Answer:
True.
Explanation:
The velocity vector is constant in magnitude but changing in direction. Because the speed is constant for such a motion, many students have the misconception that there is no acceleration. ... But the fact is that an accelerating object is an object that is changing its velocity.
definition of matter . A object which cover the place and have mass is called matter
Answer:
you have written the definition so what are you asking
Apakah yang berlaku kepada kekuatan medan magnet jika satu lagi sel kering 1.5 v ditambahkan
Answer:
The magnetic field is doubled.
Explanation:
What happens to the strength of the magnetic field if one more 1.5 v dry cell is added?
The magnetic field due to a current carrying conductor is directly proportional to the current flowing in the wire.
If we connect one more battery of 1.5 V so the voltage is doubled and according to the Ohm's law, as the resistance of wire is constant, so the current in the wire is also doubled.
When the current doubles, the magnetic field produced by the wire is also doubled.
To overcome the problems that blur images and don't provide the best resolution from Earth, astronomers have started using flexible mirrors that change shape many times each second. This technique is called:
Answer:
adaptive optics
Explanation:
simple
Give reason:
Mass and volume are called physical quantities.
a.
b
Ionath is called
Mass and volume are called physical quantities because they can be measured by using physical devices i.e. mass is measured by using beam balance and volume is measured by using metre scale .
The most powerful empire between the 1500s and 1600s was the __________ Empire.
A.
Ottoman
B.
Mauryan
C.
Roman
D.
Persian
Answer:
A
Explanation:
Answer:
Ottoman
Explanation:
siri told me after I asked
State Newton’s second law of motion. Derive this law mathematically. A 1000 kg vehicle moving with a speed of 20m/s is brought to rest in a distance of 50 metres. Find the acceleration and calculate the force acting on the vehicle.
Answer:
-4000 N
Explanation:
newton second law F=ma
m=1000kg
vi=20m/s
vf=0
d=50 m
vf^2=vi^2+2ad
0=20^2+2a×50
100 a= -400
a= -4 m/S2
F=ma = 1000×-4 = -4000 N
A stream leaving a mountain range deposits a large part of its load in a __
Answer:
(n) alluvial fan sandbar
Explanation:
Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,2.50kN of force, the tide provides 1, point, 20, k, N,1.20kN of force from the direction 30, point, 0, degrees,30.0 ∘ more northerly than the wind. Give your answer to 2 significant figures. Remember that 'an easterly wind' means a wind coming from the East
Answer:
F = 3.6 kN, direction is 9.6º to the North - East
Explanation:
The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.
Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.
Wind
X axis
F₁ = 2.50 kN
Tide
cos 30 = F₂ₓ / F₂
sin 30 = F_{2y} / F₂
F₂ₓ = F₂ cos 30
F_{2y} = F₂ sin 30
F₂ₓ = 1.20cos 30 = 1.039 kN
F_{2y} = 1.20 sin 30 = 0.600 kN
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = 2.50 +1.039
Fₓ = 3,539 kN
F_y = F_{2y}
F_y = 0.600
to find the vector we use the Pythagorean theorem
F = [tex]\sqrt{F_x^2 +F_y^2}[/tex]
F = [tex]\sqrt{ 3.539^2 + 0.600^2 }[/tex]
F = 3,589 kN
the address is
tan θ = F_y / Fₓ
θ = tan⁻¹ [tex]\frac{F_y}{F_x}[/tex]
θ = tan⁻¹ [tex]\frac{0.6}{3.539}[/tex]0.6 / 3.539
θ = 9.6º
the resultant force to two significant figures is
F = 3.6 kN
the direction is 9.6º to the North - East
A box slides down a 28.0 degree ramp with an acceleration of 1.25 m/s2. Determine the coefficient of kinetic friction between
Answer:
[tex]\mu=0.39[/tex]
Explanation:
From the question we are told that:
Angle [tex]\theta=28[/tex]
Acceleration [tex]a=1.25m/s^2[/tex]
Generally the equation for Frictional force is mathematically given by
[tex]F=\muN[/tex]
Where
[tex]N=mgcos \theta[/tex]
[tex]N=mgcos 28[/tex]
Since
Friction force is acting against move of box
Therefore
[tex]mgsin(28) - 1.25m = \mu mgcos(28)[/tex]
[tex]\mu=\frac{gsin(28) - 1.25}{gcos(28)}[/tex]
[tex]\mu=0.39[/tex]
Sí un auto viaja a 8m/s determine Tiempo en llegar a 200km de distancia Distancia que recorre en 40 minutos
Answer:
a. Tiempo = 25000 segundos
b. Distancia = 19200 metros
Explanation:
Dados los siguientes datos;
Velocidad = 8 m/s
Distancia = 200 km a metros = 200 * 1000 = 200,000
Para encontrar el tiempo para cubrir la distancia anterior;
Tiempo = distancia/velocidad
Tiempo = 200000/8
Tiempo = 25000 segundos
b. Para encontrar la distancia recorrida en 40 minutos;
Tiempo = 40 minutos a segundos = 40 * 60 = 2400 segundos
Distancia = velocidad * tiempo
Distancia = 8 * 2400
Distancia = 19200 metros
three condensers are connected in series across a 150 volt supply, the voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10^-8 c.calculate (a) the capacitance of each condenser (b)the effective capacitance of the combination
Answer:
(a) 1.5 nF, 1.2 nF, 1 nF
(b) 0.4 nF
Explanation:
V = 150 V
V' = 40 V, V'' = 50 V, V''' = 60 V, q = 6 x 10^-8 C
(a) C' = q/V' = 6 x 10^-8 / 40 = 1.5 x 10^-9 F
C'' = q/V'' = 6 x 10^-8 / 50 = 1.2 x 10^-9 F
C''' = q/V''' = 6 x 10^-8 / 60 = 1 x 10^-9 F
(b) The effective capacitance is
[tex]\frac{1}{C}=\frac{1}{C'}+\frac{1}{C''}+\frac{1}{C'''}\\\\\frac{1}{C}=\frac{10^9}{1.5}+\frac{10^9}{1.2}+\frac{10^9}{1}\\\\C = 0.4\times 10^{-9} F[/tex]
A spring attached to a mass is at rest in the initial position (not shown). The spring is compressed in position A and is then released, as shown in position B. Which equation describes conservation of energy in position A?
Answer:
Explanation:
When the spring is compressed, it is compressed to its amplitude (whereas equilibrium is the spring's natural length with no mass attached to it and displacement is the spring's reaction to a mass hung on the end of it without any "extra" pushing or pulling on the mass). It is at the amplitude where the spring experineces max potential energy, which is choice 2, E = mph
Answer:
its c
Explanation:
...
The spin of Earth creates the Coriolis effect. This effect causes which current patterns to occur? Choose all that apply
Surface ocean water in the Northern Hemisphere moves clockwise.
Surface ocean water in the Southern Hemisphere moves clockwise.
Surface ocean water in the Northern Hemisphere moves counterclockwise.
Surface ocean water in the Southern Hemisphere moves counterclockwise.
Surface ocean water moves clockwise in both hemispheres.
Surface ocean water moves counterclockwise in both hemispheres.
Answer:
The correct answer is options (1) and (4).
Explanation:
Due to the rotation of the earth Coriolis force is generated. As the earth rotates from the west to the east, it changes the pattern of global wind and ocean water movement.
This force allows the ocean surface water to move in the clockwise direction in the northern hemisphere, and anticlockwise in the southern hemisphere.
Thus, the correct answer is options (1) and (4).
please help.. i got it wrong on my last attempt
Answer:
The answer is C.
Explanation:
person has a mass of 60kg. How much do they weigh on Earth, if the gravitational field strength is 10 N/kg?
Answer:
588n is the answer may be
The core of transformer is laminated.......
a.to reduce the loss of energy in the form of heat across the transformer
b.to reduce the voltage of AC
c.to decrease the voltage of AC
d. to change the maginetic flux
Answer:
A. to reduce the loss of energy in the form of heat across the transformer
Explanation:
The core of the transformer is laminated to minimise the energy as they interfere with the efficient transfer of energy from the primary coil to the secondary one. The eddy currents cause energy to be lost from the transformer as they heat up the core - meaning that electrical energy is being wasted as heat.
En la figura, la tensión desarrollada en cada
cuerda está dada por los dinamómetros:
T1=8 N y T2=6 N, y el ángulo de inclinación
de la primera cuerda es de 45°. Determine la
masa de la caja que debe sostener y el
ángulo con respecto a la horizontal.
Answer:
Parte A
El ángulo con respecto al horizonte, de la segunda cuerda es de aproximadamente 19,47°
Parte B
La masa de la caja que se va a sostener es de aproximadamente 0,7808 kg.
Explanation:
Parte A
Los parámetros dados son;
La tensión en la cuerda, T₁ = 8 N
La tensión en la cuerda, T₂ = 6 N
El ángulo de inclinación de la primera cuerda con la horizontal, θ₁ = 45°
Sea θ₂ el ángulo de inclinación de la segunda cuerda, obtenemos;
T₁·cos (θ₁) = T₂·cos (θ₂)
∴ 8 N × cos (45°) = 6 N × cos (θ₂)
cos (θ₂) = 8 N × cos (45°) / (6 N) = (√2)/2 × (4/3) = (2·√2)/3
θ₂ = arcos ((2·√2) / 3) ≈ 19,47°
El ángulo con respecto al horizonte, de la segunda cuerda, θ₂ ≈ 19,47°
Parte B
El peso de la caja, W = T₁·sin (θ₁) + T₂·sin (θ₂)
∴ W = 8 N × sen (45 °) + 6 N × sen (19,47 °) ≈ 7,66 N
El peso de la caja que se va a sostener, W ≈ 7,66 N
La masa de la caja que se va a sostener, m ≈ 7,66 N / (9,81 m/s²) ≈ 0,7808 kg
While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for an AP shoulder radiograph with acceptable anatomical part penetration and detector element (DEL) exposure. If the radiographer desires to increase scatter absorption using a 12:1 grid, what new exposure factors should be used to maintain the same DEL exposure
Answer:
b. 12.5 mAs, 70 kVp
Explanation:
The given parameter are;
The initial exposure factors := 10 mAs and 70 kVp
The initial Grid Ratio, G.R.₁ = 8:1
The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂ = 12:1
Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;
The GCF for G.R. 8:1 = 4
The GCF for G.R. 12:1 = 5
Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;
mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs
Therefore the new exposure factors are;
12.5 mAs, 70 kVp
It takes 130 j of work to compress a certain spring 0.10m. (a) what is the force constant of this spring? To compress the spring an additional 0.10m, does it take 130
Answer:
Explanation:
The PE equation for a mass/spring system is
[tex]PE=\frac{1}{2}k[/tex]Δx² and filling in:
[tex]130=\frac{1}{2}k(.10)^2[/tex] and
[tex]k=\frac{2(130)}{(.10)^2}[/tex] so
k = 26000 N/m
If the displacement from equilibrium changes more, the PE needed to compress it will also change.
[tex]PE=\frac{1}{2}(26000)(.20)^2[/tex] gives us that
PE = 520J
How did the English bill of rights impact the colonists views of government?
The English Bill of Rights helped to shape the colonists' views of government by promoting the ideas of individual liberty, representative democracy, and limited government.
What is English bill of rights?The English Bill of Rights is a document that was passed by the Parliament of England in 1689. It was a key moment in the development of modern democratic government, as it established certain rights and protections for English citizens, such as the right to bear arms, the right to a fair trial, and the prohibition of cruel and unusual punishment.
The English Bill of Rights also limited the power of the monarchy, requiring the king or queen to obtain the consent of Parliament before levying taxes or making other important decisions. Its influence can still be seen today in many democracies around the world.
Learn more about English bill of rights, here:
https://brainly.com/question/20986147
#SPJ7
Classes. frequency
0-20 2
20-40. 2
40-60. 3
60-80 12
80-100 18
100-120 5
120-140. 2
Find mean, median and mode
Answer:
The mean is 79.[tex]\overline {54}[/tex]
The median is 80 - 100
The mode is 80 - 100
Explanation:
The given table is presented as follows;
[tex]\begin{array}{lcrc}Classes&Mid \ point &Frequency &Frequency \times Mid \ point\\0 - 20&10& 2&20\\20-40&30&2&60\\40-60&50&3&150\\60-80&70&12&840\\80-100&90&18&1620\\100-120&110&5&550\\120-140&130&2&260\end{array}[/tex]
The mean of a class of values, [tex]\overline x[/tex] = ∑(Frequency × Midpoint)/∑(Frequency)
Therefore, we get;
[tex]\overline x[/tex] = (20+60+150+840+1620+550+260)/(2+2+3+12+18+5+2) = 79.[tex]\overline {54}[/tex]
The mean, [tex]\overline x[/tex] =79.[tex]\overline {54}[/tex]
The median class = The middle value lass = The class at the 22 nd value = 80 - 100
The median = 80 - 100
The modal class = The class with the highest frequency = 80 - 100
The mode = 80 - 100
Which nuclear reaction provides energy for growing green plants?
A. Nuclear fusion
B. Nuclear fission
C. Gamma decay
D. Positron emission
Answer:
nuclear fusion
Explain: that's just my best term.
Answer:
A
Explanation:
Nuclear Fusion
Name One formula that uses joules
Answer:
[tex]{ \bf{power \: { \tt{(watts)}} = \frac{workdone \: { \tt{(joules)}}}{time \: { \tt{(seconds)}}} \: }}[/tex]
URGENT A student runs at 4.5 m/s [27° S of W] for 3.0 minutes and then he turns and runs at 3.5 m/s [35° S of E] for 4.1 minutes. a. What was his average speed? b. What was his displacement? PLEASE SHOW ALL WORK
Answer:
Explanation:
As far as the displacement goes, we have 2 displacement vectors. If we didn't have the angles to deal with, this would be a much simpler process, but then that wouldn't be any fun at all, would it? I'll deal with the average speed first, then the displacement, which is a vector addition problem.
The average speed is found by adding together the distances the student traveled and then dividing this sum by the total time he spent traveling. If we are told that the student runs at 4.5 m/s for 3.0 minutes, we can use this to find out the distance he ran during that time interval. However, the units are not the same. We will find the distance the student traveled by convering the time to seconds.
3.0 minutes = 180 seconds, and
4.1 minutes = 246 seconds.
That means that the distance he ran in 180 seconds is found by multiplying this time be the speed at which he ran:
4.5 m/s(180 s) = 810 m and
3.5 m/s(246 s) = 860 m (rounded to follow the rules of sig dig).
This makes the speed equation look like this:
[tex]s=\frac{810+861}{180+246}=\frac{1671}{426}=3.9\frac{m}{s}[/tex] That's the average speed, which is NOT at all the same as the displacement. Displacement is where he ended up in reference to where he started. The angles play a huge part in this math (that is very involved, to say the least). We begin by restating the displacement of each "leg" of this journey.
The first leg took him 810 m at 207 degrees and
the second leg took him 860 m at 325 degrees
To find the x and y components of these 2 legs, or parts, we have to use the cos and sin formulas. We will call the first leg A and the second leg B. First the x components of both A and B:
[tex]A_x=810cos207[/tex] and
[tex]A_x=-720[/tex]
[tex]B_x=860cos325[/tex] and
[tex]B_x=704[/tex] and we add these to get the x-component of the resultant vector, C:
-720
+ 704
-10 (rounded, as needed, to the tens place).
Now for the y-components of the resultant vector:
[tex]A_y=810sin207[/tex] and
[tex]A_y=-370[/tex]
[tex]B_y=860sin325[/tex] and
[tex]B_y=-490[/tex] and we add these to get the y-component of the resultant vector, C:
-370
+ -490
-860
Since the x component is negative and so is the y, we are in QIII, so when we finally find our angle, we will have to add 180 to it.
For the magnitude of the displacement vector, in m:
[tex]C_{mag}=\sqrt{(-10)^2+(-860)^2}[/tex] which gives us
[tex]C_{mag}=860m[/tex]
Now, because displacement is vector, we also need the angle. We find that is the formula
[tex]\theta=tan^{-1}(\frac{C_y}{C_x})[/tex] and filling in:
[tex]\theta=tan^{-1}(\frac{-860}{-10})=90[/tex] (rounded correctly), and then we add 180 to give us a final direction of 270 degrees.
So the final displacement of the student is 860 m at 270 degrees
A car travelling an unbanked curve of radius 200 ft notices a truckstopped on the road ahead. The driverimmediately applies brakes causing the speed of the carto decrease at the rate of 10 ft/s2. If at that instant, the stationary truckis 100 ft ahead (the distance is measured along the path) and the car is travelling at a speed of 40ft/s, whatis the magnitude of the relative velocity ofthe truck perceived by the driver of the car (i.e. from the moving frame of referenceof the car).
Answer:
u = - 40 ft / s
Explanation:
The Galilean relation for the relative velocity is
v ’= v + u
where u is the speed between the two reference frames, v is the speed of the fixed system and v 'the speed of the mobile system.
In this case the truck has a speed with respect to the ground (fixed system) 0 m / s (it is stopped), the car has a speed with respect to the ground of v = 40 ft / s,
u = v'- v
u = 0 - 40
u = - 40 ft / s
the speed perceived by the car if the system is fixed on it is -40 ft / s
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the moon is (6.10x10^11). What is the gravitational force between the two objects?
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the moon is (6.10x10^11). What is the gravitational force between the two objects?
Answer:
Explanation:
This is a simple gravitational force problem using the equation:
[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
[tex]F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}[/tex] I'm going to do the math on the top and then on the bottom and divide at the end.
[tex]F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}[/tex] and now when I divide I will express my answer to the correct number of sig dig's:
[tex]Fg=[/tex] 6.45 × 10¹⁶ N
A4.60pf capacitor that is initially uncharged is connected in series with a 7.50kohms resistor and an emf source with E=125V and negligible internal resistance. Just after the circuit is completed, what are; a. The voltage drop across the capacitor b. The voltage drop across the resistor c. The charge on the capacitor d. The current through the resistor e.
Answer:
a) [tex]V_c=0[/tex]
b) [tex]V_R=145V[/tex]
c) [tex]Q_c=0[/tex]
d) [tex]I=\frac{1}{60}A[/tex]
Explanation:
From the question we are told that:
Capacitor [tex]C=4.60[/tex]
Resistor [tex]R=7.50[/tex]
Source emf [tex]E=125V[/tex]
a)
Generally The voltage drop across the capacitor is
V_c=0
b)
Generally the equation for Voltage drop is mathematically given by
[tex]V=IR[/tex]
[tex]V=\frac{E}{R}*R[/tex]
[tex]V_R=145V[/tex]
c
Generally The Charge across the capacitor is
[tex]Q_c=0[/tex]
d)
Generally the equation for Current is mathematically given by
[tex]I=\frac{V}{R}[/tex]
[tex]I=\frac{125}{7.5*10^3}[/tex]
[tex]I=\frac{1}{60}A[/tex]
convert the following as instructed
67 kg into gram
explain step by step
please reply quickly its urgent
Answer:
1kg = 1000g
67kg therefore would be equal to
67 x 1000 = 67,000g = 6.7 x 10⁴g.
help me do the c.....
yup they r equal as mentioned above is equal magnitude and direction! so they r totally equal.
as one vector has 2 things which is used to define it ->
Magnitudedirectionso if both r equal then both vectors r equal vectors
Example Problem
The potential energy of an object is given by U(x) = 8x2 - x4, where U is in joules and x is in
(a) Determine the force acting on this object.
(b) At what positions is this object in equilibrium?
(c) Which of these equilibrium positions are stable and which are unstable?
metres.
111 Unit 2 Concepts and Definitions Prof Mark Lester
Exam Part B Example
A neutron of mass m moving with velocity v collides head-on and elastically with a stationary nucleus of mass M.
(a) Show that the velocity of the nucleus after the collision, U, is given by
U= 2m v (m+M)
(b) Hence show that the neutron loses a fraction f of its energy where
f= 4mM (m+M)
10marks 5 marks
(c) A fast neutron enters a target of carbon nuclei which may be assumed to have masses 12 times that of the neutron. How many head-on collisions will it take
before the neutron loses 95% of its energy?
4 marks
(d) Suggest one reason why in a real reactor a neutron is likely to make more
collisions with the moderator nuclei before losing this much energy
2
1 mark
Answer:
Part A
a) F = -16x + 4, b) x = 0.25 m, c) STABLE
Explanation:
Part A
a) Potential energy and force are related
F = [tex]- \frac{dU}{dx}[/tex]- dU / dx
F = - (8 2x -4)
F = -16x + 4
b) The object is in equilibrium when the forces are zero
0 = -16x + 4
x = 4/16
x = 0.25 m
c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.
In this case there is only one equilibrium point
by changing the position a bit
x ’= x + Δx
we substitute
F ’= - 16 x’ + 4
F ’= - 16 (x + Δx) + 4
F ’= (-16x +4) - 16 Δx
at equilibrium position F = 0
F ’= 0 - 16 Δx
we can see that the body returns to the equilibrium position, therefore it is STABLE
PART B
This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved
initial instant. Before the shock
p₀ = m v
final instant. After the crash
p_f = (m + M) v_f
We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic
p₀ = pf
mv = mv ’+ M v_f
in the case of the elastic collision, the kinetic energy is conserved
K₀ = K_f
½ m v² = ½ m v’² + ½ M v_f²
we write the system of equations
mv = mv ’+ M v_f (1)
m (v² -v'²) = M v_f ²
m (v - v ’) = M v_f
m (v-v ’) (v + v’) = M v_f
v + v ’= v_f
we substitute in equation 1 and solve
v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]
v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]
the mechanical energy of the neutron is
initial
Em₀ = K = ½ m v²
final moment
Em_f = K + U = ½ m v_f ² + U
U is the energy lost in the collision
total energy is conserved
Em₀ = Em_f
½ m v² = ½ m v_f ² + U
U = ½ m (v² -v_f ²)
U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex] v)² ]
U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]
U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]
U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]
Let's do the same calculations for the nucleus
initial Em₀ = 0
final Em_f = K + U = ½ M v_f ² + U
Em₀ = Em_f
0 = K + U
U = -K
U = - ½ M v_f ²
U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²
U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]
We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.
b) the fraction of energy lost
f = U / Em₀
f = 4 m M / m + M
c) let's calculate the fraction of energy lost in a collision
m = 1.67 10⁻²⁷ kg
M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg
f = 4 1.6 20 / (1.6+ 20) 10⁻²⁷
f = 5.92 10⁻²⁷ J
the energy of a fast neutron is greater than 1 eV
Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J
Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo
#_collisions = 0.95 Eo / f
#_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷
#_collisions = 2.7 10⁷ collisions
