Answer:
a) NO, YES that cable 3 should be placed to the left of cable 1
b) a = 8cm the current is UP
Explanation:
The force between two cables that carries a current is given through the magnetic field created between them, it has the expression
F₁ = μ₀ I₁ I₂ / 2π a l
where I₁ and I₂ are the current in each wire, at the separation between them and l the length of wire 1 on which the force is applied.
The direction of the force this given is given by the vector product between the current in wire 1 and the field created by 2, but we can summarize it:
* if the two currents are in the same direction the force is attractive
* if the two currents are contrary, the force is repulsive
With these relationships we can examine the different situations presented
a) Between cable 1 and 3 the force is zero. If cable 1 the current is upward in cable 3 it must be downward, so that the force is repulsive
Between wire 2 and 3. If wire 2 the current is downward, in wire 3 the current must be upward, for a repulsive force.
We see that in the two conditions the current in cable 3 has different directions, which is impossible, therefore this situation cannot occur
the answer is NO
The only way this is possible is that cable 3 should be placed to the left of cable 1
Answer YES
b) and c) Your question is unclear, I interpret that you want the position of cable 3 so that the force is zero on cable 3
we will assume that the current in cables 1 and 2 is ascending and that in cable 3 is descending
We write Newton's second law at the point of cable 3, the acceleration is zero
F13 - F23 = 0
F13 = F23
Let's calculate the forces
F₁₃ = μ₀ I₁ I₃ /2π a l
F₂₃ = μ₀ I₂ I₃ /2π (d + a) l
we substitute and simplify
I₁ / a = I₂ / (d + a)
where we have assumed that the length of all cables is the same
We solve to find
I₁ (d + a) = I₂ a
a (I₂ - I₁) = I₁ d₁
a = I₁ / (I₂ - I₁) d
let's calculate
a = 1.20 / (4.2 -1.2) 20
a = 8cm
to the left of wire 1 and the direction is the current is UP
so that it is an attractive force between 1 and 3 and a repulsive force between 2 and 3
How fast must a meter stick be moving if its length is observed to shrink to 0.57 m?
Answer:
0.8216c
Explanation:
Using the relationship
L' = L√(1 - v²/c²)
where
L = original length,
L' = observed length,
v = velocity,
c =speed.
L'/L = 0.57
Then
0.57 = √(1 - v²/c²)
1 - v²/c² = 0.57² = 0.3249
v²/c² = 1 - 0.3249 = 0.6751
v² = 0.6751c²
v = c√0.6751 = 0.8216c
Explanation:
Define fluid flow. What are the types of fluid flow?
Answer:
The different types of fluid flow are: Steady and Unsteady Flow. Uniform and Non-Uniform Flow. ... Compressible and Incompressible Flow. Rotational and Irrotational Flow.
If the momentum of a system is to be conserved, which must be true of the net external force acting on the system?
A. nonzero but constant.
B. increasing
C. decreasing
D. zero
Answer:
D. zero
Explanation:
For momentum of an isolated or closed system to be conserved (initial momentum must equal final momentum), the net external force acting on the system must be zero.
There is always external forces acting on a system, for this system’s momentum to remain constant, all the external forces acting on the system must cancel out, so that the net external force on the system is zero.
[tex]F_{ext} = 0[/tex]
Therefore, the correct option is "D"
D. zero
Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen into a beaker. After the nitrogen evaporates, how much volume does it occupy if its density is equal to that of the dry air at sea level
Answer:
The value is [tex]V_n = 2.2498 \ m^3[/tex]
Explanation:
From the question we are told that
The volume of liquid nitrogen is [tex]V_n = 3.6 \ L= 3.6 *10^{-3} \ m^3[/tex]
The density of nitrogen at gaseous form is [tex]\rho_n = 1.2929 \ kg/m^3[/tex] = The dry air at sea level
Generally the density of nitrogen at liquid form is
[tex]\rho _l = 808 \ kg/m^3[/tex]
And this is mathematically represented as
[tex]\rho_l = \frac{m}{V_l }[/tex]
=> [tex]m = \rho_l * V_l[/tex]
Now the density of gaseous nitrogen is
[tex]\rho_n = \frac{m}{V_n }[/tex]
=> [tex]m = \rho_n * V_n[/tex]
Given that the mass is constant
[tex]\rho_n * V_n = \rho_l * V_l[/tex]
[tex]1.2929* V_n = 808 * 3.6*10^{-3}[/tex]
=> [tex]V_n = 2.2498 \ m^3[/tex]
A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to accelerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him?
Answer:
70 N
Explanation:
Draw a free body diagram of the boy. There are four forces:
Weight force mg pulling down,
A 300 N normal force pushing up,
A 75 N applied force pulling right,
and a 5 N friction force pushing left.
The boy's acceleration in the y direction is 0, so the net force in the y direction is 0.
The net force in the x direction is 75 N − 5 N = 70 N.
The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The parsec is the radius of a circle for which a central angle of 1 s intercepts an arc of length 1 AU. The light-year is the distance that light travels in 1 y.
(a) How many parsecs are there in one astronomical unit?
(b) How many meters are in a parsec?
(c) How many meters in a light-year? (d) How many astronomical units in a light-year? (e) How many light-years in a parsec?
Answer:
a) How many parsecs are there in one astronomical unit?
[tex]4.85x10^{-6}pc[/tex]
(b) How many meters are in a parsec?
[tex]3.081x10^{16}m[/tex]
(c) How many meters in a light-year?
[tex]9.46x10^{15}m[/tex]
(d) How many astronomical units in a light-year?
[tex]63325AU[/tex]
(e) How many light-years in a parsec?
3.26ly
Explanation:
The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ([tex]1.496x10^{11} m[/tex]) is defined as 1 astronomical unit:
[tex]\tan{p} = \frac{1AU}{d}[/tex]
Where d is the distance to the star.
Since p is small it can be represent as:
[tex]p(rad) = \frac{1AU}{d}[/tex] (1)
Where p(rad) is the value of in radians
However, it is better to express small angles in arcseconds
[tex]p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}[/tex]
[tex]p('') = 2.06x10^5 p(rad)[/tex]
[tex]p(rad) = \frac{p('')}{2.06x10^5}[/tex] (2)
Then, equation 2 can be replace in equation 1:
[tex]\frac{p('')}{2.06x10^5} = \frac{1AU}{d}[/tex]
[tex]\frac{d}{1AU} = \frac{2.06x10^5}{p('')}[/tex] (3)
From equation 3 it can be see that [tex]1pc = 2.06x10^5 AU[/tex]
a) How many parsecs are there in one astronomical unit?
[tex]1AU . \frac{1pc}{2.06x10^5AU}[/tex] ⇒ [tex]4.85x10^{-6}pc[/tex]
(b) How many meters are in a parsec?
[tex]2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}[/tex] ⇒ [tex]3.081x10^{16}m[/tex]
(c) How many meters in a light-year?
To determine the number of meters in a light-year it is necessary to use the next equation:
[tex]x = c.t[/tex]
Where c is the speed of light ([tex]c = 3x10^{8}m/s[/tex]) and x is the distance that light travels in 1 year.
In 1 year they are 31536000 seconds
[tex]x = (3x10^{8}m/s)(31536000s)[/tex]
[tex]x = 9.46x10^{15}m[/tex]
(d) How many astronomical units in a light-year?
[tex]9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}[/tex] ⇒ [tex]63325AU[/tex]
(e) How many light-years in a parsec?
[tex]2.06x10^{5}AU . \frac{1ly}{63235AU}[/tex] ⇒ [tex]3.26ly[/tex]
A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur
Answer:
The current required winding is [tex]2.65*10^-^2 mA[/tex]Explanation:
We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil and
n= N/L (number of turns per unit length)
Given data
The number of turns n= 1200 turns
length L= 0.42 m
magnetic field B= 1*10^-4 T
μ₀= [tex]4\pi*10^-^7 T.m/A[/tex]
Applying the equation B=μ₀*n*I
I= B/μ₀*n
I= B*L/μ₀*n
[tex]I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }[/tex]
[tex]I= 2.65*10^-^2 mA[/tex]
Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of the force on a 3.0-m length of one of the wires is equal to 8.0 μN, what is the greater of the two currents?
Answer:
The greater of the two currents is 0.692 A
Explanation:
Given;
distance between the two parallel wires; r = 6 mm = 6 x 10⁻³ m
let the current in the first wire = I₁
then, the current in the second wire = 2I₁
length of the wires, L = 3.0 m
magnitude of force on the wires, F = 8 μN = 8 x 10⁻⁶ N
The magnitude of force on the two parallel wires is given by;
[tex]F = \frac{\mu_o I_1(2I_1)}{2\pi r}\\\\F = \frac{\mu_o 2I_1^2}{2\pi r}\\\\I_1^2 = \frac{F*2\pi r}{2\mu_o} \\\\I_1^2 = \frac{8*10^{-6}*2\pi (6*10^{-3})}{2(4\pi*10^{-7})}\\\\I_1^2 = 0.12\\\\I_1 = \sqrt{0.12}\\\\ I_1 =0.346 \ A[/tex]
the current in the second wire = 2I₁ = 2 x 0.346 A = 0.692 A
Therefore, the greater of the two currents is 0.692 A
A helicopter goes straight up 500m from a landing pad. It then goes north 20m. Then it goes down 452m. a) What is the displacement of the helicopter?
Express as components of a vector.
x-component_____________________
y-component_____________________
b) What is the displacement of the helicopter? Express as a vector (magnitude and direction).
Answer_____________________
Answer:
a
x-component [tex]20 \ m[/tex]
y-component [tex]500 - 452 = 48 \ m[/tex]
b
Magnitude [tex]d = 52 \ m[/tex]
direction is [tex]\theta = 67.4^o[/tex]
Explanation:
From the question we are told that
The first vertical distance is [tex]y_1 = 500 \ m[/tex]
The first horizontal distance is [tex]x = 20 \ m[/tex]
The second vertical distance is [tex]y_2 = 452 \ m[/tex]
Generally the displacement is
x-component [tex]20 \ m[/tex]
y-component [tex]500 - 452 = 48 \ m[/tex]
Generally the helicopters displacement is mathematically evaluated as
[tex]d = \sqrt{ x- component ^2 + y- component ^2 }[/tex]
[tex]d = \sqrt{ 20t ^2 + 48 ^2 }[/tex]
[tex]d = 52 \ m[/tex]
The direction is the angle the displacement of the helicopter makes with the horizontal which is mathematically evaluated as
[tex]\theta = tan ^{-1}[ \frac{48}{20}][/tex]
=> [tex]\theta = tan ^{-1}[ 2.4 ][/tex]
=> [tex]\theta = 67.4^o[/tex]
When the k. E of
the object
object is increases
by 100% the momentin
the body is
increased by
how to solve plz
[tex]\sqrt{2}[/tex]Answer:
KE2 = 2 KE1
1/2 M V2^2 = 2 * (1/2 M V1^2)
V2^2 = 2 V1^2
V2 = [tex]\sqrt{2}[/tex] V1
Since momentum = M V the momentum increases by [tex]\sqrt{2}[/tex]
you are working in a physics lab where you have made a simple circuit with a battery and bulb in which part of your circuit is the current flow maximum through the bulb filament or through the battery if you reverse the polarity would there be any difference in the intensity of the bulb
Answer:
The current moves in the terminal.
A small spherical body is tied to a string of length 1 m and revolved in a vertical circle such that the tension in the string is zero at the highest point . Find the linear speed of the body in the 1) lowest position & 2) highest position
Explanation:
At the highest point, the tension force is 0, so the only force acting on the sphere is gravity. Sum of forces on the sphere in the centripetal direction:
∑F = ma
mg = mv²/r
v = √(gr)
v = √(9.8 m/s² × 1 m)
v = 3.13 m/s
If the speed is constant, then the linear speed at the lowest point is also 3.13 m/s. Otherwise, we would need to know the tension in the string at that point.
An electrical cable consists of 125 strands of fine wire, each having 2.65 m0 resistance. The same potential difference is applied between the ends of all the strands and results in a total current of 0.750 A. (a) What is the current in each strand
Answer:
I = 6 mA
Explanation:
Given that,
Number of strands are 125
Resistance of each strand is 2.65 mΩ
The same potential difference is applied between the ends of all the strands and results in a total current of 0.750 A.
We need to find the current in each strand.
Total current is 0.75 A
Number of strands are 125
So, current in each strand :
[tex]I=\dfrac{0.75}{125}\\\\I=0.006\ A\\\\I=6\ mA[/tex]
So, 6 mA of current flows in each strand.
Gravel is __ than clay.
Answer:
more permeable
Explanation:
no idea, i just remember learning this in school lol.
... noisier when it's in a paper bag ...
A drag racer can reach a top speed of 98 m/s. How long will it take the racer to travel 1500 m?
Answer:
[tex]t=15.3s[/tex]
Explanation:
Hello,
In this case, since the speed is defined in terms of the distance over time:
[tex]V=\frac{x}{t}[/tex]
We can easily solve for the time with the given speed and distance:
[tex]t=\frac{x}{V}=\frac{1500m}{98m/s}\\ \\t=15.3s[/tex]
Regards.
If VF=Vi+AT and Vi=0,A=3,T=4 find Vf?
Answer: 12
Explanation:
Given: VF=Vi+AT
-------------------------
In this case, substitute all the given values into the equation
VF=Vi+AT
VF=0+(3)(4)
VF=0+12
VF=12
Hope this helps!! :)
Answer:
[tex]\huge \boxed{V_f=12}[/tex]
Explanation:
[tex]V_f=V_i+AT[/tex]
This is the formula for final velocity.
The values are given for initial velocity, acceleration, and time elapsed.
[tex]V_i=0, \ A=3, \ T=4[/tex]
Solve for [tex]V_f[/tex].
[tex]V_f=0+(3)(4)[/tex]
Evaluate.
[tex]V_f=12[/tex]
A motorboat starting from rest travels in a straight line on a lake. If the boat achieves a speed of 9.0 m/s in 13 s, what is the boat's average acceleration?
Answer:
Acceleration, [tex]a=0.69\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the motorboat, u = 0
Final speed off the motorboat, v = 9 m/s
Time, t = 13 s
We need to find the boat's average acceleration. It is equal to the change in velocity divided by time taken. SO,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{9-0}{13}\\\\a=0.69\ m/s^2[/tex]
So, the acceleration of the boat is [tex]0.69\ m/s^2[/tex].
Tech A says voltage drops can be measured as long as current is flowing. Tech B says voltage drops can be measured across components, connectors, or cables. Who is correct?
A. Tech A
B. Tech B
C. Both Techs A and B
D. Neither Tech A nor B
Answer:
C. Both Techs A and B
Explanation:
For voltage drop to be measured in the circuit, then there must be a voltage in the circuit. Once there is a voltage across the circuit, there will be current flowing through the the circuit, hence technician A is correct. Voltage drop is usually measured across components in the circuit. Components in a circuit are consumptive in the circuit, hence their is usually a voltage drop when current flows through them in a circuit. Technician B is correct.
Answer:
C
Explanation:
Match the story events on the left to the correct element of plot structure on the right
Victor pretends he can speak French
climax
Victor gets his school schedule.
resolution
Victor tries to get Teresa's attention
after homeroom and at lunch.
exposition
Teresa asks Victor if he will help her in
French
rising action
Victor checks out books to learn French
and help Teresa
falling action
Explanation:
- Victor pretends he can speak French > Rising action.
- Victor gets his school schedule > Exposition.
- Victor tries to get Teresa's attention after homeroom and at lunch > Rising action.
- Teresa asks Victor if he will help her in French > Falling action.
- Victor checks out books to learn French and help Teresa > Climax
Answer:
In "Seventh Grade" by Gary Soto, the story reaches its climax when Mr. Bueller stays quiet about Victor not knowing French. When Mr. Bueller asks if anyone in the class knows French and then Victor raises his hand, although he doesn't speak the language, Mr. Bueller decides not to make fun of it, and instead, he continues with the class normally. This action had a positive effect on Victor, who considers Mr. Bueller to be a good person and motivates him to do well in French, despite of his previous attempt to impress Teresa. Regarding the other options, although they occur at the beginning (Teresa sees Victor in the lunch area and smiles at him and Victor raises his hand in French to impress Teresa) and at the end (Victor assures Teresa that helping her will not be a bother), they aren't considered to be the highest point of the conflict in the storyThe length and width of a rectangle are 1.82 cm and 1.5 cm respectively. Calculate area of the rectangle and write in correct significant number.
Answer:
Hey mate ,
Area of rectangle = l×b
1.82×1.5
2.73cm2
For a certain experiment, Juan must measure the concentration of a certain substance in a solution over time. He needs to collect a measurement every 0.05 seconds. He then needs to display his data in a graph and place that graph in a text document. Select the best tools to use for this experiment. Check all that apply.
Answer:
Probeware and computer
Explanation:
Computers are more powerful and better than a graphing calculator for this situation.
are the tools he must use.
A 4.8-kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t=0 s, the block has a displacement of -0.50m, a velocity of -0.80m/s and an acceleration of +8.3m/s2 The force constant of the spring is closest to:______.
A) 62 N/m
B) 67 N/m
C) 56 N/m
D) 73 N/m
E) 80 N/m
Answer:
E) 80 N/m
Explanation:
Given;
mass of the block, m = 4.8 kg
displacement of the block, x = -0.5 m
velocity of the block, v = -0.8 m/s
acceleration of the block, a = 8.3 m/s²
From Newton's second law of motion;
F = ma
Also, from Hook's law;
F = -Kx
where;
k is the force constant
Thus, ma = -kx
k = -ma/x
k = -(4.8 x 8.3) / (-0.5)
k = 79.7 N/m
k ≅ 80 N/m
Therefore, the force constant of the spring is closest to 80 N/m
A box is sitting on a board. The coefficient of static friction between the box and the board is 0.830216. The coefficient of kinetic friction between the box and the board is 0.326245. One side of the board is raised until the box starts sliding. Give a variable legend for this problem.
a) What is the angle at which the box starts sliding? The model for this problem:
θ=__________________________________ Answer________________________________
b) What is the magnitude of its acceleration after it starts sliding? The model for this problem:
a=__________________________________ Answer________________________________
Answer:
Explanation:
Coefficient of static friction μs = .830216
Coefficient of kinetic friction μk = .326245
a ) The angle at which the box starts sliding depends upon coefficient of static friction . If θ be the required angle
tanθ = μs
tanθ = .830216
θ = 39.7°
b )
When the box starts sliding , kinetic friction will be acting on it .
frictional force on the box = μk mg cos 39.7
net force on the box
= mg sin39.7 - μk mg cos 39.7
Applying Newton's law of motion
mg sin39.7 - μk mg cos 39.7 = m a
a = g sin39.7 - μk g cos 39.7
= 9.8 x sin 39.7 - .326245 x 9.8 x cos 39.7
= 6.26 - 2.46
= 3.8 m /s² .
A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA. What amount of vars must be added to bring the pf to 0.85
Answer:
[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]
Explanation:
Given that:
A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA
Here:
the initial power factor i.e cos θ₁ = 0.7 lag
θ₁ = cos⁻¹ (0.7)
θ₁ = 45.573°
Active power P = 1500 watts
Apparent power S = 2100 VA
What amount of vars must be added to bring the pf to 0.85
i.e the required power factor here is cos θ₂ = 0.85 lag
θ₂ = cos⁻¹ (0.85)
θ₂ = 31.788°
However; the initial reactive power [tex]Q_1[/tex] = P×tanθ₁
the initial reactive power [tex]Q_1[/tex] = 1500 × tan(45.573)
the initial reactive power [tex]Q_1[/tex] = 1500 × 1.0202
the initial reactive power [tex]Q_1[/tex] = 1530.3 vars
The amount of vars that must therefore be added to bring the pf to 0.85
can be calculated as:
[tex]Q_{sh} = P( tan \theta_1 - tan \theta_2)[/tex]
[tex]Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)[/tex]
[tex]Q_{sh} = 1500( 1.0202 - 0.6197)[/tex]
[tex]Q_{sh} = 1500( 0.4005)[/tex]
[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]
A photoelectric-effect experiment finds a stopping potentialof 1.93V when light of 200nm is used to illuminate thecathode.
a) From what metal is the cathode made from?
b) What is the stopping potential if the intensity of thelight is doubled?
Answer:
a) Tantalum
b) 1.93 V
Explanation:
The energy of the incident photon= hc/λ
h= Plank's constant=6.63×10^-34 Is
c= speed of light = 3×10^8 ms-1
λ= wavelength of incident photon
E= 6.63×10^-34 × 3×10^8/ 200×10^-9
E= 0.099×10^-17
E= 9.9×10^-19 J
The kinetic energy of the electron = eV
Where;
e= electronic charge = 1.6×10^-19 C
V= 1.93 V
KE= 1.6×10^-19 C × 1.93 V
KE= 3.1 ×10^-19 J
From Einstein's photoelectric equation;
KE= E -Wo
Wo= E -KE
Wo=9.9×10^-19 J - 3.1 ×10^-19 J
Wo= 6.8×10^-19 J
Wo= 6.8×10^-19 J/1.6×10^-19
Wo= 4.25 ev
The metal is Tantalum
b) the stopping potential remains 1.93 V because intensity of incident photon has no effect on the stopping potential.
he cans have essentially the same size, shape, and mass. Which can has more energy at the bottom of the ramp
Answer:
c. both have same energy
Explanation:
The complete question is
suppose you have two cans, one with milk, and the other with refried beans. The cans have essentially the same size, shape, and mass. If you release both cans at the same time, on a downhill ramp, which can has more energy at the bottom of the ramp? ignore friction and air resistance..
a. can with beans
b. can with milk
c. both have same energy
please explain your answer
Since both cans have the same size, shape, and mass, and they are released at the same height above the ramp, they'll possess the same amount of mechanical energy. This is because their mechanical energy, which is the combination of their potential and kinetic energy are both dependent on their mass. Also, having the same physical quantities like their size and shape means that they will experience the same environmental or physical factors, which will be balanced for both.
Type your answer in the box.
An organ is a group of two or more
function.
that work together to perform a common function
Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.625×10−3 m3/s and the diameter of the nozzle you hold is 5.19×10−3 m. At what speed v does the water exit the nozzle?
Answer:
0.153 m/s
Explanation:
The flowrate Q = 0.625 x 10-3 m^3-/s
The diameter of the nozzle d = 5.19 x 10^-3 m
the velocity V = ?
The cross-sectional area of the flow A = [tex]\pi d^{2}/4[/tex]
==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2
From the continuity equation,
Q = AV
V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = 0.153 m/s
. The Moon has an average distance from the Earth of 384,403 km and takes 27.32166 days to orbit the Earth. What is the velocity of the Moon in kilometers per hour
Answer:
Velocity of moon = 586.23 km/h
Explanation:
We are given;
Distance of moon from the Earth = 384403 km
Time taken to orbit earth;t = 27.32166 days
24 hours make 1 day, thus 27.32166 days = 27.32166 × 24 = 655.72 hours
Formula for velocity is distance/time
Thus,
Velocity of moon = distance from moon to earth/time taken to orbit the earth
Velocity of moon = 384403/655.72 = 586.23 km/h
An electron from a Ti ^ + 2 hydrogen ion leaps from one orbit with radius 13.25 angstrom to another orbit with radius 2.12 angstrom. determine the energy (Joule) e produced in said transition and the wavelength (in cm)
Answer:
ΔE = 59.75 A,
Explanation:
Titanium has 3 electrons in its last shell, as it is doubly ionized, it is left with a single electron in this shell, which is why it behaves like a hydrogen-type atom, consequently we can use Bohr's atomic theory
rₙ = a₀ /Z n²
Eₙ = 1k e² / 2a₀ (Z² / n²)
Where a₀ is Bohrd's atomic radius so = 0.529 núm
Let's find out what quantum number n has each orbit
rn = 13.25 A = 1.325 nm
for Titanium with atomic number 22
n² = Z rₙ / a₀
n = √ (22 (1.325 / 0.529))
n = 7.4
since N is an entry we take
n = 7
rn = 2.12 A = 0.212 nm
n = √ (22 / 0.529) 0.212
n = 3
With these values we can calculate the energy of the transition from level ne = 7 to level no = 3
ΔE = ka e2 Z2 / 2ao (1n02 - 1 / nf2)
ΔE = 9 10⁹ 1.6 10⁻¹⁹ 22² (2 0.529 10⁻⁹) (1/3² - 1/7²)
ΔED = 6.5875 10² (0.111 - 0.0204)
ΔE = 59.75 A
let us be the Planck relation between energy and frequency
E = h f
the frequency is related to the speed of light
c = λ f
f = c / λ
we substitute
E = h c /y
E = ΔE
h c /λ = E
λ = 6.63 10-34 3 108 / 59.75
λ= 3.01939 10⁻²⁴ m
λ = 3.01939 10⁻²² cm