Internal energy is directly proportional to temperature
U1T2=U2T1U(273+410)=U2(273+202)683U=475U_2U_2=1.44U1. The sketch shows two 4 kg blocks on a horizontal surface, joined by a string S. They are attached to a 2 kg mass by another string which passes over a frictionless pulley, allowing the 2 kg mass to hang freely. Each of the blocks experiences 6 N of friction.
a) Show that the system accelerates to the right at 0.8 m/s². Calculate the tension in string S.
b) 4 kg S 4 kg 2 kg [4] [4]
Answer:
M g - T = M a acceleration of hanging block (M = 2)
2 g - T = 2 a where M = 2
T = 2 g - 2 a
T = 8 * a + 12 N acceleration of blocks on table
2 g - 2 a = 8 a + 12
10 a = 2 g - 12
a = .8 m/s^2 (assuming g = 10 m/s^2)
T = 2 g - 2 a = 2 * 10 - 2 * .8 = 20 - 1.6 = 18.4 N
Check: For blocks on table
a = (18.4 - 12) / 8 = 6.4 / 8 = .8 m/s^2
Derive the Schrödinger's Time independent wave equation using kinetic energy and momentum.
Consider,
[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =m\langle v(t)\rangle=m\int_{-\infty}^{\infty}x\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial t}\Psi (x,t)+\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial t}\bigg\}dx}[/tex]
Multiply both sides by ih and simplification will yield
[tex]{:\implies \quad \displaystyle \sf ih\langle p\rangle =m\int_{-\infty}^{\infty}x\bigg[\Psi (x,t)\bigg\{\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi^{*}(x,t)}{\partial x^2}-V(x)\Psi^{*}(x,t)\bigg\}+\Psi^{*}(x,t)\bigg\{V(x)\Psi (x,t)-\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}\bigg\}\bigg]dx}[/tex]
Some simplification, Then Integrate by parts and then knowing the fact that the wave function vanishes for [tex]{\bf x\to \pm \infty}[/tex] will yield:
[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =\dfrac{ih}{2}\int_{-\infty}^{\infty}\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial x}\Psi (x,t)-\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}\bigg\}dx}[/tex]
Integrating by parts and knowing the same fact by some simplification will yield:
[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =-ih\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}dx}[/tex]
The momentum is thus contained within the wave function, so we can then deduce that:
[tex]{:\implies \quad \sf p\rightarrow -ih\dfrac{\partial}{\partial x}}[/tex]
[tex]{:\implies \quad \sf p^{n}\rightarrow \bigg(-ih\dfrac{\partial}{\partial x}\bigg)^{n}}[/tex]
[tex]{:\implies \therefore \quad \displaystyle \sf \langle p^{2}\rangle =-h^{2}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]
Now the kinetic energy
[tex]{:\implies \quad \displaystyle \sf \langle K\rangle =\dfrac{\langle p^{2}\rangle}{2m}=\dfrac{-h^2}{2m}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]
The classical formula for the total energy
[tex]{:\implies \quad \sf \dfrac{p^2}{2m}+V(x)=E}[/tex]
Multiplying this equation by [tex]{\sf \Psi (x,t)=\psi (x)exp\bigg(\dfrac{-iEt}{h}\bigg)}[/tex] and use the above equations and simplify it we will be having
[tex]{:\implies \quad \boxed{\bf{\dfrac{-h^2}{2m}\dfrac{d^{2}\psi (x)}{dx^{2}}+V(x)\psi (x)=E\psi (x)}}}[/tex]
This is the Famous Time-Independent Schrödinger wave equation
Note:- If I write all the explanation then the Answer box willn't allow me to submit the answer
1. The object or point from which movement is determined is called a
of
Answer: reference point
Explanation: A reference point could be a put or question utilized for comparison to decide in case something is in movement.
A small object with a mass of 1 kg moves in a circular path with a constant speed of 5 m/s. What is the change in momentum during 1⁄2 of period; one period?
The change in momentum of the object during the half period is -10 kgm/s.
The change in momentum of the object during the one period is 0.
Change in momentum of the object during ¹/₂ period
The change in momentum of the object during the half period is calculated as follows;
ΔP(¹/₂) = Pf - Pi
where;
Pf is final momentumPi is the initial momentumΔP(¹/₂) = (-5 x ) - (5 x 1)
ΔP(¹/₂) = -10 kgm/s
Change in momentum of the object during 1 periodΔP(1) = Pf - Pi
ΔP(1) = (5 x 1) - (5 x 1) = 0
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Two satellites some distance apart exert a gravitational force on each other of 20 newtons (N). Each satellite has a mass of 300 kilograms (kg). If the satellites remained tho same
distance apart but each had a mass of 600 kg, what would the force of gravity between them be?
•5N
•10N
•40N
•80N
If the masses of the objects are both doubled to 600 kg each, the force between them will be 40N.
Newton's law of Universal GravitationThe Newton's law of universal gravitation states that, force of gravity between two objects is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
As such;
F = Km1m2/r^2
If the masses of the objects are doubled but the distance of separation remains the same, it follows that the force between the masses will also be doubled. Hence, if the masses of the objects are both doubled to 600 kg each, the force between them will be 40N.
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4) The specific heat of aluminum is 0.88 J/g °C. How much heat is released when a 10 g piece of aluminum foil is taken out of the oven and cools from 100° to 50°?
4.40 J
568 J
440 J
Answer:
See below
Explanation:
The units of the specific heat hint at how to solve these types of problems
.88 J/(g-C) * 10 g * (100- 50 C) = 440 J
( see how the g and the C 'cancel out' ?)
Compare the level of energy of radio and microwaves when compared to other waves on the spectrum.
A sphere of mass 15 kg possess kinetic energy of 1200 J. Calculate its
velocity
Answer:
12.65 Meters / Second (m/s)
Explanation:
the problem is really easy to solve, we begin by stating the kinetic energy formula which is:
KE = (1/2)(mass)(velocity)^2
We equate 1200 J to the formula, substitute the mass, and solve for velocity either using a CAS, app, or by hand
1200J = (0.5)(15kg)v^2
v = 12.65 m/s
Waves
Please need help fast
-10m/s² and vertical height of 0.6m.How will we use this information to find the velocity of the falling object?
1. Some ripples spread across a pond with a frequency of 5 Hz. How much would be the period of these ripples.
2. A sound wave has a time period of 0.01 seconds. What is its frequency?
3. A radio wave has a time period of 0.0000002 seconds. What is its frequency?
1)The period of these ripples will be 0.2 sec.
2)The time period of sound wave will be 100 Hz.
3)The frequency of the sound wave will be 5×10⁶ Hz.
What is frequency?Frequency is defined as the number of repetitions of a wave occurring waves in 1 second.
Frequency is given by the formula as,
Frequency of ripples (f₁)=5 Hz
The period of these ripples will be;
T₁=1/f₁
T₁=1/5
T₁=0.2 sec
The time period of sound wave (T)=0.01 seconds
The time period of sound wave;
f=1/T
f=1/0.01
f=100 Hz
Time period of A radio wave = 0.0000002 seconds
The frequency of the sound wave is;
f=1/T
f=1/0.0000002
f=5×10⁶ Hz
Hence,the period of these ripples will be 0.2 sec.
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Name two ways to get more voltage from a generator.
Answer:
1. The faster the coil rotates the faster will it cut the magnetic field and the bigger will be the output voltage.
2. You could increase the output voltage of a generator by rotating it faster, increasing the number of turns on the coil or using stronger magnets,
Explanation:
Hopefully this helps if not im terribly sorry
A 95kg fullback was running downfield at 8m/s. A defensive player (110kg) was running upfield at 6m/s straight at the fullback. What was the momentum of each player before they impacted?
A)Fullback: 760kgm/s, Defender: 660kgm/s
B)Fullback: 760kgm/s, Defender: -660kgm/s
C)Fullback: 570kgm/s, Defender: 880kgm/s
D) Fullback: 570kgm/s, Defender: -880kgm/s
A 95 kg fullback was running downfield at 8 m/s. A defensive player (110 kg) was running upfield at 6 m/s straight at the fullback. The momentum of each player before they impacted Fullback 760 kgm/s, Defender: 660 kgm/s.
What is impulse?In classical mechanics, the integral of a force, F, over the time period over which it acts, t, is referred to as the "impulse." Impulse is a vector quantity because force is one as well. When an item receives an impulse, it experiences an equivalent vector change in its linear momentum and also in the direction that results.
momentum = mass *velocity
momentum = 95*8 = 760 kg-m/sec Fullback
momentum = 110*6 = 660 kg-m/sec Defender
A 95 kg fullback was running downfield at 8 m/s. A defensive player (110 kg) was running upfield at 6 m/s straight at the fullback. The momentum of each player before they impacted Fullback 760 kgm/s, Defender: 660 kgm/s.
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An object of mass 20 kg is raised vertically through a distance of 8 m above ground level. If g = 10 m/s2, what is the gravitational potential energy of this object at this position?
Answer:
1600 joules
Explanation:
Potential Energy = m * g * h
= 20 kg * 10 m/s^2 * 8 m = 1600 J
A force of 400N acts on a body for time 10ms impulse will be
Answer:
4000Ns
Explanation:
I=F×T=400N×10ms
As a pan is sitting on a stove it begins to heat up. Eventually it may get hot enough that it glows red, meaning it is radiating electromagnetic waves within the visible portion of the spectrum. What does this situation show us about electromagnetic radiation?
Electromagnetic radiation can only be emitted in the form of visible light.
The hotter an object is, the faster the electromagnetic radiation will travel.
The electromagnetic radiation emitted by an object is not affected by its temperature.
The hotter an object is, the greater the energy of the electromagnetic radiation in emits.
The magnetic field lines of a bar magnet spread out from the
A. north end to the south end.
B. south end to the north end.
C. edges to the center.
D. center to the edges.
Answer:
I think the letter B. But I'm not sure pero I try mo Lang po baka Yan po ang tamang sagot kac Yan po Yung nilagay na sagot saakin e Kaya try mo Lang po
The diagram shows two liquid-in-glass thermometers.
Which has the greatest range? Explain how can you tell?
(Image attached)
Answer:
thermometer A
Explanation:
The range of a thermometer is the difference between the minimum and maximum temperatures that the thermometer can read.
∴ Range of thermometer A = 110° - (-10)° = 120°
Range of thermometer B = 50° - (-10)° = 60°
∴ Range of thermometer A is greatest.
Hope this helps!
An object weighing 75 N is dropped from the top of a building and falls a distance of 28 m to the ground. How much work does gravity do on the object from the time it is dropped to the time it hits the ground?
A rifle recoils while firing a bullet. The speed of the rifle's recoil is small because the
A rifle recoils while firing a bullet. The speed of the rifle's recoil is small because the rifle has more mass than the bullet.
[tex]p_{1} =-p_2}[/tex]
[tex]m_{1} V_{1} =-M_{2} v_{2}[/tex]
Answer:
Explanation:
A rifle recoils while firing a bullet. The speed of the rifle's recoil is small and typically fast because the rifle has more mass than the bullet.
What is specific heat capacity??
Specific heat capacity is the unit mass of a substance is defined as the amount of heat absorbed (kg) (J), when its temperature rises 1 K (or 1 ° C), and its units J / ( kg K) or J / happen. (Kg ° C).Hope it helps u
Stay safe, stay healthy and blessed.Have a good day !Thank you
#Jennifer
Answer:
The specific heat capacity is defined as the quantity of heat (J) absorbed per unit mass (kg) of the material when its temperature increases 1 K (or 1 °C), and its units are J/(kg K) or J/(kg °C).
light frequency of 3.6 x 10^14 hz what is the wavelength
8.3×10 to the power minus 7
Explanation:
lambda equals velocity over frequency.
How much work would a child do while pulling 12kg wagon distance of 4.3m with a 22N force?
[tex]\huge{\color{red}{\fbox{\textsf{\textbf{Answer}}}}} [/tex]
94.6J
Explanation:
Force (F) = 22N
mass (m) = 12kg
distance (S) = 4.3m
While
[tex] \sf Work \: done = Force \times distance[/tex]
This means
W = FS
while mass has no role in equation
Now substituting the required values
[tex] \sf W = 4.3 \times 22 \\ \\ \sf W = 94.6J[/tex]
86. What is the average velocity if the initial velocity is at rest and the final velocity of 16 m/s? a) 16 m/s b) 12 m/s c) 8 m/s d) 4 m/s
Answer:
8 m/s
Explanation:
(16-0)/2 = 8 m/s
what is the main function of a diode
Answer:
acts as a one-way switch for current
!!!
MULTIPLE CHOICE QUESTION
If a simple machine has a MA of 5 and you lift an object UP 2 m, how far must you push the
object along the ramp's surface?
0.5 m
10 m
0.2 m
1 m
Answer:
0.5
Explanation:
I think it is this beacause if u have the answer you have that
3. A bottle of vitamin C contains 100 tablets and weighs 80 g. If the
bottle alone weighs 23 g. (a) What is the mass of one tablet? (b) What is
the mass of 36 tablets? (c) What is the weight of 50 tablets?
Answer:
a
0.57g
b
20.52
c
28.5
Explanation:
a
The bottle weighs 80g with tablets
If the bottle alone weighs 23g, the tablets weigh 57g
100 tablets weigh 57g, 1 tablet weighs 0.57g
b
0.57g is one tablet, so to achieve 36 tablets we must multiply by 36
0.57 multiplied by 36 is 20.52
c
To find 50 tablets we can use the same method we used before or a slightly faster method
100 tablets is 57g, so all we have to do is halve to find 50
57 divided by 2 is 28.5
. True or False: A bystander intervention is the term used to define when an individual actively moves in and changes the outcome of a situation.
Answer:
true
Explanation:
How to calculate magnitude of tension if given an angle of 10°
Answer:
Mag isip ka po wag Kang tanong ng tanong kac kami po Yung nahihirapan mag hanap ng answer
A body of mass 200gram is executing SHM with amplitude of 20mm. The magnitude of maximum force which acts upon it is 0.6N. Calculate its maximum velocity?
Given ,
Mass of the body (m) = 200g
Amplitude (A) = 20mm
Maximum force (F) = 0.6 N
Maximum velocity (vmax) =?
We know that
Fmax = mv2max/A
or , 0.6 = 200×10-3xv2max/20×10-3
or , 0.6×20/200= v2max
or , vmax = 0.24 m/s
∴ Maximum velocity = 0.24 m/s2
The value of maximum velocity will be 0.3464 m/s². Energy or work is equal to the product of force and displacement.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
Mass of the body is,(m= 200g)
Amplitude is,(A =20 mm)
Maximum force is,F= 0.6 N
To find;
Maximum velocity
Energy or work is equal to the product of force and displacement.
[tex]\rm KE = \frac{1}{2} mV_{max}^2 \\\\ \rm F_{max} \times A = \frac{1}{2} mV_{max}^2 \\\\ F_{max}= \frac{1}{2}\frac{ mV_{max}^2}{A} \\\\ 0.6= \frac{1}{2}\times \frac{ 0.2 \times V_{max}^2}{20 \times 10^{-3}} \\\\ V_{max}=0.3464 \ m/sec^2[/tex]
Hence,the value of maximum velocity will be 0.3464 m/s².
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