The closest object that can be photographed is 81.63mm and the magnification of this closest object is -0.404.
The focal length of a lens is determined when the lens is focused at infinity. It is obtained from the reciprocal of objects' distance and image distance. Magnification is the enlarged image that is formed over the object size.
From the given,
focal length (f) = 23.5mm
object's distance (u) = 33mm
imagen distance(v) =?
Focal length, (1/f) = 1/u + 1/v
1/v = 1/f - 1/u
=1/23.5 - 1/33
1/v = 12.2mm
v = 1/12.2 mm
= 81.96mm
Thud, the image distance is v= 81.96mm.
Magnification (M) = -v/u
M = -33 / 81.96
= - 0.402.
Thus, the magnification is -0.402.
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A box with a mass of 25 kg rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.20. What horizontal force must be applied to the box for it to start s
To start the box sliding along the surface in the positive x direction, a horizontal force greater than 49 N in the positive x direction must be applied.
The maximum static friction force can be calculated using the equation:
f_static_max = μ_static * N
where μ_static is the coefficient of static friction and N is the normal force acting on the box. In this case, since the box is on a horizontal surface, the normal force is equal to the weight of the box:
N = m * g
Substituting the given values:
N = 25 kg * 9.8 m/s² = 245 N
Now, we can determine the maximum static friction force:
f_static_max = 0.20 * 245 N = 49 N
This is the maximum force that can be exerted before the box starts sliding. Therefore, to overcome the static friction and initiate sliding in the positive x direction, a horizontal force greater than 49 N in the positive x direction must be applied. The exact value of the force will depend on the magnitude of the static friction and the force applied.
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Complete Question:
A box with a mass of 25 kg rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.20. What horizontal force must be applied to the box for it to start sliding along the surface in the positive x direction? Use g = 9.8 m/s². O A horizontal force greater than 49 N in the positive x direction. O A horizontal force equal to 49 N in the positive x direction. O A horizontal force less than 49 N in the positive x direction. O A horizontal force that is either equal to or greater than 49 N in the positive x direction. O None of the other answers
Show Attempt History Current Attempt in Progress A proton initially has = (18.0)i + (-490) + (-18.0) and then 5.20 s later has = (7.50)i + (-4.90)j + (13.0) (in meters per second). (a) For that 5.20 s, what is the proton's average acceleration av in unit vector notation, (b) in magnitude, and (c) the angle between ag and the positive direction of the xaxis? (a) Number Units (b) Number Units (c) Number Units eTextbook and Media,
(a) The proton's average acceleration av in unit vector notation is (-2.50)i + (197)j + (6.70)k m/s^2.
(b) The magnitude of the proton's average acceleration av is 198 m/s^2.
(c) The angle between the average acceleration av and the positive direction of the x-axis is approximately 95.4 degrees.
Explanation to the above given short answers are written below,
(a) To find the average acceleration av, we need to calculate the change in velocity and divide it by the time interval. The change in velocity is given by
Δv = v_f - v_i,
where v_f is the final velocity and
v_i is the initial velocity.
Subtracting the initial velocity from the final velocity, we get
Δv = (7.50 - 18.0)i + (-4.90 - (-490))j + (13.0 - (-18.0))k = (-10.5)i + (485.1)j + (31.0)k.
Dividing Δv by the time interval of 5.20 s, we get the average acceleration av = (-2.50)i + (197)j + (6.70)k m/s^2.
(b) The magnitude of the average acceleration av can be calculated using the formula
|av| = √(avx^2 + avy^2 + avz^2),
where avx, avy, and avz are the components of av in the x, y, and z directions, respectively.
Substituting the values, we get |av| = √((-2.50)^2 + (197)^2 + (6.70)^2) = 198 m/s^2.
(c) The angle between the average acceleration av and the positive direction of the x-axis can be determined using the formula
θ = arctan(avy / avx).
Substituting the values, we get θ = arctan(197 / (-2.50)) ≈ 95.4 degrees.
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The plates have (20%) Problem 3: Two metal plates form a capacitor. Both plates have the dimensions L a distance between them of d 0.1 m, and are parallel to each other. 0.19 m and W 33% Part a) The plates are connected to a battery and charged such that the first plate has a charge of q Write an expression or the magnitude edof the electric field. E, halfway between the plates. ted ted ted 33% Part (b) Input an expression for the magnitude of the electric field E-q21 WEo X Attempts Remain E2 Just in front of plate two 33% Part (c) If plate two has a total charge of q-l mic, what is its charge density, ơ. n Cim2? Grade Summary ơ-1-0.023 Potential 96% cos) cotan)asin acos(O atan acotan sinh cosh)tan cotanh) . Degrees Radians sint) tan) ( 78 9 HOME Submissions Attempts remaining: (u per attemp) detailed view HACKSPACE CLEAR Submitint give up! deduction per hint.
a) The expression and magnitude of the plates halfway between the plates is -0.594 × 10⁶ V/m. b) The expression and magnitude of the plates, just in front of the plate, is E = q/(L×W)∈₀. c) the charge density is
-0.052×10⁻⁶ C/m².
Given information,
Distance between the plates, d = 0.1 m
Area, L×W = 0.19 m
Q = -1μC
a) The expression for the electric field,
E = q/(L×W)∈₀
E = -1×10⁻⁶/(0.19)8.85× 10⁻¹²
E = -0.594 × 10⁶ V/m
Hence, the electric field is -0.594 × 10⁶ V/m.
b) The expression for the magnitude of the electric field, in front of the plates,
E = q/(L×W)∈₀
Hence, the expression for the magnitude of the electric field, in front of the plates is E = q/(L×W)∈₀.
c) The charge density σ,
σ = Q/A
σ = -1×10⁻⁶/0.19
σ = -0.052×10⁻⁶ C/m²
Hence, the charge density is -0.052×10⁻⁶ C/m².
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A billiard ball of mass 0.28 kg hits a second, identical ball at a speed of 5.8 m/s and comes to rest as the second ball flies off. The collision takes 250 μs.
A.) What is the average force on the first ball?
B.) What is the average force on the second ball?
The average force on the first ball is 0 N. The average force on the second ball is 0 N.
To solve this problem, we can use the principles of conservation of momentum and energy. Let's start by calculating the velocity of the second ball after the collision using the conservation of momentum:
Initial momentum = Final momentum
(mass_1 * velocity_1) + (mass_2 * velocity_2) = 0
(0.28 kg * 5.8 m/s) + (0.28 kg * velocity_2) = 0
velocity_2 = -(0.28 kg * 5.8 m/s) / 0.28 kg
velocity_2 = -5.8 m/s. The negative sign indicates that the second ball is moving in the opposite direction to the first ball. Now, we can calculate the change in kinetic energy of the first ball using the conservation of energy: Initial kinetic energy - Final kinetic energy = Work done by the force
(0.5 * mass_1 * velocity_1^2) - 0 = Average force * distance.
0.5 * 0.28 kg * (5.8 m/s)^2 = Average force * 0.
Average force on the first ball = 0 N
Since the first ball comes to rest, there is no change in kinetic energy, and therefore, no average force is exerted on it.
Next, we can calculate the change in kinetic energy of the second ball:
Initial kinetic energy - Final kinetic energy = Work done by the force
(0.5 * mass_2 * velocity_2^2) - 0 = Average force * distance
0.5 * 0.28 kg * (-5.8 m/s)^2 = Average force * 0
Average force on the second ball = 0 N.
Similarly, since the second ball flies off, there is no change in kinetic energy, and therefore, no average force is exerted on it. In conclusion:
A) The average force on the first ball is 0 N.
B) The average force on the second ball is 0 N.
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A(n) asymmetric encryption algorithm requires the use of a secret key known to both the sender and receiver.
True/False
Statement : A(n) asymmetric encryption algorithm requires the use of a secret key known to both the sender and receiver, is False.
In asymmetric encryption, also known as public-key encryption, there are two different keys: a public key and a private key. The public key is available to anyone and is used for encryption, while the private key is kept secret and is used for decryption. The sender uses the recipient's public key to encrypt the message, and the recipient uses their private key to decrypt it.
Asymmetric encryption does not require the use of a shared secret key between the sender and receiver. It relies on the use of different key pairs, where the public key can be freely shared while the private key remains confidential. This property makes asymmetric encryption more secure and suitable for various applications such as secure communication and digital signatures.
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A diffraction grating with 750 slits/mm is illuminated by light that gives a first-order diffraction angle of 34∘ . What is the wavelength of the light?
When a diffraction grating having a specified number of slits per unit length is illuminated by a beam of light, a pattern of bright spots or dark lines is produced on a screen placed perpendicular to the beam. Therefore, the wavelength of the light diffracted by the grating is 0.00072516 mm.
A pattern of this kind is called a diffraction pattern. A diffraction grating is a device that divides light into its component colors and produces diffraction patterns. It is used for analyzing light and determining the wavelengths of the different colors that make up the light.
The equation used to find the wavelength of light diffracted by a grating is
`d*sin(theta) = n*lambda`.
Here, d is the distance between two successive slits on the grating, theta is the angle of diffraction, n is the order of the diffraction, and lambda is the wavelength of the light. To determine the wavelength of the light in this case, we will use the given data and the above equation. The first-order diffraction angle is 34° and the diffraction grating has 750 slits/mm. Therefore, the distance between two successive slits on the grating is d = 1/750 mm = 0.001333 mm. The order of diffraction is 1.Using the above equation, we have`0.001333*sin(34) = 1*lambda`
Simplifying, we get `lambda = 0.00072516 mm`
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i
need the answer to the upper control limit and lower control limit
for the r-chart. i know the x-chart answers are correct
Ross Hopkins is attempting to monitor a filling process that has an overall average of 725 mL. The average range R is 4 mL. For a sample size of 10, the control limits for 3-sigma x chart are: Upper C
The control limits for 3-sigma x chart are 718.5 mL and 731.5 mL.
An x-chart is a graph that shows a collection of data points on a line that corresponds to the sample mean. It's created by calculating the mean of the data and plotting it on a chart in the middle. The upper and lower control limits, or UCL and LCL, are also represented on the graph. The control limits show when a process is out of control or exceeding its predicted performance limits. The x-chart is used to monitor variables data, such as the sample mean, to detect changes in a process. The average range R is a measure of process variability. The average range R is a measure of process variability. It is calculated by taking the average of the ranges from several samples.
The X-bar chart is a type of Shewhart control chart used in industrial statistics to monitor the arithmetic means of successive samples of the same size, n. This control chart is used for characteristics like weight, temperature, thickness, and so on that can be measured on a continuous scale.
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Two external forces are applied to a particle: F1→=11 N i^+-5 N
j^ and F2→=18 N i^+-2.5 N j^.
A) Find the force F3→ that will keep the particle in
equilibrium.
Enter the x and y components separ
The force F3→ that will keep the particle in equilibrium is: F3→ = -29 N i^ + 7.5 N j^.
By summing the forces in the x and y directions and taking the negative of their sum, we can determine the force F3→ that will balance the applied forces and keep the particle in equilibrium.
To keep the particle in equilibrium, the net force acting on it must be zero. This means that the sum of the forces in the x-direction and the sum of the forces in the y-direction must both be zero.
F1→ = 11 N i^ - 5 N j^
F2→ = 18 N i^ - 2.5 N j^
To find the force F3→ that will keep the particle in equilibrium, we need to find the negative of the vector sum of F1→ and F2→.
Summing the forces in the x-direction:
F1x = 11 N
F2x = 18 N
F3x = -(F1x + F2x) = -(11 N + 18 N) = -29 N
Summing the forces in the y-direction:
F1y = -5 N
F2y = -2.5 N
F3y = -(F1y + F2y) = -(-5 N + (-2.5 N)) = 7.5 N
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Which of the following is NOT an NGO? a) CARE b) Red Cross c) UNICEF d) World Vision e) Oxfam
Option c) UNICEF is not an NGO, while options a) CARE, b) Red Cross, d) World Vision, and e) Oxfam are all NGOs.
Which of the following is NOT an NGO?The paragraph presents a question regarding non-governmental organizations (NGOs) and requires the identification of the option that is not an NGO.
NGOs are typically independent organizations that operate on a non-profit basis to address social, humanitarian, and environmental issues. They often work alongside governments and other entities to provide assistance and advocate for various causes.
Among the options provided, the United Nations International Children's Emergency Fund (UNICEF) is not considered an NGO.
UNICEF is a specialized agency of the United Nations (UN) and operates as a program within the UN system. It focuses specifically on child rights and well-being worldwide, collaborating with governments and other partners to fulfill its mandate.
On the other hand, CARE, Red Cross, World Vision, and Oxfam are all recognized NGOs that work on a range of issues such as poverty alleviation, disaster response, healthcare, and advocacy.
Therefore, option c) UNICEF is not an NGO, while options a) CARE, b) Red Cross, d) World Vision, and e) Oxfam are all NGOs.
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One carat is equivalent to a mass of 0.200 g. Use the fact that 1 kg (1000 g) has a weight of 2.205 lb under certain conditions, and determine the weight of a 1876 carat diamond in pounds (lb). Number
Diamonds are evaluated based on their carat weight. The weight of 1876 carat diamond in pounds will be approximately 0.826 pounds If 1 kg (1000 g) has a weight of 2.205 lb
Carat weight, on the other hand, refers to the mass of a diamond. A carat is the unit of weight used to weigh a diamond. Carat weight is a significant consideration when selecting a diamond. One carat is equivalent to a mass of 0.200 g.
Therefore, 1876 carats would weigh:1876 carats × 0.200 g/carats = 375.2 gNow we need to convert the weight from grams to pounds. 1 kg (1000 g) has a weight of 2.205 lb under certain conditions. Therefore,375.2 g × (1 kg/1000 g) × (2.205 lb/1 kg) = 0.826 lb A 1876-carat diamond would weigh approximately 0.826 pounds (lb).It is crucial to realize that carat weight is not the same as size.
Carat weight merely refers to the mass of a diamond, while size refers to the dimensions of the diamond when viewed from above. A 1-carat diamond, for example, may appear large or tiny depending on how it is cut. As a result, carat weight should not be the sole factor considered when selecting a diamond.
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determine the average emissivity of the filament at (a) 1500 k and (b) 2500 k. also, determine the absorptivity and reflectivity of the filament at both temperatures.
At 1500 K, the average emissivity of the tungsten filament is 0.325, the absorptivity is 0.325, and the reflectivity is 0.675. At 2500 K, the average emissivity of the tungsten filament is 0.325, the absorptivity is 0.325, and the reflectivity is 0.675.
At 1500 K and 2500 K, we can calculate the tungsten filament's average emissivity, absorptivity, and reflectivity.
Emissivity is 0.5 for photons with wavelengths < 1 μm. Emissivity is 0.15 for radiation > 1 μm.
(a) 1500 K
We must compute average emissivity (_avg), absorptivity (α), and reflectivity () at this temperature.
The blackbody's spectral range at 1500 K determines the average emissivity. Emissivity is 0.5 for < 1 μm and 0.15 for > 1 μm.
Stefan-Boltzmann law calculates average emissivity:
(A1 + A2)/(A1 + A2) = _avg.
Where: 0.5 (emissivity for < 1 μm) 0.15 (emissivity for > 1 μm).
A1 and A2 are spectral range regions.
Calculate the average emissivity assuming equal regions for both spectral bands (A1 = A2 = 0.5):
ε_avg = (0.5 * 0.5 + 0.15 * 0.5) / (0.5 + 0.5) = 0.325
For opaque materials, absorptivity (α) equals emissivity. Thus, α = 0.325_avg.
Reflectivity is the counterpart of absorptivity: - α = 1 - 0.325 = 0.675.
Thus, at 1500 K, the average tungsten filament emissivity, absorptivity, and reflectivity are 0.325, 0.325, and 0.675, respectively.
2500 K
We can determine 2500 K average emissivity, absorptivity, and reflectivity using the same method.
(0.5 * 0.5 + 0.15 * 0.5) / (0.5 + 0.5) = 0.325.
The average tungsten filament emissivity at 2500 K is 0.325.
Since absorptivity equals emissivity, α = _avg = 0.325.
Reflectivity is the counterpart of absorptivity: - α = 1 - 0.325 = 0.675.
Thus, at 2500 K, the average tungsten filament emissivity, absorptivity, and reflectivity are 0.325, 0.325, and 0.675, respectively.
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Find the rest energy, in terajoules, of a 17.1 g piece of chocolate. 1 TJ is equal to 1012 J .
rest energy:
TJ
The rest energy of a 17.1 g piece of chocolate is 485.3 terajoules.
According to the formula E = mc², the energy (E) of an object is equal to its mass (m) multiplied by the speed of light (c) squared. The rest energy (E₀) of an object is its energy at rest. The rest energy of a 17.1 g piece of chocolate can be found as follows:
$$E₀ = mc²$$
Where m = 17.1 g = 0.0171 kg and c = speed of light = 2.998 × 10⁸ m/s.
Plugging in these values, we get:
$$E₀ = (0.0171 kg) × (2.998 × 10⁸ m/s)² = 4.853 × 10¹⁴ J$$
To convert joules to terajoules, we divide by 10¹²:
$$E₀ = \frac{4.853 × 10¹⁴ J}{10¹² J/TJ} = 485.3 TJ
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wo asteroids are flying through space towards one another.Comet A has a mass of 147kg and is moving at 80m/s [R]. Comet B is moving at 29m/s [L] and has a mass of 147kg. a. Calculate the total kinetic energy and momentum of the system just before the two asteroids collide.4 Marks,C:1 b. The two asteroids collide head-on in a perfectly elastic collision.Show the steps that you would follow in order to calculate/determine the velocity of each(3 Marks,C:1
(a) The total kinetic energy and momentum of the system before collision is 532,213.5 J and 16,023 kgm/s respectively.
(b) The final velocity of Comet A after the collision is 0 m/s and the final velocity of Comet B is 51 m/s.
What is the total momentum and kinetic energy of the asteroids?(a) The total kinetic energy and momentum of the system just before the two asteroids collide is calculated by applying the following formula.
Momentum of the system;
P = (147 kg x 80 m/s) + ( 147 kg x 29 m/s)
P = 16,023 kgm/s
Kinetic energy of the system;
K.E = ¹/₂ x 147 x 80² + ¹/₂ x 147 x 29²
K.E = 532,213.5 J
(b) The velocity of the each asteroid after the perfectly elastic collision is calculated by applying the principle of conservation of linear momentum as follows;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
m₁ is the mass of Comet Am₂ is the mass of Comet Bu₁ is the initial velocity of Comet Au₂ is the initial velocity of Comet Bv₁ is the final velocity of Comet Av₂ is the final velocity of Comet B147 x 80 - 147 x 29 = 147v₁ + 147v₂
7497 = 147(v₁ + v₂)
v₁ + v₂ = 7497 / 147
v₁ + v₂ = 51 -------- (1)
Since the collision of the system occurred in one direction, our second equation is;
u₁ + v₁ = u₂ + v₂
80 + v₁ = 29 + v₂
v₁ = v₂ - 51 --------- (2)
Substitute (2) into (1);
v₁ + v₂ = 51
v₂ - 51 + v₂ = 51
2v₂ = 51 + 51
2v₂ = 102
v₂ = 102/2
v₂ = 51 m/s
The value of v₁ becomes;
v₁ = v₂ - 51
v₁ = 51 - 51
v₁ = 0 m/s
Thus, the final velocity of Comet A is 0 m/s and the final velocity of Comet B is 51 m/s.
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for the following exothermic reaction at equilibrium: h2o (g) co (g) co2(g) h2(g) decide if each of the following changes will increase the value of k (t = temperature)
For the given exothermic reaction at equilibrium:H2O(g) + CO(g) ⇌ CO2(g) + H2(g)Changes in pressure, temperature, or concentration may shift the equilibrium position, but they do not affect the value of Kc, which is constant for a given reaction at a given temperature. Hence, Kc is independent of any changes in the concentrations of reactants and products, as well as changes in the reaction conditions, as long as the temperature remains constant.To assess the effect of each change on the equilibrium constant, we must use Le Chatelier's principle to predict which direction the reaction will proceed to reestablish equilibrium. The shift in the equilibrium can cause Kc to vary when the system comes to equilibrium at the new conditions.A change in pressure will influence the equilibrium position of a gaseous reaction since gases are extremely responsive to pressure. If the pressure is increased on one side of an equilibrium reaction, the reaction will shift to the opposite side of the equation to balance the pressure. The equilibrium constant (Kc) will not change, but the pressure will influence the mole fractions of reactants and products, which will have an impact on the direction of the equilibrium shift and the rate at which it occurs. Increasing the pressure by decreasing the volume of the container in which the equilibrium reaction is occurring will result in a shift towards the side of the equation with fewer gas molecules, and the system will attempt to balance the pressure. Therefore, the reaction will shift to the left, resulting in a decrease in Kc. Since the reverse reaction, which is exothermic, is favored at lower temperatures, an increase in the value of Kc is not expected as the temperature is lowered. This means that the first option will not result in an increase in Kc. If the volume is increased, the reaction will shift towards the side with more gas molecules to compensate, resulting in an increase in Kc. This means that the second option will lead to an increase in Kc.
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Exothermic reactions at equilibrium: In an exothermic reaction, the energy is released to the surrounding as heat. An exothermic reaction always has a negative sign for ΔH. An exothermic reaction at equilibrium means that the reactants and products are still reacting, but at the same rate. The reaction quotient, Qc, is equal to the equilibrium constant, Kc. The given exothermic reaction is: H2O (g) + CO (g) ⇌ CO2(g) + H2(g)The balanced equation is as follows: H2O(g) + CO(g) ⇌ CO2(g) + H2(g)Decide if each of the following changes will increase the value of K (T = temperature): Increasing the temperature The given reaction is exothermic.
An increase in temperature will favor the backward reaction and oppose the forward reaction to attain equilibrium. According to Le Chatelier’s principle, if stress is applied to an equilibrium system, it will react to counteract the effect of that stress. Hence, an increase in temperature will cause the equilibrium to shift towards the reactants, as it is an endothermic process. Therefore, the value of Kc will decrease. Decreasing the pressure CO and H2 are gaseous reactants, whereas CO2 and H2O are gaseous products. A decrease in pressure will favor the side of the reaction with more number of gaseous molecules to oppose the change. Therefore, the equilibrium will shift towards the reactants to balance the pressure. Hence, the value of Kc will increase. Adding a catalyst A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative pathway for the reaction with a lower activation energy. A catalyst does not affect the equilibrium position of the reaction, but it helps in achieving the equilibrium state at a faster rate. Hence, adding a catalyst will not affect the value of Kc, as it is independent of the rate of the reaction. The following changes will increase the value of K (T = temperature): Decreasing the temperature Increasing the pressure Therefore, the decrease in temperature and increase in pressure will increase the value of Kc.
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determine the amplitude a and the phase angle γ (in radians), and express the displacement in the form x(t)=acos(ωt−γ), with x in meters.
The displacement function is x(t) = 0.4 cos(3πt - 0.93) m, expressed in the given form. Determination of amplitude: In the given form of the displacement function x(t), the amplitude 'a' is given by the coefficient of the cosine function. Therefore, a = 0.4 m.
Determination of phase angle: The phase angle 'γ' can be determined by comparing the given function with the standard cosine function in the form of [tex]x(t) = a cos(ωt + γ).[/tex]
Here, we need to note that in the given function, the argument of the cosine function is (ωt - γ).
Therefore, [tex]γ = (ωt - arc cos (x/a))[/tex]
We know that [tex]cos(γ) = x/a[/tex]
∴ arc cos(x/a)
= γ= arc cos(0.4/0.6)
= 0.93 rad (approx)
Hence, the phase angle is γ = 0.93 rad.
Expressing displacement in the given form: Given that the displacement function is
x(t) = 0.4 cos(3πt - 0.93)
The angular frequency is ω = 3π rad/s and the phase angle is γ = 0.93 rad. Thus, the displacement function is x(t) = 0.4 cos(3πt - 0.93) m, expressed in the given form.
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b) [3 marks] Two 10 nC charges are located at x= - 4 cm and x= 4.0 cm. (i) Calculate the electric potential, V at point P, x=0 cm. Calculate the work required to bring a 20 nC charge from infinity to
The electric potential at point P (x=0 cm) due to two 10 nC charges located at x= - 4 cm and x= 4.0 cm is 4.5 × 10⁵ volts. The work required to bring a 20 nC charge from infinity to point P is 9 × 10⁻⁴joules.
Calculation of electric potential at point P (x=0 cm):
Charge 1: q1 = 10 nC
Charge 2: q2 = 10 nC
Distance from Charge 1 to point P: r1 = 4 cm
P: r1 = 0.04 m
Distance from Charge 2 to point P: r2 = 4 cm
P: r2 = 0.04 m
Electric potential (V) at a point due to a point charge is given by the equation:
V = k * q / r
where:
k is the electrostatic constant (k = 9 × 10⁹ N m²/C²)
q is the charge
r is the distance from the charge to the point
Let's calculate the electric potential at point P due to each charge:
For Charge 1:
V1 = k * q1 / r1
Substituting the values:
V1 = (9 × 10⁹ N m²/C²) * (10 × 10⁻⁹ C) / (0.04 m)
V1 = 2.25 × 10⁵ V
For Charge 2:
V2 = k * q2 / r2
Substituting the values:
V2 = (9 × 110⁹ N m²/C²) * (10 × 10⁻⁹ C) / (0.04 m)
V2 = 2.25 × 10⁵ V
Since the electric potentials are scalar quantities, the electric potential at point P due to both charges is the algebraic sum of the potentials due to each charge:
V = V1 + V2
V = (2.25 × 10⁵ V) + (2.25 × 10⁵ V)
V = 4.5 × 10⁵ V
Therefore, the electric potential at point P (x=0 cm) is 4.5 × 10⁵volts.
Calculation of work required to bring a 20 nC charge from infinity to point P:
To calculate the work required, we need to consider the change in potential energy of the 20 nC charge as it moves from infinity to point P.
The work done (W) is given by the equation:
W = ΔPE
W = q * ΔV
where:
ΔPE is the change in potential energy
q is the charge
ΔV is the change in electric potential
As the charge moves from infinity to point P, the change in potential energy is given by:
ΔPE = q * (V - 0)
where V is the electric potential at point P.
Substituting the values:
ΔPE = (20 × 10⁻⁹) C) * (4.5 × 10⁵ V - 0 V)
ΔPE = 9 × 10⁻⁴ J
Therefore, the work required to bring a 20 nC charge from infinity to point P is 9 × 10⁻⁴ joules.
The electric potential at point P (x=0 cm) due to two 10 nC charges located at x= - 4 cm and x= 4.0 cm is 4.5 × 10⁵ volts. The work required to bring a 20 nC charge from infinity to point P is 9 × 10⁻⁴ joules.
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The cholesterol content of large eggs of a particular brand is normally distributed with a mean of u = 195 mg and a standard deviation of o= 12 mg. Suppose we take a random sample of 50 eggs. What is
The sampling distribution of the mean of the cholesterol content of 50 eggs is normal with mean 195 and standard deviation 1.697 (rounded to 3 decimal places).
The central limit theorem states that the sampling distribution of the mean of a sufficiently large sample size from any population has a normal distribution, regardless of the population's distribution. As n >= 30, the sample size is large enough to use the central limit theorem in this case.The standard deviation of the sampling distribution, also known as the standard error of the mean (SEM), can be calculated using the formula: SEM = σ/√n, where σ is the population standard deviation and n is the sample size. Plugging in the given values, we get SEM = 12/√50 = 1.697 (rounded to 3 decimal places). Therefore, the sampling distribution of the mean of the cholesterol content of 50 eggs is normal with mean 195 and standard deviation 1.697 (rounded to 3 decimal places).
A measure of how dispersed the data are in relation to the mean is called the standard deviation (or ). Data with a low standard deviation are grouped around the mean, while data with a high standard deviation are more dispersed.
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Footprints on the Moon (Adapted from Bennett, Donahue, Schneider, and Voit)
It has been estimated that about 25 million micrometeorites impact the surface of the Moon daily. (This estimate comes from observing the number of micrometeorites that impact the Earth’s atmosphere daily.) Assuming that these impacts are distributed randomly across the surface of the Moon, estimate the length of time which a footprint left on the Moon by the Apollo astronauts will remain intact, given that it takes approximately 20 micrometeorite impacts to destroy a footprint. (Hint: this is an order of magnitude type calculation, and requires you to make some estimates. Be sure to clearly explain what you are doing at each step of your calculation, and determine if the resulting answer is reasonable!)
Escape Velocity
a) Gravitational Potential energy V = -GMm/r, Kinetic Energy K = 1/2 mv2 Derive the escape velocity for a planet of mass M and radius R. Calculate this value for the surfaces of Earth and Jupiter.
b) Temperature is the average kinetic energy of a group of particles. For an idea gas, K = 3/2 kBT, where K is the kinetic energy, kB is Boltzmann’s constant, and T is temperature. Derive the average velocity of a gas molecule as a function of its mass and Temperature. Calculate this value for a molecule of Oxygen (O2) and Hydrogen (H2).
c) Why does the Earth’s atmosphere have so little Hydrogen, while Jupiter’s atmosphere is full of it?
25 million micrometeorites hit the surface of the moon daily. The Apollo astronauts' footprint will stay on the surface of the moon if it takes around 20 micrometeorites to damage it.
So, to calculate the duration, we'll need to find the number of footprints that have been damaged. We don't know how many footprints there are, so let's estimate that. Assume the average person walks at a rate of 1 step per second. Assume that each step is one foot in length. Assume the average person walks for 2 hours. Then, each person walks for 7200 seconds. The number of footprints per individual is 7200 x 1 = 7200. If we presume 12 people in total, the total number of footprints is 7200 x 12 = 86400.
Therefore, assuming that the footprints are uniformly distributed on the surface of the moon and that 25 million micrometeorites hit the moon's surface daily, the footprints are destroyed at a rate of 25,000,000/20 = 1,250,000 footprints per day.
The duration for the Apollo astronaut's footprints on the moon to remain intact:86400/1,250,000 = 0.06912 days, or roughly 1 hour and 40 minutes.
To calculate how long an Apollo astronaut's footprint would stay on the surface of the Moon, given that it takes around 20 micrometeorites to destroy a footprint, and given that 25 million micrometeorites hit the Moon's surface every day, we'll need to do some calculations. We'll begin by assuming that the footprints were uniformly distributed on the surface of the moon. We'll also assume that each person took 1 step per second, that each step is one foot in length, and that the average person walked for 2 hours. That means each person walked for 7200 seconds, or took 7200 steps. If we assume that there were 12 people on the Apollo mission, then the total number of footprints left by the astronauts would be 12 x 7200 = 86400.
Now, we need to figure out how quickly these footprints are being destroyed. Given that it takes around 20 micrometeorites to destroy a footprint, and given that 25 million micrometeorites hit the Moon's surface every day, we can calculate that the footprints are being destroyed at a rate of 25,000,000/20 = 1,250,000 footprints per day.
So, to find out how long it would take for the footprints to be destroyed, we divide the total number of footprints by the rate at which they are being destroyed:86400/1,250,000 = 0.06912 days, or roughly 1 hour and 40 minutes. Therefore, the length of time for the footprint to remain intact is approximately 1 hour and 40 minutes.
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The fundamental frequency of a pipe that is open at both ends is 594 Hz .
How long is this pipe?
If one end is now closed, find the wavelength of the newfundamental.
If one end is now closed, find the frequency of the newfundamental.
When one end is closed, the new wavelength is 1.154 m and the new fundamental frequency is 297 Hz.
The fundamental frequency of a pipe that is open at both ends is 594 Hz. In order to calculate the length of this pipe, we will use the formula v = fλ where v is the speed of sound, f is the frequency and λ is the wavelength.
The speed of sound in air is approximately 343 m/s.
We will therefore have: 594 = (343/λ)λ = (343/594)m = 0.577m or 57.7cm.
If one end of the pipe is now closed, it will act as a closed-end resonator which means that the wavelength will now be twice the length of the pipe.
Therefore, the new wavelength will be 2(0.577) = 1.154 m or 115.4 cm.
Using the formula v = fλ and substituting the new wavelength and speed of sound, we have 343 = f(1.154) which gives us the new fundamental frequency f as:
f = 297 Hz.
Thus, the length of the pipe that is open at both ends is 57.7 cm. When one end is closed, the new wavelength is 1.154 m and the new fundamental frequency is 297 Hz.
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hello please show all work
and solutions, formulas etc. please try yo answer asap for huge
thumbs up!
12. A 2.5 x 10¹8 Hz x-ray photon strikes a metal foil and frees an electron. After the collision a lower energy 2.3 x 1018 Hz x-ray photon emerges. What is the speed of the electron? [P4]
An x-ray photon at 2.5 x 10¹⁸ Hz strikes a metal foil, releasing an electron. The resulting photon has a frequency of 2.3 x 10¹⁸ Hz, and the electron's speed is determined to be 1.24 x 10⁸ m/s when its energy matches that of the photon.
The energy of a photon is given by the equation:
E = hν
where h is Planck's constant and ν is the frequency of the photon.
The energy of the electron is given by the equation:
[tex]E = \frac{1}{2} m v^2[/tex]
where m is the mass of the electron and v is the speed of the electron.
We can set these two equations equal to each other to find the speed of the electron:
[tex]h\nu = \frac{1}{2} m v^2[/tex]
We can rearrange this equation to solve for v:
[tex]v = \sqrt{\frac{2h\nu}{m}}[/tex]
We know the value of h, ν, and m. Plugging these values into the equation, we get:
[tex]v = \sqrt{\frac{2 \times (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (2.5 \times 10^{18} \, \text{Hz})}{9.11 \times 10^{-31} \, \text{kg}}}[/tex]
v = 1.24 x 10⁸ m/s
Therefore, the speed of the electron is 1.24 x 10⁸ m/s.
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the force per meter between the two wires of a jumper cable being utilized to start a stalled van is 0.215 n/m.
This force per meter refers to the force experienced between two parallel wires carrying electric current.
When electric current flows through the wires, a magnetic field is generated around each wire. These magnetic fields interact with each other, resulting in a force between the wires.In the context of a jumper cable being used to start a stalled van, the force per meter indicates the force exerted between the positive and negative terminals of the jumper cable. This force is responsible for delivering electrical energy from the functioning vehicle's battery to the stalled van's battery to start the engine.
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1. (a) In reaching equilibrium, how much heat transfer occurs from 1.1 kg of water at 40°C when it is placed in contact with 1.1 kg of 20°C water? Specific heat of water c=4186 J/(kg°C) Hint: If th
The heat transfer that occurs from 1.1 kg of water at 40°C to 1.1 kg of water at 20°C is 92,270 J.
To calculate the heat transfer that occurs when two substances reach thermal equilibrium, we can use the equation Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat, and ΔT is the change in temperature.
In this case, we have two equal masses of water, each weighing 1.1 kg. The specific heat of water, c, is given as 4186 J/(kg°C).
First, we need to calculate the change in temperature, ΔT, which is the difference between the final equilibrium temperature and the initial temperature. Since the masses are equal, the equilibrium temperature will be the average of the initial temperatures, which is (40°C + 20°C) / 2 = 30°C.
Next, we can calculate the heat transfer for each mass of water using the equation Q = mcΔT. For the water at 40°C, the heat transfer is Q₁ = (1.1 kg) * (4186 J/(kg°C)) * (30°C - 40°C) = -45,530 J (negative because heat is transferred out of the water). Similarly, for the water at 20°C, the heat transfer is Q₂ = (1.1 kg) * (4186 J/(kg°C)) * (30°C - 20°C) = 137,800 J.
The total heat transfer is the sum of the individual heat transfers: Q_total = Q₁ + Q₂ = -45,530 J + 137,800 J = 92,270 J.
Therefore, the heat transfer that occurs from 1.1 kg of water at 40°C to 1.1 kg of water at 20°C is 92,270 J.
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Complete Question:
(a) In reaching equilibrium, how much heat transfer occurs from 1.1 kg of water at 40€ when it is placed in contact with 1.1 kg of 20€ water? Specific heat of water c=4186 J/(kg) Hint: If the masses of water are equal, what is the equilirium temperature of the water mixture?
A voltaic cell consists of an Mn/Mn2+ half-cell and a Caicd2+ half-cell. The standard reduction potential for Mn2+ is -1.18V and for Cd2+ is -0.40 V. Calculate Ecell at 25 °C when the concentration of [Cd2+] = 8.84 x 10-0 M and [Mn2+1=9.57 x 10-5 M. (value + 0.02) Selected Answer: [None Given] Correct Answer: 0.93 +0.02
The value of E-cell at 25 °C when the concentration of [Cd2+] = 8.84 × 10⁻⁰ M and [Mn2+1=9.57 × 10⁻⁵ M is 0.93 + 0.02 V.
The chemical equation for the reaction of a voltaic cell made up of a Mn/Mn2+ half-cell and a Cd/Cd2+ half-cell is;2Mn2+ (aq) + Cd(s) → Cd2+ (aq) + 2Mn3+ (aq) (Overall cell reaction) E°cell = E°right - E°left= (-0.40) - (-1.18) = 0.78 V (The positive value indicates that the reaction is spontaneous)From the Nernst equation, Ecell = E°cell - (RT/nF) * ln Q
where; R = gas constant = 8.31 J/mol. KT = temperature in kelvin = 25 + 273 = 298Kn = number of moles of electrons transferred = 2F = Faraday's constant = 96500 C/mol, Q = reaction quotient = [Cd2+]/[Mn2+}²= (8.84 × 10⁻⁰) / (9.57 × 10⁻⁵)²= 97.3Ecell = 0.78 - [(8.31 × 298) / (2 × 96500)] * ln 97.3Ecell = 0.93 + 0.02 V (rounded off to 2 decimal places).
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How much work does the electric field do in moving a -6.4x10-6 charge from ground to a point whose potential is 92 V higher?
The work done by the electric field in moving a -6.4x10^-6 charge from ground to a point 92 V higher is -5.888x10^-4 J.
The work done by an electric field in moving a charge can be calculated using the formula:
Work = q * ΔV
Where:
Work is the work done (in joules)
q is the charge (in coulombs)
ΔV is the change in potential (in volts)
q = -6.4x10^-6 C
ΔV = 92 V
Substituting these values into the formula, we get:
Work = (-6.4x10^-6 C) * (92 V)
= -5.888x10^-4 J
The work done by the electric field in moving a -6.4x10^-6 charge from ground to a point whose potential is 92 V higher is -5.888x10^-4 J. The negative sign indicates that the electric field does work against the motion of the charge, as the charge is moving to a higher potential.
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Two long straight current-carrying wires run parallel to each other. The current in one of the wires is 8 A, their separation is 5.5 cm and they repel each other with a force per unit length of 2.6 x104 N/m. Determine the current in the other wire.
The current in the other wire is 0.225 A. When two long, straight current-carrying wires run parallel to each other, they experience a force that is repulsive. This is due to the interaction of magnetic fields produced by the current-carrying wires.
The magnetic fields around each wire interact, resulting in a force of repulsion between them. This force is proportional to the product of the current in each wire, the length of the wires, and inversely proportional to the distance between them.
Let us consider the situation in which two parallel wires are placed at a separation of 5.5 cm. The current in one wire is 8 A, and they repel each other with a force per unit length of 2.6 x104 N/m.
We can determine the current in the second wire by using the formula for the force per unit length between the two wires:
F/l = μ0*I1*I2/(2π*d)
where:
F is the force per unit length between the two wires
l is the length of each wire
μ0 is the magnetic constant (4π x 10-7 T m/A)
I1 is the current in the first wire
I2 is the current in the second wire
d is the distance between the two wires
Substituting the given values in the formula, we get:
2.6 x104 N/m = 4π x 10-7 T m/A * 8 A * I2 / (2π * 0.055 m)
Simplifying this expression, we get:
I2 = (2.6 x104 N/m * 2π * 0.055 m) / (4π x 10-7 T m/A * 8 A),I2 = 0.225 A
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for an electromagnetic wave the direction of the vector e x b gives
The speed of an electromagnetic wave is 299,792,458 meters per second (m/s) or the speed of light.
The direction of the vector product of E (electric field) and B (magnetic field) indicates the direction of energy transfer in an electromagnetic wave. This direction is perpendicular to both the E and B fields. The wave propagates in this direction as well. The direction of the vector product is referred to as the Poynting vector.
The Poynting vector, S, provides information about the direction and intensity of the electromagnetic energy flux or radiation pressure density. Its SI unit is watt per square meter (W/m²). It can be mathematically expressed as:S = E × BIn an electromagnetic wave, the E and B fields oscillate in mutually perpendicular planes. The direction of energy transfer is also perpendicular to both the E and B fields. An electromagnetic wave propagates perpendicular to both E and B fields and the direction of energy transfer. It has both electric and magnetic properties and carries energy. Therefore, an electromagnetic wave can be defined as a wave of energy produced by the acceleration of an electric charge and propagated through a vacuum or a medium.
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Name formula mol. Eq mw mmol amount imine intermediate 1. 00 280 mg nabh4 70 mg thf - - - 10 ml product
The number of millimoles of the product produced is: = 0.846 mmol. The equation for the imine intermediate 1 is as follows: C₁₉H₂₁N₃O₂ + NaBH₄ + THF → C₁₉H₂₃N₃O₂ + NaBH₃CN + NaCl + THF
The formula for imine intermediate 1 is C₁₉H₂₁N₃O₂. The molecular weight (MW) of imine intermediate 1 is 331.4 g/mol.
The molecular weight of NaBH₄ is 37.83 g/mol.
The molecular weight of THF is 72.11 g/mol.
Therefore, the amount of NaBH₄ is 70 mg, and the amount of THF is 10 mL.
The number of millimoles of NaBH₄ can be calculated as follows: 70 mg × 1 mol/37.83 g × 1000 mg/1 g
= 1.85 mmol
The number of millimoles of THF is: 10 mL × 0.088 g/mL × 1 mol/72.11 g × 1000 mg/1 g
= 1.22 mmol
The number of millimoles of imine intermediate 1 can be calculated as follows:
280 mg × 1 mol/331.4 g × 1000 mg/1 g
= 0.846 mmol
The number of millimoles of NaBH₃CN produced can be calculated as follows:
1.85 mmol × 1 mol/1 mol
= 1.85 mmol
The number of millimoles of the product produced is:
0.846 mmol × 1 mol/1 mol
= 0.846 mmol
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An electron situated at point P experiences an electrostatic force of 4.8 x 10-14 N acting on it. What is the electric field strength at P? 3.0 x 10^5 N/C 7.7 x 10^-33 N/C 3.3 x 10^-6 N/C 6.4 x 10^-14
Based on the information provided in the question, we cannot determine the electric field strength at point P.
The electric field strength at point P can be calculated using the formula:
Electric Field Strength = Force / Charge
In this case, the given force acting on the electron is 4.8 x 10^-14 N. However, the charge of the electron is not provided in the question. Without knowing the charge, we cannot accurately calculate the electric field strength.
The electric field strength is defined as the force experienced by a unit positive charge. Since the charge of the electron is negative, we would need to consider the magnitude of the charge to calculate the electric field strength correctly.
Therefore, based on the information provided in the question, we cannot determine the electric field strength at point P.
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an air-track glider attached to a spring oscillates between the 14.0 cm c m mark and the 65.0 cm c m mark on the track. the glider completes 11.0 oscillations in 37.0 s s .(a) period, (b) frequency. (c) amplitude, and (d) maximum speed of the glider?
The maximum speed of the oscillatory motion is 1.52 m/s. a) Period = 3.36 s b) Frequency = 0.30 Hz c) Amplitude = 0.255 m d) Maximum speed = 1.52 m/s
Given Data: Length of oscillations in air track, L = 65 cm – 14 cm = 51 cm = 0.51 m. Number of oscillations, n = 11Time taken for n oscillations, t = 37 s. We can obtain different properties of the oscillatory motion using these values.
(a) Period of the oscillatory motion. The period of the oscillatory motion is defined as the time taken for one complete oscillation. We can calculate the period using the following formula: T = t/n = 37/11 s = 3.36 s. Therefore, the period of the oscillatory motion is 3.36 s.
(b) Frequency of the oscillatory motion. The frequency of the oscillatory motion is defined as the number of oscillations completed in one second. It is the reciprocal of the period and is given by the following formula: f = 1/T = 1/3.36 Hz. Therefore, the frequency of the oscillatory motion is 0.30 Hz.
(c) Amplitude of the oscillatory motion. The amplitude of the oscillatory motion is defined as half the distance between the extreme positions of the motion. It is given by the following formula: A = (L/2) = (0.51/2) m = 0.255 m. Therefore, the amplitude of the oscillatory motion is 0.255 m. (d) Maximum speed of the oscillatory motion. The maximum speed of the oscillatory motion occurs at the mean position (center). At the extreme positions, the velocity is zero. Therefore, we can calculate the maximum speed using the following formula: vmax = 2πA/T where A is the amplitude and T is the period. Substituting the given values, we get: vmax = (2π × 0.255)/3.36 m/s≈ 1.52 m/s.
Therefore, the maximum speed of the oscillatory motion is 1.52 m/s. Answer: Period = 3.36 s Frequency = 0.30 Hz Amplitude = 0.255 m Maximum speed = 1.52 m/s.
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QUESTION 6: ELECTRICITY 1 Explain what is meant when a substance is referred to as a bad conductor of electricity and give ONE example. 2 THREE equal resistors are connected in parallel. The total res
When a substance is referred to as a bad conductor of electricity, it means that it does not allow electric current to flow easily through it. This is because the substance has high resistance to the flow of electric charge.
In bad conductors, the electrons are tightly bound to their atoms or molecules, making it difficult for them to move freely and carry the electric current. As a result, only a small amount of current can pass through the substance.
Example: One example of a bad conductor of electricity is rubber. Rubber has high resistance to the flow of electric charge and is commonly used as an insulating material to prevent the flow of current in electrical wires and cables.
2. When three equal resistors are connected in parallel, the total resistance (R_total) of the combination can be calculated using the formula:
1/R_total = 1/R_1 + 1/R_2 + 1/R_3
Where R_1, R_2, and R_3 are the resistances of the individual resistors.
Since the three resistors are equal, the formula simplifies to:
1/R_total = 1/R + 1/R + 1/R = 3/R
We can invert both sides of the equation for value of R_total :
R_total = R/3
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