When a parallel beam of α particles with fixed kinetic energy is normally incident on a piece of gold foil,
a) If 100 α particles per minute are detected at 20°, 3.200 α particles, 9.960 α particles, 2048 α particles, 320000 α particles will be counted at 40°, 60°, 80°, and 100° respectively.
b) If the kinetic energy of the incident α particles is doubled, 50.0 alpha particles per minute will be observed at 20.
c) If the same parallel beam of alpha particles with fixed kinetic energy is normally incident on a copper foil of the same thickness, 197.4 alpha particles per minute would be detected at 20°.
In 1911, Ernest Rutherford conducted an experiment in which he bombarded a thin sheet of gold foil with alpha particles and observed their scattering pattern. This experiment provided evidence for the existence of the atomic nucleus and helped to establish the structure of the atom. In this question, we will use the principles of Rutherford scattering to determine the number of scattered alpha particles at various angles for a fixed kinetic energy and for different materials.
(a) The number of scattered alpha particles at an angle θ can be calculated using the Rutherford scattering formula:
dN/dΩ = (N1 * Z2² * e^4)/(16πε0² * E^2 * sin⁴(θ/2))
where dN/dΩ is the number of scattered alpha particles per unit solid angle, N1 is the number of incident alpha particles per unit time, Z2 is the atomic number of the target material, e is the elementary charge, ε0 is the electric constant, E is the kinetic energy of the incident alpha particles, and θ is the scattering angle.
For a fixed kinetic energy, N1 is constant, so we can compare the number of scattered alpha particles at different angles by comparing the values of sin^4(θ/2) for each angle. Using this formula, we can calculate the number of scattered alpha particles at 40°, 60°, 80°, and 100°, given that 100 alpha particles per minute are detected at 20°. The calculations are as follows:
dN/dΩ(20°) = 100 alpha particles per minute
sin^4(20°/2) = 0.03125
dN/dΩ(40°) = dN/dΩ(20°) * sin⁴(20°/2) / sin⁴(40°/2) = 100 * 0.03125 / 0.98438 = 3.200 alpha particles per minute
dN/dΩ(60°) = dN/dΩ(20°) * sin⁴(20°/2) / sin⁴(60°/2) = 100 * 0.03125 / 0.31641 = 9.960 alpha particles per minute
dN/dΩ(80°) = dN/dΩ(20°) * sin⁴(20°/2) / sin⁴(80°/2) = 100 * 0.03125 / 0.01563 = 2048 alpha particles per minute
dN/dΩ(100°) = dN/dΩ(20°) * sin⁴(20°/2) / sin⁴(100°/2) = 100 * 0.03125 / 0.00098 = 320000 alpha particles per minute
(b) If the kinetic energy of the incident alpha particles is doubled, the Rutherford scattering formula becomes:
dN/dΩ = (N1 * Z2² * e⁴)/(16πε0² * 4E² * sin⁴(θ/2))
The number of scattered alpha particles at 20° can be calculated using this formula with N1 doubled. The calculation is as follows:
dN/dΩ(20°) = (2 * 79² * (1.6022 x 10⁻¹⁹)⁴)/(16π(8.8542 x 10⁻¹²)^2 * 4 * (2E6)² * sin⁴(20°/2)) = 50.0 alpha particles per minute.
c) dN/dΩ = (N1 * Z2² * e⁴)/(16πε0² * E² * sin⁴(θ/2)) * (ρAu/ρCu)²
where ρAu is the density of gold and ρCu is the density of copper.
Since the thickness of the foil is the same, we can assume that the number of atoms per unit area is the same for both gold and copper foils. Therefore, N1 is the same for both cases.
Using the given values of ρAu = 19.3 g/cm³ and ρCu = 8.9 g/cm³, the ratio (ρAu/ρCu)²is:
(ρAu/ρCu)² = (19.3/8.9)² = 8.031
Substituting the values of N1, Z2, e, ε0, E, θ, and (ρAu/ρCu)² into the modified Rutherford scattering formula, we can calculate the number of scattered alpha particles at 20° for the copper foil:
dN/dΩ(20°) = (100 * 29² * (1.6022 x 10⁻¹⁹)⁴)/(16π(8.8542 x 10⁻¹²)² * (2E6)² * sin⁴(20°/2)) * 8.031 = 197.4 alpha particles per minute
Learn more about energy at: https://brainly.com/question/2003548
#SPJ11
What is the velocity of a 0.8kg ball that has a momentum of 3 kg*m/s?
the unit of acceleration is derived unit why
Answer:
Mass is a base unit, acceleration is a derived unit calculated as a function of length and time which are both base units. Any unit definition that is a function of more basic units is considered a derived unit.
A 1-kg book is at rest on a desk. Determine the force the desk exerts on the book
[tex]\displaystyle F_n = 9.8 \ N[/tex]
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to RightPhysics
Forces
SI Unit: Newtons N
Free Body Diagrams
Gravitational Force: [tex]\displaystyle F_g = mg[/tex]
m is mass (in kg)g is Earth's gravity (9.8 m/s²)Normal Force: [tex]\displaystyle F_n[/tex]
Newton's Law of Motions
Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motionNewton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)Newton's 3rd Law of Motion: For every action, there is an equal and opposite reactionExplanation:Step 1: Define
1 kg book at rest
Step 2: FBD
See Attachment
Draw a free body diagram to label the forces acting upon the book. We see that we would have gravitational force from Earth pointing downwards and normal force from the surface of the desk pointing upwards.
Since the book is not moving, we know that ∑F = 0 (sum of forces equal to 0).
Step 4: Find Normal Force
Define Forces [Newton's Law of Motions]: [tex]\displaystyle \sum F = 0[/tex][Newton's Law of Motions] Substitute in forces: [tex]\displaystyle F_g - F_n = 0[/tex][Newton's Law of Motions] [Addition Property of Equality] Isolate [tex]\displaystyle F_n[/tex]: [tex]\displaystyle F_g = F_n[/tex][Newton's Law of Motions] Substitute in [tex]\displaystyle F_g[/tex]: [tex]\displaystyle mg = F_n[/tex][Newton's Law of Motions] Rewrite: [tex]\displaystyle F_n = mg[/tex][Newton's Law of Motions] Substitute in variables: [tex]\displaystyle F_n = (1 \ kg)(9.8 \ \frac{m}{s^2})[/tex][Newton's Law of Motions] Multiply: [tex]\displaystyle F_n = 9.8 \ N[/tex]An object moving in the air at a constant speed is considered to be in a balanced state. Explain Why
Answer:
and there is no friction
Explanation:
An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force. An object in motion with a velocity of 2 m/s, East will (in the absence of an unbalanced force) remain in motion with a velocity of 2 m/s, East. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force. Objects resist changes in their velocity.
A hockey puck, with an initial velocity of 65 km/h [W], ricochets off the boards. After 0.76 s in contact with the boards, its final velocity is 47 km/h [E]. Determine the acceleration of the puck.
Answer:
a = 40.937 m / s²
Explanation:
For this exercise let's use the relationship between momentum and momentum variation
I = Δp
F t = m v_f - mv₀
F = m (v_f -v₀) / t
let's reduce the magnitudes to the SI system
v_f = 47 km / h (1000m / 1 km) (1h / 3600 s) = 13.056 m / s
v₀ = - 65 km / h = -18.056 m / s
the negative sign is bearing the speed is west
let's calculate
F = m (13.056 + 18.056) / 0.76
F = m 40.937
now we can use Newton's second law
F = m a
m 40.937 = m a
a = 40.937 m / s²
An object with a mass 4.0 kg has a momentum of 64 kgm/s . How fast is the object traveling ?
Answer:
The object will travel at the speed of 16 m/s.
Explanation:
Given
Mass m = 4.0 kg Momentum p = 64 kgm/sTo determine
How fast is the object traveling?
Important Tip:
The product of the mass and velocity of an object — momentum.
Using the formula
[tex]p = mv[/tex]
where
m = massv = velcityp = momentumThus, in order to determine the speed of the object, all we need to do is to substitute p = 64 and m = 4 in the formula
[tex]p = mv[/tex]
[tex]64\:=\:4\times v[/tex]
switch the equation
[tex]\:4\times \:v\:=64[/tex]
divide both sides by 4
[tex]\frac{4v}{4}=\frac{64}{4}[/tex]
simplify
[tex]v=16[/tex] m/s
Therefore, the object will travel at the speed of 16 m/s.
6th grade science pls help
Answer:
D
Explanation:
I hope you get a good grade!
What is the only part in our body we do not stretch?
Answer:
The quadriceps consists of four muscles: the skinny rectus femoris and the three huge “vasti” — vastus lateralis/intermedius/medialis. The vasti are only elongated by knee flexion, which is limited to about 120˚ when the calf hits the hamstrings. The vasti cannot be stretched strongly.
Answer:Masseter and temporalis
Why it’s unstretchable: The jaw can only open so far.
Why it’s a dang shame: Jaw tension is epidemic, and trigger points in these muscles cause a wide array of strange face and head pains, including toothaches, headaches, and earaches.
The acceleration of an object is ____________________ related to the net force exerted upon it and _____________________ related to the mass of the object.
Determine the amount of work you would need to do in order to stop a 1100kg car with 1400J of kinetic energy.
Answer:
W = -1400 J
Explanation:
Let's use the relationship between work and kinetic energy
W = ΔK
kinetic energy is
K = ½ m v²
therefore in this case
K₀ = 1400 J
Thus
W = 0 -1400
W = -1400 J
The negative sign indicates that the work is done against the energy, that is, in the opposite direction to the movement
A ray diagram shows an object placed between 2F and F of a convex lens.
The image produced is
smaller than the object.
upright.
larger than the object.
virtual
Answer:
larger than object
Explanation:
the image formed will be on the opposite side of lens and magnified
Answer:
C
Explanation:
In a bar magnet the magnetic force seems to come from each end. The ends of a bar magnet are
called the___
What one word completes the sentence?
What is a Sport car transformation
Answer:
when a car transforms...
Explanation:
Explanation:
Vivimos en un puente en lo que el diseño de coches se refiere, consecuencia de la gran disrupción de la automación que ocurriría en los próximos años.
A horse has a momentum of 25 kg*m/s and a velocity of 2.5 m/s. What is the horse mass?
==================================================
Work Shown:
We have these variables
p = momentum = 25 kg*m/sm = mass = unknownv = velocity = 2.5 meters per secondSolving for the mass gets us...
p = m*v
25 = m*2.5
25/2.5 = m
10 = m
m = 10
The mass of the horse is 10 kg.
Is it resistant to mud fire
Dirt that you can not burn typically does no longer have any materials that are flammable at decrease temperatures. Mostly due to the fact it is full of water and non-burnable minerals. But a lot of what's in grime can certainly burn if it is dried. Dirt is made of many unique things.
yes
Explanation:
since they are extremely dense they lack the ability to trap air due to its structure
An outfielder throws a baseball to home plate with a velocity of 27 m/s at 35o. It is caught by the first baseman. How long was it in the air
Answer:
Explanation:
Initial velocity of the ball = 27 m /s .
Its vertical component = 27 sin 35 = 15.48 m /s
Considering vertical displacement to the top point ,
initial velocity u = 15.48 m /s
acceleration due to gravity g = 9.8 ,
velocity at the top v = 0
v = u - gt
0 = 15.48 - 9.8 t
t = 15.48 / 9.8 = 1.579 s
So time of ascent = 1.579 s
Similarly , time of descent = 1.579
Total time of flight = 2 x 1.579
= 3.16 s
How many straight edges does a cube have
Answer:
12
Explanation:
Guys please help me I am dead confused
Answer:
A
Explanation
The diagram in A. contain a variable resistor which can be use to vary the pd from the 6v battery
Light described as what two things?
Answer:modeled as an electromagnetic wave. In this model, a changing electric field creates a changing magnetic field.
Explanation:
15. Which of the following is the correct electron configuration for neon (10Ne)?
A. 1s 2s22p
C. 184 2s2 3s2 2p?
B. 1s 2s2 3s2 2p4
D. 152 282 3s2p5
Answer:
1s2,2s2,2p6 or (He) 2s2 2p6
how was youll day still in school by the way.
My day is alright! I just finished school for the day.. and I’m excited for the weekend! How was your day?
1. How much electrical energy is needed to bring two charges Q1= 5.5 x 10-7C and Q2= 1.7 x 10-6 C from infinity to where they are 1 meter apart.
Answer:
The electrical energy needed is 8.415*10⁻³ N
Explanation:
Energy is the ability of a body to make changes or work.
By separating or joining two electric charges a distance (for example, a radius r) within their electric fields, you are taking away or giving the electric charges energetic potentials, relative to each other. By releasing these charges, they will attract or repel each other, releasing that acquired electrical energy.
In other words, electric potential energy is linked to the particular configuration of a conglomerate of point charges in a defined system.
That is, it calculates the capacity of an electrical system to carry out a task based exclusively on its position or configuration. So, it is a kind of energy stored in the system, or the amount of energy that it is capable of delivering.
Thus, a charge will exert a force on any other charge and the potential energy is the result of the set of charges.
The electric potential energy that has a point charge q in the presence of another point charge Q that are separated by a certain distance r is:
[tex]Ep=K*\frac{Q1*Q2}{r^{2} }[/tex]
where:
Ep is the electric potential energy. It is measured in Newton (N). Q1 and Q2 are the values of the two point charges. They are measured in Coulombs (C). r is the value of the distance that separates them. It is measured in meters (m). K is the constant of Coulomb's law. For vacuum its value is approximately 9*10⁹ N*m²/C²In this case:
Q1=5.5*10⁻⁷ CQ2=1.7*10⁻⁶ Cr=1 mReplacing:
[tex]Ep=9*10^{9}\frac{N*m^{2} }{C^{2} } *\frac{5.5*10^{-7}C *1.7*10^{-6}C }{(1m)^{2} }[/tex]
Solving:
Ep= 8.415*10⁻³ N
The electrical energy needed is 8.415*10⁻³ N
The electrical energy is needed to bring two charges Q1 and Q2 from infinity to where they are 1 meter apart is 8.415 x 10⁻³ N.
What is the electric potential?An electric potential can be defined as the amount of work or energy needed to move a charge from one point to another against the electrical field.
We know that in order to know electrical energy is needed to bring two charges Q1= 5.5 x 10⁻⁷C and Q2= 1.7 x 10⁻⁶ C from infinity to where they are 1 meter apart, we need to calculate electric potential energy between the two charged particles,
[tex]E_P = k\dfrac{Q_1Q_2}{d^2}[/tex]
Substitute the values,
[tex]E_P = 9 \times 10^9 \dfrac{5.5 \times 10^{-7} \times 1.7 \times 10^{-6}}{1^2}\\\\E_P = 8.415 \times 10^{-3}\rm\ N[/tex]
Hence, the electrical energy is needed to bring two charges Q1 and Q2 from infinity to where they are 1 meter apart is 8.415 x 10⁻³ N.
Learn more about Electrical Potential:
https://brainly.com/question/9383604
pls help i’ll give brainliest if you give a correct answer!!
Answer:
C. The distance traveled by an object at a certain velocity.
Explanation:
YW!
Answer:
third option is the most accurate
Two charged spheres 10 cm apart attract each other with a force of 3.0 x 10^-6 N. What electrostatic force will result if both charges are doubled and the distance remains the same?
Answer:
The electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N
Explanation:
Coulomb's law of electricity states that the magnitude of the force of attraction or repulsion between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance of of separation between them.
In formula; F = KQ₁Q₂/d²
Using the formula for electrostatic force of attraction above to determine the force of between the two charged spheres
Let the charges be Q₁ and Q₂; distance of separation be d, K is a constant
Initially, F₁ = KQ₁Q₂/d² ---- (1)
F₁ = 3.0 * 10⁻⁶ N
when the charges are doubled, Q₁ = 2Q₁; Q₂ = 2Q₂; K and d remains constant
F₂ = 2KQ₁Q₂/d² ----(2)
Dividing equation (2) by (1) to find the ratio of their forces
F₂/F₁ = (2KQ₁Q₂/d²) / KQ₁Q₂/d²
F₂/F₁ = 2
Thus, F₂ is twice F₁.
Since F₁ = 3.0 * 10⁻⁶ N; F₂ = 2 * 3.0 * 10⁻⁶ N
F₂ = 6.0 * 10⁻⁶ N
Therefore, the electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N
Electrostatic force is a force imposed by one charge on another as a result of the field.Electrostatic force will be 12.0 x 10⁻⁶ N if both charges are doubled and the distance remains the same.
What is the electrostatic force?It is a force imposed by one charge on another as a result of the field.
The electrostatic force produced by one line charge on another line charge separated by distance d is determined by the charge potency of each charge as well as the separation distance between them.
Hence the electrostatic force is given by
[tex]\rm F=\frac{Kq_1q_2}{d^2}[/tex]
The given data in the question ,
d is the distance between the charge
F₁ is the electric force for case 1= 3.0 x 10⁻⁶ N
F₂ is the electric force for case 2= ?
Conditions for case 2;
(q₁=2q₁),(q₂=2q₂),d₂=d₁
For case 1,
[tex]\rm F_1=\frac{Kq_1q_2}{d_1^2}[/tex]
For case 2,
[tex]\rm F_2=\frac{K(2q_1)(2q_2)}{d_1^2}[/tex]
[tex]\rm F_2=4\times\frac{Kq_1q_2}{d_1^2}\\\\\rm F_2=4F_1[/tex]
[tex]\rm F_2=4\times\frac{Kq_1q_2}{d_1^2}\\\\\rm F_2=4\times3.0\times 10^{-6}\\\\\rm F_2=12.0\times10^{-6}[/tex]
Hence electrostatic force will be 12.0 x 10⁻⁶ N. if both charges are doubled and the distance remains the same.
To learn more about the electrostatic force refer to the link;
https://brainly.com/question/9774180
balance the equation
P + 02 → P205
Answer:
4P + 502 → 2P205
Explanation:
Hence the above equation is balanced.
Explanation:
[tex] \underline{ \underline{ \large{ \text{Given \: chemical \: equation}}}} : [/tex]
[tex] \tt{P \: + O _{2} \: \: \: \: ➞ \: \: P_{2} \: O_{5}}[/tex]Here , Multiply P2O5 by 2 and O2 by 5 to equalize oxygen :
[tex] \tt{P+ 5O_{2} \: \: ➞\: \: 2P_{2} \:O_{5}}[/tex]
Now , To equalize P atoms , Multiply P by 4
[tex] \boxed{ \large{\tt{ 4P + 5O _{2} \: ➞ \: \: \: \: 2P_{2} \: O_{5}}}}[/tex]
Here , On the left side we have 4P and On the right side ,we have 4P. Again , On the left side , we have 10 O and On the right side , we have 10 O. And yippie , we got the balanced chemical equation !!!
Hope I helped ! ♡
Have a wonderful day / night ! ツ
▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁
If Jay pushes on a box with a force of 20 N to the right and Bradley pushes on a box with a force of 15 N to the left, what is the net force on the box?
Answer:
Net force = 5N
Explanation:
Jay = 20N to the right
Bradley = 15N to the left
To find the net force;
Since the forces are being applied to the box in opposite direction i.e acting in opposite direction, we would subtract them.
Net force = 20 - 15
Net force = 5N
Therefore, the net force on the box is 5 Newton.
A 100-newton object is lifted 100 meters in 100 seconds. What is the
power generated in this situation?
Theoretically, what would be the mass of an object accelerated
to 100% the speed of light?
I know most of the answers will say infinity, but still this needs deeper look about its physical meaning and does it consider a practical logic?
Infinity usually used for something we could not measure, however the reality may be different!
The denominator of the equation would become 0 and the mass would become infinite if the mass's velocity ever surpassed the speed of light.
Speed of light and mass:The amount of energy needed to accelerate an infinite mass would be infinite as well. The fact that light travels at the speed of c indicates that it has no rest mass.
When a mass particle approaches the speed of light, its energy grows and becomes infinite at that speed, which is why it can never be accelerated to that speed.
Experiments have confirmed this, and it has been demonstrated that nothing goes faster than the speed of light.
Find out more information about 'Speed of light'.
https://brainly.com/question/394103?referrer=searchResults
An island reports that in the year 2000 there were 240 babies born. That same year, 100 individuals died.
What is the growth rate for this island?(Show your work and indicate if it is positive or negative.)
Answer:
+ 140
Explanation:
You can show the natural growth rate by subtracting the death rate from the birth rate during one year and converting this into a percentage.
Here it would be:
240 - 100 = + 140
// if you want to convert it to percentage, you need to know the size of the population
it would be
140 / (population size) * 100 %
A concrete dam holds back a large reservoir of water
potential or kinetic energy
Answer:
The water in a reservoir behind a hydropower dam is another example of potential energy. The stored energy in the reservoir is converted into kinetic energy as the water flows down a large pipe called a penstock and spins a turbine.
Explanation:
Hope this helps :))
A concrete dam holds back a large reservoir of water potential energy, the stored potential energy is converted into the kinetic energy which is converted into mechanical energy and then further converted into electrical energy.
What is hydropower?A dam or other construction that alters the natural flow of a river or other body of water is used to generate hydropower, often known as hydroelectric power.
In order to generate energy, hydropower uses the perpetual, never-ending water cycle, which uses water as a fuel and leaves no waste products behind. Although there are many different kinds of hydropower plants, they are always propelled by the kinetic energy of water moving downstream.
Thus, a concrete dam holds back a large reservoir of water potential energy.
To learn more about hydropower here, refer to the link given below ;
https://brainly.com/question/22258411
#SPJ2