, y, u) = 400 -0.4y -0.6z-0.2(5 - x)^2 \ ^o C[/tex]
The time considered is t = 10 \ seconds
The distance that the bird flew is x = 1 m
Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird
Generally the change in the bird temperature with time is mathematically represented as
[tex]\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 * (5-x)] [-\frac{dx}{dt} ][/tex]
Here the negative sign in [tex]\frac{dx}{dt}[/tex] is because of the negative sign that is attached to x in the equation
So
[tex]\frac{dT}{dt} = -0.4v_y -0.6v_z -0.2[2 * (5-x)][ -v_x][/tex]
From the given equation of velocity field
[tex]v_x = 0.6x[/tex]
[tex]v_y = 0.2t[/tex]
[tex]v_z = -1.4 [/tex]
So
[tex]\frac{dT}{dt} = -0.4[0.2t] -0.6[-1.4] -0.2[2 * (5-x)][ -[0.6x]][/tex]
substituting the given values of x and t
[tex]\frac{dT}{dt} = -0.4[0.2(10)] -0.6[-1.4] -0.2[2 * (5-1)][ -[0.61]][/tex]
[tex]\frac{dT}{dt} = -0.8 +0.84 + 0.976[/tex]
[tex]\frac{dT}{dt} = 1.016 ^oC/m[/tex]
Related Questions
A
6. All other changeable factors that must
be kept the same to ensure a fair test
(what you keep the same).
Answers
Answer:
a constant variable?
Explanation:
A constant variable is any aspect of an experiment that a researcher intentionally keeps unchanged throughout an experiment.
Experiments are always testing for measurable change, which is the dependent variable. You can also think of a dependent variable as the result obtained from an experiment. It is dependent on the change that occurs
An object moving 20 m/s
experiences an acceleration of 4 m/s' for 8
seconds. How far did it move in that time?
Variables:
Equation and Solve:
Answers
Answer:
We are given:
initial velocity (u) = 20m/s
acceleration (a) = 4 m/s²
time (t) = 8 seconds
displacement (s) = s m
Solving for Displacement:
From the seconds equation of motion:
s = ut + 1/2 * at²
replacing the variables
s = 20(8) + 1/2 * (4)*(8)*(8)
s = 160 + 128
s = 288 m
Matching type. Send help please. ASAP!
Answers
Answer:
46-D
47-C
48-F
49-A
50-B
I am not very sure I am right about those answers though.
Calculate the ratio of the mechanical energy at B and mechanical energy at a (eb,ea) and (ec,ea). What do these ratios tell you about the conservation of energy?
A) is the mechanical conserved between a and b? explain
B) is the mechanical energy conserved between b and c ?explain
Answers
Answer:
Yes at A the mechanical energy is conserved.
Yes at B the part of mechanical energy is conserved potential energy and kinetic energy and some is lost as frictional force.
Explanation:
Ratio = Eb/ Ea= 1058.3 J/2940 J= 0.3599
Ratio = Ec/ Eb= 0J/ 1058.3 J= 0
At point A the skater is at rest or it is the starting point and the whole energy is due to the position of the skater i.e= mgh = 50 *9.8*6= 2940 J
Since there's no movement there is no Kinetic energy = 0 J
Yes at A the mechanical energy is conserved.
At point B the skater has traveled for some of the distance . It has potential energy and kinetic energy.
Yes at B the part of mechanical energy is conserved as potential energy and kinetic energy.
The total Mechanical energy = 1058.3 J
At point B Total Mechanical energy = PE+ KE
1058.3J = 980 J + 78.3 J
1058.3 J = mgh + 1/2mv²
= 50*2*9.8 + 1/2 *50*(8.85)²
= 980 J + 78.3 J
As the total energy of the system must remain the same some of the mechanical energy is lost as frictional force at point B .
2940 J-1058.3 J= 1881.7
At Point C the skater has arrived at the end point and the height , speed, PE, KE and ME all are zero.
(a) The ratio of the mechanical energy at B and mechanical energy at A is 0.36.
(b) The ratio of the mechanical energy at C and mechanical energy at A is 0.
(c) mechanical energy is conserved between a and b.
(d) mechanical energy is not conserved between b and c.
The given parameters;
mechanical energy at A, [tex]E_a = 2,940 \ J[/tex]mechanical energy at B, [tex]E_b =1,058.3 \ J[/tex]mechanical energy at C, [tex]E_c = 0[/tex]The ratio of the mechanical energy at B and mechanical energy at A;
[tex]ratio = \frac{E_b}{E_a} = \frac{1058.3}{2940} = 0.36[/tex]
The ratio of the mechanical energy at C and mechanical energy at A;
[tex]ratio = \frac{E_c}{E_a} = \frac{0}{2940} = 0[/tex]
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_b - h_a) - \frac{1}{2}m(v_b^2 - v_a^2)\\\\ \Delta E = 50\times 9.8(2-6) \ - \ \frac{1}{2} \times 50(8.85^2 - 0)\\\\\Delta E =- 1960 + 1960\\\\\Delta E = 0 \ J[/tex]
Thus, we can conclude that mechanical energy is conserved between a and b.
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_c - h_b) - \frac{1}{2}m(v_c^2 - v_b^2)\\\\ \Delta E = 50\times 9.8(0-2) \ - \ \frac{1}{2} \times 50(0^2 - 8.85^2)\\\\\Delta E = -980 + 1960 \\\\\Delta E = 980 \ J[/tex]
Thus, we can conclude that mechanical energy is not conserved between b and c.
Learn more here:https://brainly.com/question/19969393
Determine the electrical force of attraction between two balloons
that are charged with the opposite type of charge but the same
quantity of charge. The charge on the balloons is 6.0 x 10-7 C and they
are separated by a distance of 0.50 m.
Answers
Answer:
F=1.3x10^-2N
Explanation:
Fe= k(6x10^-7C)^2/(0.5)^2
Electrical force of attraction between the balloons is F=1.3x10^-2N
The electric force of attraction between two balloons should be F=1.3x10^-2N.
Calculation of the electric force;Since The charge on the balloons is 6.0 x 10-7 C and they are separated by a distance of 0.50 m.
So, here the electric force is
Fe= k(6x10^-7C)^2/(0.5)^2
F=1.3x10^-2N
hence, The electric force of attraction between two balloons should be F=1.3x10^-2N.
Learn more about force here: https://brainly.com/question/19848845
Which factor affects the amount of runoff that occurs in an area?
land use
the water table
the saturation zone
amount of nutrients in soil
Answers
Answer:
A. land use
Explanation:
Answer:
a
Explanation:
A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?
Answers
Answer:
The answer is 45 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
distance = 3 meters
force = 15 newtons
We have
workdone = 15 × 3
We have the final answer as
45 JHope this helps you
a motor boat is traveling at 25 knots towards 340 degree on a river flowing at 20 knots towards 175 degrees. What is the actual speed of the boat as seen by a helicopter piolet hovering above?
Answers
Answer:
Vbx = 25 * cos 340 = 23.5 knot x-component of boats speed
Vrx = 20 cos 175 = -19.9 knot x-component of rivers speed
Vx = 3.58 knot x-component of boat and river speed
Vby = 25 sin 340 = -8.55 knot y-component of boat speed
Vry = 20 sin 175 = 1.74 knot y-component of river speed
Vy = -6.81 knot y-component of boat and river speed
V = (Vx^2 + Vb^2)^1/2 = (3.58^2 + 6.81^2)^1/2 = 7.69 knots
Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.
Answers
The question is incomplete. Here is the complete question.
A current in the long, straight wire, which lies in the plane of rectangular loop, that also carries a current, as shown in the figure.
Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.
Answer: Net Force = [tex]50.215.10^{-7}[/tex]N
Explanation: Force and Magnetic field are related through the following formula:
F = I.L.B.sinθ
Magnetic field (B) in a straight long wire is given by
[tex]B=\frac{\mu_{0}.I}{2.\pi.r}[/tex]
in which
[tex]\mu_{0}[/tex] is permeability of free space and is [tex]4.\pi.10^{-7}[/tex]T.m/A
I is current in the wire;
r is distance to the wire;
Examining the square loop and using the right hand rule, the top, which we will name it F₂, and the bottom, named F₄, have angle θ = 0, giving sin(0) = 0 and therefore, F₁ = F₃ = 0.
So, for the net force, the relevant forces will be on the sides parallel to the wire.
For the other forces, angle is 90°, sin(90°) = 1, then:
F = I.L.B
Replacing magnetic field:
F = [tex]\frac{\mu_{0}.I_{w}.L.I_{l}}{2.\pi.r}[/tex]
Note: The side closest to the wire is F₁, while the farthest is F₃.
Note2: As the constant unit is in meters, distance and length of side of the square loop are also in meters.
Calculating forces:
F₁ = [tex]\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.082}[/tex]
F₁ = [tex]278.975*10^{-7}[/tex]N
Current in F₃ is flowing thoruhg the negative side of the referential, so:
F₃ = [tex]-\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.1}[/tex]
F₃ = [tex]-228.76*10^{-7}[/tex]N
Net force is total force:
[tex]F_{net} = F_{1}+F_{3}[/tex]
[tex]F_{net}=(278.975-228.76).10^{-7}[/tex]
[tex]F_{net}=50.22.10^{-7}[/tex]
The total force acting on the square loop is [tex]F_{net}=50.22.10^{-7}[/tex]N.
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
Answers
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.
Pls help pls pls pls pls
Answers
2.activity log
3.specific warm up
4.activities of daily living
5.planned exercise
6.general warm up
if you are driving 110 km/h along a straight road and you look to the side for 2.0 s , how far do you travel during this inattentive period ? explain.
Answers
Explanation:
hope this helps, have a good one :D
Answer:
60.12m
Explanation:
Distance = Velocity x Time
To use this formula we must first convert 110km/h to m/s, which we can do by dividing the value by 3.6:
110/3.6 = 30.56m/s (2dp)
Velocity = 30.56m/s
Time = 2s
Distance = 30.56x2
Distance = 61.12m
You travel 60.12m during this inattentive period.
Hope this helped!
Grass and plants get energy from
А
the sun.
B
eating food.
с
windmills.
D
electrons.
Answers
Answer:
From the Sun
Explanation:
Plants can't eat any food. They don't ue or need windmills to get energy. They are plants so they don't have any electrons. The only way that they can recive energy from is the sun. Sometimes plants die when they don't get enough sun because they don't have any energy to live.
1
2
3
4
5
6
8
9
10
Dressing appropriately for exercise includes
A. wearing the same clothes for all exercises
B. choosing dark colored clothing when exercising at night
C. wearing sunscreen when exercising outside
D. making sure you wear the best brand-name clothes
Please select the best answer from the choices provided.
A
B.
C.
D.
The answer is C.
Answers
Answer:
the answer is C
Explanation:
you said its c
What must be true if a wave's wavelength is short?
A
Its frequency is low.
B
It does not carry energy.
с
Its frequency is high.
D
The waves are visible.
Answers
Answer:
im sure its A
Explanation:
How is the voltage V across the resistor related to the current I and the resistance R of the resistor? (Use I for current and R for resistance.)
Answers
Answer:
This relationship is explained by Ohm's law
Explanation:
Ohm's law states that the current flowing through a circuit or a resistor is directly proportional to the voltage across the resistor and inversely proportional to the resistance. Where current is i, voltage is v and resistance is r, Ohm's law can be represented mathematically as
V= IR
What does the principle of superposition help scientists determine?
A) The super powers of a rock layer
B) The exact and absolute age of a rock layer
C) The relative age of a rock layer
D) The position of a fossil
Answers
Answer:
B
Explanation:
the exact and absolute age of a rock layer
Answer:
The relative age of a rock layer.
Explanation:
The answer is C.
An object is dropped from a 45 m high building. At the same time, another object is thrown
upward with a velocity of 8.5 ms 1. How high above the ground will the two objects meet?
(With work please)
Answers
Answer:
-92.33 (meaning the objects will not meet above the ground).
Explanation:
We can use the kinematic equation displacement = initial velocity*time + 1/2*acceleration*time^2.
We can plug in the known values of the 2 objects into the equation, where t is the time and x is the displacement:
x = 0*t + 1/2*(-9.8)*t^2+45
x = 8.5*t + 1/2*(-9.8)*t^2
We need to first solve for t to solve for x. Since both equations are equal to x, we can set them equal to each other and solve for t:
0*t + 1/2*(-9.8)*t^2+45 = 8.5*t + 1/2*(-9.8)*t^2
-4.9*t^2 +45 = 8.5*t + -4.9*t^2
45 = 8.5*t
t = 45/8.5 ≈5.294
Now, we can plug t as 5.294 into any of the equations above to solve for x:
x = 0*5.294 + 1/2*-9.8*(5.294)^2+45 ≈ -92.33
That means, the objects will not meet above the ground.
Victor has 100 trading cards, each with a distinct power level between 101 and 200 inclusive. Whenever he heads out, he always randomly selects 21 cards to bring with him just in case he meets a fellow collector. Prove that no matter which 21 cards he brings, Victor will always be able to select 4 of those cards that exhibit the following property:
Let the average power level of all 4 cards bep. The cards can be split into two pairs, each of which also has an average power level of p.
Answers
Answer:
Victor will always be able to select 4 of those cards with the following property
Explanation:
Number of trading cards = 100
victor selects 21 cards
let the 4 cards be labelled : A,B,C and D
The average power level of : A,B,C,D = ( A + B + C + D )/ 4 = P
let the two pairs be : ( A + B ) and ( C + D )
note: average power of each pair = P and this shows that
( A + B ) = ( C + D ) for Victor to select 4 cards out of the 21 cards that exhibit the same property
we have to check out the possible choices of two cards out of 21 cards yield distinct sums.
= C(21,2)=(21x20)/2 = 210.
from the question the number of distinct sums that can be created using 101 through 200 is < 210 .
hence it is impossible to get 210 distinct sums therefore Victor will always be able to select 4 of those cards
A certain superconducting magnet in the form of a solenoid of length 0.300 m can generate a magnetic field of 8.90 T in its core when its coils carry a current of 95 A. Find the number of turns in the solenoid.
Answers
Answer:
The number of turns in the solenoid is 22366.
Explanation:
The number of turns in the solenoid can be found using the following equation:
[tex] B = \mu_{0} I\frac{N}{L} [/tex]
Where:
B: is the magnetic field = 8.90 T
L: is the solenoid's length = 0.300 m
N: is the number of turns =?
I: is the current = 95 A
μ₀: is the magnetic constant = 4π×10⁻⁷ H/m
By solving equation (1) for N we have:
[tex] N = \frac{BL}{\mu_{0} I} = \frac{8.90 T*0.300 m}{4\pi \cdot 10^{-7} H/m*95 A} = 22366 turns [/tex]
Therefore, the number of turns in the solenoid is 22366.
I hope it helps you!
A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 555 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tension in the rope to the left of the mountain climber.
Answers
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The tension in the rope on the left of the mountain climber is [tex] T_a = 1106 \ N [/tex]
Explanation:
From the question we are told that
The weight of the mountain climber is m = 555 N
Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as
[tex]T_a* cos 65 -555 + T_b * cos(85) = 0[/tex]
Here [tex]T_a, T_b[/tex] are the tension of the rope on the left and on the right hand side
So
[tex]0.423T_a + 0.0871T_b = 555[/tex]
=> [tex] 0.0871T_b = 555 - 0.423T_a[/tex]
=> [tex] T_b = \frac{555 - 0.423T_a}{0.0871}[/tex]
Generally from the diagram , the total amount of force acting on the rope along the horizontal axis at equilibrium is mathematically represented as
[tex]T_a* sin 65 - T_b * sin(85) = 0[/tex]
=> [tex] 0.9063T_a - 0.9962T_b = 0[/tex]
=> [tex] 0.9063T_a = 0.9962T_b [/tex]
=> [tex] 0.9063T_a = 0.9962[\frac{555 - 0.423T_a}{0.0871}] [/tex]
=> [tex] 0.9063T_a = [\frac{552.891 - 0.421T_a}{0.0871}] [/tex]
=> [tex] 0.0789T_a = [552.891 - 0.421T_a[/tex]
=> [tex] 0.4999T_a = 552.891 [/tex]
=> [tex] T_a = 1106 \ N [/tex]
If Mary runs 5 miles in 50 minutes, what is her speed with the correct
label?
Answers
A diffusion couple, made by welding a thin onecentimeter square slab of pure metal A to a similar slab of pure metal B, was given a diffusion anneal at an elevated temperature and then cooled to room temperature. On chemically analyzing successive layers of the specimen, cut parallel to the weld interface, it was observed that, at one position, over a distance of 5000 nm, the atom fraction of metal A, NA, changed from 0.30 to 0.35. Assume that the number of atoms per m3 of both pure metals is 9 x 10^28. First determine the concentration gradient dnA/dx. Then if the diffusion coefficient, at the point in question and annealing temperature, was 2 10^-14 m^2/s.
Required:
Determine the number of A atoms per second that would pass through this cross-section at the annealing temperature.
Answers
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]
how much power is used if it takes frank (a 450 N boy ) 3 seconds to run 2 meters ?
Answers
Answer:
300
Explanation:
450Newton × 2Meter ÷ 3sec
Why wouldn't carbon dating work to determine the age of the earth?
A) Carbon dating works best on other planets
B) The half life of carbon is too short
C) The age of the earth cannot be determined
D) The half life of carbon is too long.
Answers
Answer:
The half-life of carbon is too short.
Explanation:
The answer is B.
The bending of rocks due to the compression of tectonic plates is called
Ofaulting
O folding
subduction
plyometrics
Answers
Answer:
Folding
Explanation:
g A child bounces a 50 g super ball on the sidewalk. The velocity change of the super bowl is from 27 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answers
Answer:
The average force exerted on the superball by the sidewalk is 0.00122 N.
Explanation:
Given;
mass of the super ball, m = 50 g = 0.05 kg
initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)
final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)
time of motion, t = 1800 s
The average force exerted on the superball by the sidewalk is given by;
[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{0.05(17-(-27))}{1800}\\\\ F = \frac{0.05(44)}{1800}\\\\F = 0.00122 \ N[/tex]
Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.
What is the change in internal energy (in J) of a system that absorbs 0.523 kJ of heat from its surroundings and has 0.366 kcal of work done on it
Answers
Answer:
The change in internal energy of the system is 2,054 J
Explanation:
The first law of thermodynamics relates the work and the transferred heat exchanged in a system through internal energy. This energy is neither created nor destroyed, it is only transformed.
Taking into account that the internal energy is the sum of all the energies of the particles that the system has, you have:
ΔU= Q + W
where U is the internal energy of the system (isolated), Q is the amount of heat contributed to the system and W is the work done by the system.
By convention, Q is positive if it goes from the environment to the system, or negative otherwise, and W is positive if it is carried out on the system and negative if it is carried out by the system.
In this case:
Q= 0.523 kJ (because the energy is absorbed, this is,it goes from the environment to the system)W= 0.366 kcal= 1.531 kJ (because the work is done on the system, and being 1 kcal= 4.184 kJ)Replacing:
ΔU= 0.523 kJ + 1.531 kJ
Solving:
ΔU= 2.054 kJ = 2,054 J (being 1 kJ=1,000 J)
The change in internal energy of the system is 2,054 J
An object is dropped from rest and falls freely 20 m to Earth. When is the speed of the object 9.8 m/s?
At the end of the first second of its fall
At the end of the first second of its fall
During the entire time of its fall
During the entire time of its fall
At no time is the speed 9.8 m/s
At no time is the speed 9.8 m/s
During the entire first second of its fall
During the entire first second of its fall
After it has fallen 9.8 meters
Answers
3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage dropped across each phase?
Answers
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