The question is incomplete. The complete question is :
Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 6 in a one-dimensional box 34.0 pm in length.
Solution :
In an one dimensional box, energy of a particle is given by :
[tex]$E=\frac{n^2h^2}{8ma^2}$[/tex]
Here, h = Planck's constant
n = level of energy
= 6
m = mass of particle
a = box length
For n = 6, the energy associated is :
[tex]$\Delta E = E_6 - E_1 $[/tex]
[tex]$\Delta E = \left( \frac{n_6^2h_2}{8ma^2}\right) - \left( \frac{n_1^2h_2}{8ma^2}\right) $[/tex]
[tex]$=\frac{h^2(n_6^2 - n_1^2)}{8ma^2}$[/tex]
We know that,
[tex]$E = \frac{hc}{\lambda} $[/tex]
Here, λ = wavelength
h = Plank's constant
c = velocity of light
So the wavelength,
[tex]$= \frac{hc}{E}$[/tex]
[tex]$=\frac{hc}{\frac{h^2(n_6^2 - n_1^2)}{8ma^2}}$[/tex]
[tex]$=\frac{8ma^2c}{h(n_6^2 - n_1^2)}$[/tex]
[tex]$=\frac{8 \times 9.109 \times 10^{-31}(0.34 \times 10^{-10})^2 (3 \times 10^8)}{6.626 \times 10^{-34} \times (36-1)}$[/tex]
[tex]$= \frac{ 8 \times 9.109 \times 0.34 \times 0.34 \times 3 \times 10^{-43}}{6.626 \times 35 \times 10^{-34}}$[/tex]
[tex]$=\frac{25.27 \times 10^{-43}}{231.91 \times 10^{-34}}$[/tex]
[tex]$= 0.108 \times 10^{-9}$[/tex] m
= 108 pm
How is energy transferred when the
torch is switched on?
What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?
Answer:
The new force becomes 4 times the initial force.
Explanation:
The force of attraction or repulsion is given by the relation as follows :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
Where
d is the distance between the interacting charges
F is inversely proportional to the distance between charges.
If the distance is halved, d'=(d/2), new force is given by :
[tex]F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F[/tex]
So, the new force becomes 4 times the initial force.
While investigating Kirchhoff's Laws, you begin observing a blackbody, such as a star, from Earth using advanced technology that can analyze spectra. While pointing it at the star with nothing between you and the star, you observe a full spectrum. You come back and repeat this same experiment a year later using the same star, except this time you observe an absorption spectrum. What is the most likely explanation for this
Answer:
the second time there is a gas between you and the star,
Explanation:
When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.
When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,
The information obtained from the two spectra is the same, the type of atoms that make up the star
Name two types of mechanical weathering in NewBedford
Answer:
Explanation:
Mechanical weathering types
Mechanical weathering is the breaking down of rocks into smaller pieces without changing the composition of the minerals in the rock. This can be divided into four basic types – abrasion, pressure release, thermal expansion and contraction, and crystal growth.
A parallel-plate capacitor with plate area 2.6 cm^2 and air-gap separation 0.25 mm is connected to a 30 V battery, and fully charged. The battery is then disconnected.
I Solved most of the parts of this q's but not this one:
The plates are now pulled to a separation of 0.55 mm. What is the charge on the capacitor now?
note:
I found 125.5pC and the answer unit is pC
Answer:
276.12 pC
Explanation:
We are given that
Area,A=[tex]2.6 cm^2=2.6\times 10^{-4}m^2[/tex]
Where [tex]1 cm^2=10^{-4} m^2[/tex]
[tex]d=0.25 mm=0.25\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
Potential difference, V=30 V
We have to find the charge on the capacitor when the plates are pulled to a separation of 0.55 mm.
We know that
Charge ,[tex]Q=\frac{\epsilon_0 A V}{d}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
Using the formula
[tex]Q=\frac{8.85\times 10^{-12}\times 2.6\times 10^{-4}\times 30}{0.25\times 10^{-3}}[/tex]
[tex]Q=2.7612\times 10^{-10} C[/tex]
[tex]Q=276.12p C[/tex]
[tex]1 pC=10^{-12} C[/tex]
When the plates are now pulled to a separation of 0.55 mm.Then, the charge on the plates remain same because the battery has been disconnected.
Therefore, charge on the capacitor=276.12 pC
A 3 kg ball moving to the right at 4 m/sec collides with a 4 kg ball moving to the right at 2 m/sec. Find the final velocities of the balls in m/sec if the coefficient of restitution is 0.6.
A. 2.2, 3.4
B. 1, 2
C. 4, 5
D. 6, 8
Answer:
Option A
Explanation:
To solve this problem we need to apply the momentum conservation, and analyze the data.
For this problem, I will call the initial velocities as V₁ and V₂, while the final velocities will be V₃ and V₄.
According to the momentum principle, this states the following:
m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄ (1)
From this equation we can write an expression in function of V₃ and V₄. We also know that coefficient of restitution is 0.6. Knowing this, we can write the expression that will help us to solve for the final velocities:
e = V₄ - V₃ / 2 (2)
With both expressions we can solve for the final velocities. Let's use (1) first and see what we can simplify first by replacing the given data:
(3*4) + (4*2) = 3V₃ + 4V₄
12 + 8 = 3V₃ + 4V₄
20 = 3V₃ + 4V₄ (3)
This is all we can do for now. Let's use (2) now:
0.6 = V₄ - V₃ / 2
1.2 = V₄ - V₃
V₄ = 1.2 + V₃ (4)
Now, we can replace (4) into (3), and then, solve for V₃:
20 = 3V₃ + 4(1.2 + V₃)
20 = 3V₃ + 4.8 + 4V₃
15.2 = 7V₃
V₃ = 15.2 / 7
V₃ = 2.17 m/sWe have the value of one final velocity, let's see the other one.
V₄ = 1.2 + V₃
V₄ = 1.2 + 2.17
V₄ = 3.37 m/sThe closest values to these results are in option A, so this will be the correct option.
Hope this helps
what is the total distance traveled by a bike rider who rides for three hours at 40 km/hr and then two more hours at 50 km/hr?
Answer:
220 km
Explanation:
3 hours at 40 km/hr
40 km = 1 hour
40 km times 3 = 120 km
2 hours at 50 km/hr
50 km = 1 hour
50 times 2 = 100 km
Total distance
120 km + 100 km = 220 km
What is Ex(P), the value of the x-component of the electric field produced by by the line of charge at point P which is located at (x,y) = (a,0), where a = 8.7 cm?
Answer:
The answer is below
Explanation:
We are going to use Gauss’ law to find the electric field equation. Since electric field is coming from an infinite line of charge, hence it is going out in a radial direction.
Therefore we use the area of the electric field which passes through, forming a Gaussian cylinder. We neglect the ends of the area.
Hence:
[tex]\int\limits {E} \, dA=\frac{Q_{enc}}{\epsilon_o}\\\\E(2\pi rL)= \frac{\lambda L}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r\epsilon_o} \\\\Given \ that:\\\\r=a=8.7\ cm=0.087\ m, \lambda=-2.3 \mu C/cm=-2.3*10^{-4}\ C/m,\epsilon_o=8.85*10^{-12}F/m.\\\\Hence:\\\\E=\frac{-2.3*10^{-4}}{2\pi *0.087*8.85*10^{-12}}=-4.75*10^7\ N/C[/tex]
The value of the x-component of the electric field is -475213.968 newtons per coulomb.
Procedure - Determination of the magnitude of an electric field at a given pointIn this question we shall apply Gauss' Law to determine the magnitude of the electric field ([tex]E_{x}[/tex]), in newtons per coulomb, rapidly and based on the assumptions of uniform charge distribution and cylindrical symmetry.
[tex]\frac{Q_{enc}}{\epsilon_{o}} = \oint\,\vec E\,\bullet d\vec A[/tex] (1)
Where:
[tex]Q_{enc}[/tex] - Enclosed charge, in coulombs.[tex]\epsilon_{o}[/tex] - Vacuum permitivity, in quartic second-square amperes per kilogram-cubic meter.[tex]\vec E[/tex] - Electric field vector, in newtons per coulomb.[tex]\vec A[/tex] - Area vector, in square meters.Based on all assumptions, we simplify (1) as follows:
[tex]\frac{\lambda\cdot l}{\epsilon_{o}} = E \cdot (2\pi\cdot r\cdot l)[/tex]
And the equation of the x-component of the electric field is:
[tex]E = \frac{\lambda}{2\pi\cdot \epsilon_{o}\cdot r}[/tex] (2)
Where [tex]\lambda[/tex] is the linear charge density, in coulomb per meter.
If we know that [tex]\lambda = -2.3\times 10^{-6}\,\frac{C}{m}[/tex] and [tex]a = 0.087\,m[/tex], then the electric field produced by the line of charge at point P is:
[tex]E = \frac{\left(-2.3\times 10^{-6}\,\frac{C}{m} \right)}{2\pi\cdot \left(8.854\times 10^{-12}\,\frac{s^{4}\cdot A^{2}}{kg\cdot m^{3}} \right)\cdot (0.087\,m)}[/tex]
[tex]E_{x} = -475213.968 \,\frac{N}{C}[/tex]
The value of the x-component of the electric field is -475213.968 newtons per coulomb. [tex]\blacksquare[/tex]
RemarkThe figure is missing, we present the corresponding image in the file attached below.
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When monochromatic light passes through the interface between two unknown materials at an angle θ where 0∘<θ<90∘, no changes in the direction of propagation of light are observed. What can be said about the two materials? Check all that apply. View Available Hint(s) Check all that apply. The two materials have matching indexes of refraction. The second material through which light propagates has a lower index of refraction. The second material through which light propagates has a higher index of refraction. The two materials are identical.
Answer:
the correct one is the first, the refractive index of the two materials must be the same
Explanation:
When a beam of light passes through two materials, it must comply with the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of each medium.
In this case, it indicates that the light does not change direction, so the input and output angle of the interface must be the same,
θ₁ = θ₂ = θ
substituting
n₁ = n₂
therefore the refractive index of the two materials must be the same
When reviewing the answers, the correct one is the first
Which is larger: 65 mph (miles per hour) or 120 kph (kilometers per hour)? As a percentage, how much faster is one than the other?
To Find :
Which is larger: 65 mph (miles per hour) or 120 kph (kilometers per hour).
Solution :
We know, 1 mph = 1.61 kph
So, 65 mph = 1.61 × 65 kph
65 mph = 104.65 kph
Since, 65 mph is 104.65 kph which is smaller than 120 kph.
Therefore, 120 kph is faster than 65 mph by ( 120 - 104.65 ) = 15.35 kph.
Select the Moon and use the Info view to determine which of the following statements is correct. The Last Quarter Moon.... rises near noon and sets near midnight. rises at about 6am and sets at about 6pm. rises near midnight and sets near midday. rises and sets at the same time as the Sun. Submit Your Answer
Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).
Explanation:
Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.
There are 8 main types of the moon phases these includes:
--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.
--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.
--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.
--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.
--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.
--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker
--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.
--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.
Statement A: 2.567 km, to two significant figures. Statement B: 2.567 km, to three significant figures. Determine the correct relationship between the statements. View Available Hint(s) Determine the correct relationship between the statements. Statement A is greater than Statement B. Statement A is less than Statement B. Statement A is equal to Statement B.
Answer:
Statement A is greater than Statement B.
Explanation:
Statement A: 2.567 km, to two significant figures..
To 2 sig figures means only 2 whole numbers should be left after approximation. Thus, 2.567 to 2 significant figures is 2.6 km
Statement B: 2.567 km, to three significant figures. To 3 sig figures means only 3 whole numbers should be left after approximation. Thus, 2.567 to 3 significant figures is 2.57 km
Comparing both values, statement A is obviously greater than Statement B
A 0.5kg football thrown by Tony Romo with a velocity of 15 m/s is caught by a stationary receiver and brought to rest in 0.02 seconds. a) what impulse is delivered to the ball? b) how much force must be exerted in order to stop the ball?
The answers are -7.5kg m/s and F = -375N, but I don't know how they got them. Steps please! Thank you!!
Answer:
a.-7.5 kg m/s
b.-375 N
Explanation:
We are given that
Mass of football, m=0.5 kg
Initial velocity ,u=15m/s
Final speed ,v=0
Time, t=0.02 s
a. We have to find the impulse delivered to the ball.
We know that
Impulse=Change in momentum
I=m(v-u)
Using the formula
[tex]I=0.5(0-15)=-7.5 kg m/s[/tex]
Hence, the impulse delivered to the ball=-7.5 kg m/s
(b)
We know that
Force,[tex]F=\frac{|mpulse}{time}[/tex]
Using the formula
[tex]F=\frac{-7.5}{0.02}=-375 N[/tex]
the luminous flux of a torch of intensity 50 cd is?
Answer:
i dont know i am right but here Luminous intensity is defined as dI=dΨλ / dΩ, where dΨλ is the luminous flux (light energy flux in watts per m2) emitted within a solid angle dΩ. The light energy flux may be expressed in terms of the incident x-ray energy flux and the x-ray absorption and conversion properties of the scintillator(7,8,9).
Explanation:
a wooden block is cut into two pieces, one with three times the mass of the other. a depression is made in both faces of the cut so that a fire cracker can be placed in it and the block is reassembled. the reassembled block is set on rough surface and the fuse is lit. when the fire cracker explodes, the two blocks separate. what is the ratio of distances traveled by blocks?
Answer:
1/9
Explanation:
Let A denote the bigger piece and let B denote the smaller piece.
We are told that one with three times the mass of the other.
Therefore, we have;
M_a = 3M_b
Firecracker is placed in the block and it explodes and thus, momentum is conserved.
Thus;
V_ai = V_bi = 0
Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.
Since initial momentum equals final momentum, we have;
P_i = P_f
Thus;
0 = (M_a × V_af) + (M_b × V_bf)
Since M_a = 3M_b, we have;
(3M_b × V_af) + (M_b × Vbf) = 0
Making V_af the subject, we have;
V_af = -⅓V_bf
The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;
W_f = -½M_b•(v_bf)²
Now, let's express the work is in terms of the force and the distance.
Thus;
W_f = F_f × Δx × cos 180°
Frictional force is also expressed as μmg
Thus;
W_f = -μM_b × g × Δx
Earlier, we saw that;
W_f = -½M_b•(v_bf)²
Thus;
-½M_b•(v_bf)²= -μM_b × g × Δx
Δx = (v_bf)²/2μg
Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b
Thus;
Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)
Δx_a/Δx_b = ((v_af)²/((v_bf)²)
Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²
Δx_a/Δx_b = 1/9
What is a response of an organism when the temperature drops? A. Panting, which exchanges cold air for warm air in the body raising the temperature Sweating, which pulls moisture to the surface to trap heat close to the skin Salivating, which lets go of extra liquid in the body releasing energy as heat Shivering, where muscles contract and expand quickly and the body temperature rises
Shivering is a response of an organism when the temperature drops. In shivering, muscles contract and expand quickly and the body temperature rises.
What is Shivering ?Warm-blooded animals shiver (also known as shudder) as a physiological reaction to cold and acute fear. To preserve homeostasis, the shivering reaction is activated when the body's core temperature lowers.
Small skeletal muscle movements start, producing warmth as the muscles use up their energy. Another sign of a fever is shivering because the person may feel cold.
The hypothalamic set point for temperature rises during a fever. Up until the new set point is reached, the patient experiences cold until the increased set point raises body temperature (pyrexia).
Rigors are the medical term for extremely cold symptoms accompanied by ferocious shivering. Rigors happen as a result of the patient's body shivering in an effort to physiologically raise body temperature to the new set point.
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Which of the following is a vector quantity?
speed
distance
acceleration
Compare and Contrast the Following:
1. Pitch and Loudness
2. Infrasonic and Ultrasonic
3. Luminous and Nonluminous
It's okay if you answer only one of these.
Thanks in Advance
Answer:
pitch and loudnesss
thanks for your question
An automobile follows a circular road whose radius is 50 m. Let x and y respectively denote the eastern and northern directions, with origin at the center of the circle. Suppose the vehicle starts from rest at x = 50 m heading north, and its speed depends on the distance s it travels according to v = 0.5s − 0.0025s 2 , where s is measured in meters and v is in meters per second. It is known that the tires will begin to skid when the total acceleration of the vehicle is 0.6g. Where will the automobile be and how fast will it be going when it begins to skid? Describe the position in terms of the angle of the radial line relative to the x axis.
Answer:
The automobile is running at speed of 23.806 meters per second.
Explanation:
From Kinematic we remember that acceleration ([tex]a[/tex]) can be defined by this ordinary differential equation in terms of distance:
[tex]a = v\cdot \frac{dv}{ds}[/tex] (1)
Where:
[tex]v[/tex] - Speed of the automobile, measured in meters per second.
[tex]s[/tex] - Distance travelled by the automobile, measured in meters.
If we know that [tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex], then the equation of acceleration is:
[tex]a = (0.5\cdot s - 0.0025\cdot s^{2})\cdot \left(0.5-0.0050\cdot s\right)[/tex]
[tex]a = s\cdot (0.5-0.0025\cdot s)\cdot (0.5-0.0050\cdot s)[/tex]
[tex]a = s\cdot (0.0025\cdot s - 0.5)\cdot (0.0050\cdot s-0.5)[/tex]
But distance covered by the vehicle is defined by the following formula:
[tex]s = \theta \cdot r[/tex] (2)
Where:
[tex]\theta[/tex] - Arc angle, measured in radians.
[tex]r[/tex] - Radius, measured in radians.
Then, we expand (1) by means of this result:
[tex]a = \theta\cdot r \cdot (0.0025\cdot \theta\cdot r -0.5)\cdot (0.0050\cdot \theta \cdot r-0.5)[/tex]
[tex]a = \theta\cdot r \cdot (1.25\times 10^{-5}\cdot \theta^{2}\cdot r^{2}-3.75\times 10^{-3}\cdot \theta\cdot r +0.25)[/tex]
[tex]a = 1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r[/tex]
And finally we get the following third order polynomial:
[tex]1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r - a = 0[/tex] (3)
If we know that [tex]r = 50\,m[/tex], [tex]a = 0.6\cdot g[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the polynomial becomes into this:
[tex]1.5625\cdot \theta^{3}-9.375\cdot \theta^{2} +12.5\cdot \theta - 5.886 = 0[/tex] (3b)
This polynomial can be solved analytically by Cardano's Method or by numerical methods. The roots of the polynomial are, respectivelly:
[tex]\theta_{1} \approx 4.365\,rad[/tex], [tex]\theta_{2} \approx 0.818+i\,0.441\,rad[/tex], [tex]\theta_{3}\approx 0.818 -i\,0.441\,rad[/tex], [tex]\theta_{4} \approx 1.563\,rad[/tex]
Both first and fourth roots are physically reasonable solution, but the latter represents the angle where automobile begins to skid first. Then, the automobile begins to skid at an angle of 1.563 radians relative to x axis.
The distance travelled by the automobile is: ([tex]r = 50\,m[/tex], [tex]\theta \approx 1.563\,rad[/tex])
[tex]s = (1.563\,rad)\cdot (50\,m)[/tex]
[tex]s = 78.15\,m[/tex]
Lastly, the speed of the automobile at this location is: ([tex]s = 78.15\,m[/tex])
[tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex] (4)
[tex]v = 0.5\cdot (78.15)-0.0025\cdot (78.15)^{2}[/tex]
[tex]v = 23.806\,\frac{m}{s}[/tex]
The automobile is running at speed of 23.806 meters per second.
2. Mrs. Stern is standing still on rollerblades on a frictionless floor in the middle of the A-gym while
carrying heavy textbooks. How can she use the textbooks to get herself moving?
Answer:
If she bends forward
Explanation:
because the equilibrium of gravity will not stay the same causing her to move forward
A long copper bar of rectangular cross-section, whose width w is much greater than its thickness L, is maintained in contact with a heat sink at its lower surface, and the temperature throughout the bar is approximately equal to that of the sink, To. Suddenly, an electric current is passed through the bar and an airstream of temperature T is passed over the top surface, while the bottom surface continues to be maintained at To. Obtain the differential equation and the top surface boundary condition that could be solved to determine the temperature as a function of position and time in the bar.
If the speed of an object does NOT change, the object is traveling at a
constant speed
increasing speed
decreasing speed
Answer:
If the speed does not change at all, the object would be moving at a constant speed.
Welding requires extensive training.
True
False
Answer:
True.
Explanation:
Welding requires extensive training because welding involves fire and we need to use fire safety measurements. A normal man can't just simply go and weld so a person must require extensive training for welding.
The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. The freight train is traveling at 15.0m/s in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant accelaration of -0.100m/s2, while the freight train continues with constant speed. Take x=0 at the location of the front of the passenger train when the engineer applies the brake.
a) Will the cows nearby witness a collision?
b) If so, where does it take place
c) On a single graph, sketch the positions of the frong of the passenger train and the back of the freight train.
Answer:
a) if the two trains are going to collide
b) x = 538 m
Explanation:
To solve this exercise we use the kinematics relations in one dimension.
We set a reference system at the starting point of the passenger train
Freight train, we use index 1 for this train
x₁ = x₀ + v₁ t
passenger train
x₂ = v₂ t - ½ a₂ t²
as we are using the same system to measure the position of the two trains, at the meeting point the position must be the same
x₁ = x₂
we substitute
x₀ + v₁ t = v₂ t - ½ a₂ t²
½ a₂ t² + t (v₁ -v₂) + x₀ = 0
we subjugate the values
½ 0.1 t² + t (15-25) + 200 = 0
0.05 t² - 10 t + 200 = 0
t² - 200 t + 4000 = 0
we solve the second degree system
t = [200 ±[tex]\sqrt{200^2 - 4 \ 4000}[/tex] ]/ 2
t = [200 ± 154.9] / 2
t₁ = 177.45 s
t₂ = 22.55 s
therefore for the smallest time the two trains must meet t₂ = 22.55 s
merchandise train
x = 200+ 15 22.55
x = 538 m
passenger train
x = 25 22.55 -1/2 0.100 22.55²
x = 538 m
we see that the trains meet for this distance
a) if the two trains are going to collide
b) x = 538 m
c) see attachment for schematic graphics
2.
Which is the value of a vector quantity?
A 200V
B 100kg/m
C 20m/s, east
D 50J/(kg°C)
А
B
C
D
3.
The diagrams show three uniform beams P Q and Reach pivoted at its centre
Answer:
c
Explanation:
a vector quantity has both magnitude and direction
The value of 20m/s, east is a vector quantity is Hence, option (C) is correct.
What is vector quantity?A physical quantity that has both directions and magnitude is referred to as a vector quantity.
A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.
Given values 200V, 100kg/m, 50J/(kg°C) are denoting magnitude of different physical quantity. Hence, they and scalar quantity ( Physical quantities with merely magnitude and no direction are referred to as scalar quantities. These physical quantities can be explained just by their numerical value without any further guidance.).
But The value of 20m/s, east has a magnitude of 20 m/s and a direction along east. Hence, 20m/s, east denotes a vector quantity is Hence, option (C) is correct.
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What is the SI unit of energy and Work?
Answer:
joule is the answer
Explanation:
I hope it helps you
The diagram shows two balls before they collide.
2 balls with grey arrows pointing to them from the outside. The left ball has below it m subscript 1 = 0.6 kilograms v subscript 1 = 0.5 meters per second. The right ball has below it m subscript 2 = 0.5 kilograms v subscript 2 = negative 0.2 meters per second.
What is the momentum of the system after the collision?
1. 0.0 kg • m/s
2. 0.2 kg • m/s
3. 0.3 kg • m/s
4. 0.4 kg • m/s
Answer:
The Answer is B)0.2 kg • m/s
Explanation:
I made a 100 on my test. Sorry if I'm late but hope I helped.
Answer:
B. 0.2 kg x m/s
Explanation:
Passage of an electric current through a long conducting rod of radiusriand thermalconductivitykrresults in uniform volumetric heating at a rate ofq. The conducting rodis wrapped in an electrically nonconducting cladding material of outer radiusroandthermal conductivitykc, and convection cooling is provided by an adjoining fluid. Forsteady-state conditions, write appropriate forms of the heat equations for the rod andcladding. Express appropriate boundary conditions for the solution of these equations.
Answer:
a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface; [tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0
b) For constant temperature; [tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])
c) The heat transfer in the conducting rod and the cladding material is the same, i.e; [tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] = [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]
d) The convection surface conduction by cooling fluid will be;
[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )
Explanation:
Given the data in question;
we write the general form of the heat conduction equation equation in cylindrical coordinates with internal heat generation.
1/r[tex]\frac{d}{dr}[/tex]( kr[tex]\frac{dT}{dr}[/tex] ) + 1/r² [tex]\frac{d}{d\beta }[/tex]( ( k[tex]\frac{dT}{dr}[/tex] ) + [tex]\frac{d}{dz}[/tex]( k[tex]\frac{dT}{dr}[/tex]) + q = 0
where radius of cylinder is r, thermal conductivity of the cylinder is k, and q is heat generated in cylinder.
Now, Assume one dimensional heat conduction
lets substitute the condition for conducting rod with steady state condition.
[tex]k_{y}[/tex]/r [tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{y} }{dr}[/tex] ) + q = 0
Apply the conditions for cladding by substituting 0 for q
[tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{r} }{dr}[/tex] ) = 0
Apply the following boundary conditions;
a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface;
[tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0
b) For constant temperature
[tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])
c) The heat transfer in the conducting rod and the cladding material is the same, i.e
[tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] = [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]
d) The convection surface conduction by cooling fluid will be;
[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )
Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.169 m to the right of Q2. The force on Q2 due to its interaction with Q3 is directed to the.....
1.Left if the two charges have opposite signs. t/f
2.Right if the two charges are negative. t/f
3.Left if the two charges are positive. t/f
4.Left if the two charges are negative. t/f
5. Right if the two charges have opposite signs. t/f
In the above problem, Q1= 1.90·10-6 C, Q2= -2.84·10-6 C, and Q3= 3.03·10-6 C.
1. Calculate the total force on Q2. Give with the plus sign for a force directed to the right.
2. Now the charges Q1= 1.90·10-6 C and Q2= -2.84·10-6 C are fixed at their positions, distance 0.301 m apart, and the charge Q3= 3.03·10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.
Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called Electrostatic Force.
If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.
So,
1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;
2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;
Sentences 3 and 4 are also TRUE due to the reasons described above;
5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;
1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:
[tex]F=\frac{k.q.Q}{r^{2}}[/tex]
where k is a constant that equals 9 x 10⁹ N.m²/C²
Calculating force between 1 and 2:
[tex]F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}[/tex]
[tex]F_{12}=536.02.10^{-3}[/tex] N
Force between 2 and 3:
[tex]F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}[/tex]
[tex]F_{23}=2711.63.10^{-3}[/tex] N
Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:
[tex]F_{T}=2711.63.10^{-3}-536.02.10^{-3}[/tex]
[tex]F_{T}=2175.61.10^{-3}[/tex] N
The total force on Q2 is 2175.61 x 10⁻³ N
2. For net force to be 0, [tex]F_{13}=F_{23}[/tex]. Suppose distance from 1 to 3 is x, then from 2 to 3 is [tex]x-0.301[/tex]
Calculating:
[tex]\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}[/tex]
[tex]\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}[/tex]
[tex]\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}[/tex]
[tex]x^{2}=0.67x^{2}-0.40x+0.061[/tex]
[tex]0.33x^{2}+0.40x-0.061=0[/tex]
roots = 0.14 or -1.35
Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.
The distance of Q3 relative to Q1 is 0.14 m
In what way did the cloud model resemble the Bohr model
Answer:
While Bohr's atomic model hypothesizes that electrons move in particular energy levels around the nucleus, the electron cloud model suggests that electrons move in an unpredictable pattern but are more likely to be in certain regions than others.
Explanation: