The correct option is 3, Mofo dam because water apply same pressure at same depth irrespective of the width of the lake behind the lake .
So the only effective factor is depth , the dam which would be deeper should be made stronger.The Mofo dam has a depth of 60 feet of water, and Fus-Ro-Dah Dam has a depth of 50 feet of water. Hence, the Mofo dam is constructed to be the strongest.
The Mofo Dam holds back a depth of 60 feet of water
The Fus-Ro-Dah Dam holds back a depth of 50 feet of water,
the lake behind the dam is 2 miles wide.
Generally, The main independent factor to be considered is the depth of a dam, as its the depth of water that applies the most pressure on dams, So the only effective factor is depth.
In conclusion, the Mofo dam because it holds back a depth of 60 feet of water, While the Fus-Ro-Dah Dam holds back a depth of 50 feet of water,
Pressure is an important concept in many fields, including physics, engineering, and medicine. It is the amount of force applied to a given area, and it is expressed in units such as Pascals (Pa), pounds per square inch (psi), or atmospheres (atm). Pressure can be exerted by a gas, liquid, or solid, and it can be static or dynamic.
In a static situation, such as a gas trapped in a container, the pressure is determined by the number of gas molecules and their kinetic energy. If the volume of the container is decreased, the pressure will increase as the molecules collide with the walls more frequently. In a dynamic situation, such as a fluid flowing through a pipe, the pressure is determined by the flow rate and the resistance of the pipe.
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Complete Question: -
The Mofo Dam holds back a depth of 60 feet of water, but the lake bchind the dam is 100 feet wide. The Fus-Ro-Dah Dam holds back a depth of 50 feet of water, but the lake behind the dam is 2 miles wide. If the dams are to be constructed in the same way, which dam had to be constructed to be strongest? (The water levels do not vary seasonally.) 1. The Fus-Roh-Dah Dam 2. Both dams would have to be constructed to be the same in strength. 3. The Mofo Dam 4. Insufficient information has been supplied to give an answer.
a pump is to move water from a lake into a large, pressurized tank as shown in the figure at a rate of 1000 gal in 10 min or less. will a pump that adds 3 hp to the water work for this purpose? support your answer with appropriate calculations. repeat the problem if the tank were pressurized to 3, rather than 2, atmospheres.
A 3 hp pump would be used to move water from a lake into a large, pressurized tank.
To solve,
P = F × V,
where P is the power, F is the force, and V is the velocity of the water.
We know the power is 3 hp and the velocity is 1000 gal/10 min, so we can solve for F:
F = P ÷ V = 3 hp ÷ 1000 gal/10 min
= 0.003 hp/gal/min.
Now, if the tank is pressurized to 3 atmospheres, the pressure will increase the force needed to move the water.
So, the equation for pressure is P = F × A, where P is the pressure, F is the force, and A is the area.
We know the pressure is 3 atmospheres and the force is 0.003 hp/gal/min, so we can solve for A:
A = P ÷ F = 3 atmospheres ÷ 0.003 hp/gal/min
= 1000 gal/10 min/3 atmospheres.
Therefore, a 3 hp pump will work for this purpose, even if the tank is pressurized to 3 atmospheres.
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a vhf television station assigned to channel 22 transmits its signal using radio waves with a frequency of 518 mhz. calculate the wavelength of the radio waves. round your answer to significant digits.
The wavelength of the radio waves is approximately 0.579 m or 57.9 cm
Wavelength is the distance covered by an electromagnetic wave while propagating through space. The relationship between the wavelength and the frequency of an electromagnetic wave is given by the formula;
Wavelength = speed of light / frequency = c / f
where c is the speed of light and f is the frequency of the wave.
To calculate the wavelength of a VHF television station assigned to channel 22 that transmits its signal using radio waves with a frequency of 518 MHz, we substitute the known values into the equation above.
Wavelength = c / f = (3 x 10⁸ m/s) / (518 x 10⁶ Hz) = 0.579 m or 57.9 cm (rounded to three significant digits)
Therefore, the wavelength of the radio waves transmitted by the VHF television station assigned to channel 22 is 0.579 m or 57.9 cm (rounded to three significant digits).
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a series circuit has a capacitor of 1.25x10-5 farad, a resistor of 260 ohms and an inductor of 0.2 henry. the initial charge on the capacitor is 2x10-6 coulomb and there is no initial current. find the charge q(t) on the capacitor at any time t.
The final expression for the charge Q(t) at any time t is given as:Q(t) = CV(t) = 2.5 × 10^-11 e- t/RC
To find the charge on the capacitor at any time t, we need to find the total current in the circuit and then find the charge using the formula Q = CV, where V is the potential difference across the capacitor.Let's find the total current in the circuit using the formula:
I = (1/LC)½ x (e- Rt/2L) sin(wt - φ)
where, L = inductance C = capacitance R = resistance ω = (1/LC)½ = 5000 sinφ = RωL = 260 × 5000 × 0.2 = 2600
Let's now substitute the given values into the formula and simplify:I = (1/(0.2 × 1.25 × 10^-5))½ x (e- 260t/2 × 0.2) sin(5000t - φ)I = 10^5 x (e- 130t) sin(5000t - φ). Let's now find the charge Q on the capacitor using the formula:
Q = CV where, C = capacitance V = potential difference across the capacitor. To find the potential difference across the capacitor, we need to find the current passing through it, which is given as the total current minus the current passing through the inductor. Let's find the current passing through the inductor using the formula:
I L = I x sin(wt - φ)IL = I x sin(5000t - φ).The potential difference across the capacitor can be calculated using the formula:V C = V 0 × e- t/RC where, V0 = initial potential difference across the capacitor R = resistance of the circuit C = capacitance of the circuit. Let's now find the current passing through the capacitor:I C = (I - I L)I C = I - I L
Now we have all the necessary formulas to find the charge Q(t) at any time t. Let's substitute the given values into the formulas and simplify:
I = 10^5 x (e- 130t) sin(5000t - φ)IL = I x sin(5000t - φ)IC = I - I LVC = V0 × e- t/RCQ = CVCI = I - I L = 10^5 x (e- 130t) sin(5000t - φ) - I sin(5000t - φ)V C = V 0 × e- t/RC = 2 × 10^-6 e- t/RCQ = C × V C = (1.25 × 10^-5) × (2 × 10^-6) e- t/RC = 2.5 × 10^-11 e- t/RC
Now, let's substitute the values of I and V C into the formula for IC to obtain:IC = 10^5 × (e- 130t) sin(5000t - φ) - 10^5 sin(5000t - φ) × e- t/RC. Therefore final expression for the charge Q(t) at any time t is given as:Q(t) = CV(t) = 2.5 × 10^-11 e- t/RC
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We can use the equation [tex]q(t) = C.V(t)[/tex] to calculate the charge q (t) on the capacitor at any time t: [tex]q(t) = 1,25 . 10-5 Farad.V(t)[/tex].
The charge on a capacitor in a series circuit at any time t is given by the equation [tex]q(t) = C.V(t)[/tex], where C is the capacitance of the capacitor and V(t) is the voltage across the capacitor at time t.
In the given circuit, the capacitance of the capacitor is 1.25 x 10-5 Farad, and the initial charge on the capacitor is 2 x 10-6 Coulomb. Therefore, to find the charge q(t) on the capacitor at any time t, we need to find the voltage V(t) across the capacitor at time t.
To do this, we must first calculate the total inductance and resistance in the circuit. The total inductance is the sum of the inductances of each inductor, so the total inductance in this circuit is 0.2 Henry. The total resistance is the sum of the resistances of each resistor, so the total resistance in this circuit is 260 Ohms.
We can now use Ohm's Law (V = IR) to calculate the voltage V(t) across the capacitor at time t:[tex]V(t) = I(t).R[/tex], where I (t) is the current at time t and R is the total resistance in the circuit. Since the inductance of the circuit is 0.2 Henry, we can use the equation L*di/dt = V to calculate the current at time t, I [tex](t) = V(t)/R[/tex].
Substituting this into Ohm's Law, we get: V(t) = (V(t)/R)*R. Solving for V(t), we get V(t) = V(t). Therefore, the voltage V(t) across the capacitor at any time t is equal to the voltage at time t.
Finally, we can use the equation [tex]q(t) = C.V(t)[/tex]to calculate the charge q(t) on the capacitor at any time t: [tex]q(t) = 1,25 . 10-5 Farad.V(t)[/tex].
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if two identical resistors are connected in series to a battery, does the battery have to supply more power or less power than when only one of the resistors is connected? explain
The battery has to supply more power when two resistors are connected in series than when only one resistor is connected. This is because the power dissipated in a series circuit is equal to the sum of the power dissipated in each resistor.
When two identical resistors are connected in series to a battery, the battery has to supply more power than when only one of the resistors is connected. This is because the resistors offer resistance, which results in the dissipation of energy as heat. The higher the resistance of a resistor, the more power it requires to operate.Resistors consume energy as they offer resistance to the flow of current. The power supplied by the battery is converted to heat energy in the resistor, and the amount of heat energy dissipated is determined by the resistance of the resistor. The greater the resistance of the resistor, the more power it requires to function.
As a result, when two identical resistors are connected in series to a battery, the battery has to supply more power than when only one of the resistors is connected, to produce the same current through the circuit. Therefore, if two resistors of equal value are connected in series, the total power dissipated is twice that of when a single resistor is connected.
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!!! If each compound undergoes electrophilic aromatic substitution, where should the substituent be added? Phenol?
Benzaldehyde?
Benzoic Acid?
Bromobenzene?
Nitrobenzene?
Toluene?
The substituent in Phenol is added to the ortho and para positions of the benzene ring. The substituent in Benzaldehyde is added to the ortho and para positions of the benzene ring.
The substituent in Bromobenzene is added to the ortho and para positions of the benzene ring. The substituent in Nitrobenzene is added to the meta position of the benzene ring. The substituent in Toluene is added to the ortho and para positions of the benzene ring.
Substituents on different aromatic compounds. The substituent is added to different positions for each of the aromatic compounds if they undergo electrophilic aromatic substitution. The positions where the substituents are added to Phenol, Benzaldehyde, Benzoic Acid, Bromobenzene, Nitrobenzene, and Toluene are described below:
Phenol- The substituent in Phenol is added to the ortho and para positions of the benzene ring.
Benzaldehyde- The substituent in Benzaldehyde is added to the ortho and para positions of the benzene ring.
Benzoic Acid- The substituent in Benzoic acid is added to the meta position of the benzene ring.
Bromobenzene- The substituent in Bromobenzene is added to the ortho and para positions of the benzene ring.
Nitrobenzene- The substituent in Nitrobenzene is added to the meta position of the benzene ring.
Toluene- The substituent in Toluene is added to the ortho and para positions of the benzene ring.
Thus, we can see that the positions of the substituent in each aromatic compound depend on the particular compound that undergoes electrophilic aromatic substitution.
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1. I’m in the 2nd column, 4th row, and I’m a metal. Who am I? ________________ 2. I’m a very lonely nonmetal. Who am I? ____________ 3. I’m the only metal who is a liquid at room temperature. Who am I? ____________ 4. I’m named after the person who created the 1st Periodic Table. Who am I? ___________ 5. I have 92 protons. Who am I? _____________ 6. I’m the only nonmetal who is a liquid at room temperature. Who am I? ___________ 7. I’m named after a very famous scientist. Who am I? ___________ 8. I have 46 electrons. Who am I? ____________ 9. My atomic mass is 183. 84. Who am I? _____________ 10. My chemical symbol is Ag. Who am I? ________________ 11. I’m the only metalloid in period 3. Who am I? ___________ 12. I’m the only element that is solid and a nonmetal in group 14. Who am I? _____________ 13. I have 5 neutrons. Who am I? ____________ 14. I’m the only gas at room temperature that is in group 16. Who am I? ___________ 15. I have 68 protons. Who am I? __________ 16. What element has the chemical symbol of Ir? ______________ 17. Which element is in group 7 and has 30 neutrons. Who am I? ___________ 18. I’m the only metal in group 15. Who am I? ____________ 19. I have 88 electrons. Who am I? ___________ 20. I’m the only gas at room temperature and in period 5. Who am I? ____________ 21. My symbol is Am. Who am I? ______________ 22. I’m the only nonmetal in period 6. Who am I? ____________ 23. My atomic number is 69. 723. Who am I? _________________ 24. I have 159 neutrons. Who am I? ________________ 25. I’m the only metalloid in group 17. Who am I? ______________ 26. I have 50 electrons. Who am I? __________________ 27. I’m in the 1st group and the 4th period. Who am I? ________________ 28. I’m a metalloid whose symbol is Sb. Who am I? ______________ ©JFlowers2017 Name: ______________________________ Date: ___________Class: ________ Periodic Table Scavenger Hunt Directions: You will use the Periodic Table to answer the questions. 1. I’m in the 17th column, a nonmetal, & a solid at room temperature. Who am I? ________________ 2. I have 79 electrons. Who am I? ____________ 3. I’m the only gas in period 6. Who am I? ____________ 4. My atomic mass is 257. Who am I? ___________ 5. My chemical symbol is Hs. Who am I? _____________ 6. I have 114 neutrons. Who am I? ___________ 7. I’m in the 18th group and 2 nd period. Who am I? ___________ 8. I have 67 protons. Who am I? ____________ 9. I’m a nonmetal who is solid at room temperature & has 2 letters for my symbol. Who am I? _________ 10. I’m in the 1 st group & 7 th period. Who am I? ________________ 11. I’m the only metalloid in group 13. Who am I? ___________ 12. I have 97 electrons. Who am I? _____________ 13. I am the only gas in column 15. Who am I? ____________ 14. My name is similar to Mickey Mouse’s best friend. Who am I? ___________ 15. I’m in group 11 & period 4. Who am I? __________ 16. I have 62 protons. Who am I? ______________ 17. My name fits really well with doctors because they try to do this. Who am I? ___________ 18. My name reminds me of where we all live. Who am I? ____________ 19. I’m the only nonmetal in period 2. Who am I? ___________ 20. My atomic number is 87. 62. Who am I? ____________ 21. My symbol is Mt. Who am I? ______________ 22. I’m in group 17 & the only metalloid. Who am I? ____________ 23. I have 71 electrons. Who am I? _________________ 24. My symbol is Pd. Who am I? ________________ 25. I’m Dorothy’s friend who needed a heart. Who am I? ______________ 26. I have 41 protons. Who am I? __________________ 27. I have 125 neutrons. Who am I? ________________ 28. My name comes from the 8th planet. Who am I? ______________
The Periodic Table of Elements served as the inspiration for this scavenger hunt. The exercise consists of two sets of questions, each of which has 28 questions that must be answered using the Periodic Table.
Students are tasked with identifying elements in the first set of questions using information from their attributes, such as the element's position on the periodic table, atomic mass, or quantity of electrons, protons, or neutrons. The objectives of the questions are to familiarise students with the properties of various elements and the structure of the Periodic Table. The second series of questions is comparable to the first, but more difficult because it asks students to identify components using less obvious cues, like their chemical symbol or a chemical formula. In order to succeed in their future studies of chemistry and other related sciences, students will benefit from being more familiar with the structure of the periodic table and the characteristics of various elements.
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when a 2.75-kg fan, having blades 18.5 cm long, is turned off, its angular speed decreases uniformly from 10.0 rad/s to 6.30 rad/s in 5.00 s. (a) what is the magnitude of the angular acceleration of the fan?
The angular acceleration of the fan is 0.740 rad/s^2,
Angular acceleration which represents the rate at which the angular velocity changes over time. The unit used to measure angular acceleration is radians per square second (rad/s2), according to the International System of Units. The Greek alphabet symbol alpha (α) is used to denote angular acceleration.
To calculate the angular acceleration of the fan, the formula α = Δω/Δt is used. Here, α represents angular acceleration, Δω represents the change in angular speed, and Δt represents the change in time.
In this scenario, Δω is equal to 10.0 - 6.30 = 3.70 rad/s, and Δt is equal to 5.00 s. By substituting these values into the formula, we obtain α = 3.70/5.00 = 0.740 rad/s^2.
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What allowed the Voyager 2 spacecraft to make a "tour" of all four of the jovian planets in the late 1970's and the 1980's?
1) NASA had developed a new kind of rocket that could propel the craft from planet to planet
2) the four planets were approximately aligned on one side of the Sun and we used the gravity of each planet to speed up the spacecraft to get to the next one in its path
3) the spacecraft stopped off to collect fuel on the satellites of each planet before proceeding to the next one
4) we used laser beams to propel the spacecraft into the outer solar system, where sunlight is dim
5) you can't fool me, no single spacecraft has ever explored four different planets
Answer:
The four planets were approximately aligned on one side of the Sun and we used the gravity of each planet to speed up the spacecraft to get to the next one in its path
Explanation:
All the Options 1, 2, 3, 4 are true about the Voyager 2 spacecraft to make a "tour" of all four of the jovian planets in the late 1970's and the 1980's.
The Voyager 2 spacecraft was able to make a "tour" of all four of the jovian planets in the late 1970's and the 1980's due to the following:
NASA had developed a new kind of rocket that could propel the craft from planet to planet.The four planets were approximately aligned on one side of the Sun and we used the gravity of each planet to speed up the spacecraft to get to the next one in its path.The spacecraft stopped off to collect fuel on the satellites of each planet before proceeding to the next one.We used laser beams to propel the spacecraft into the outer solar system, where sunlight is dim.Learn more about "NASA and spacecraft" at: https://brainly.com/question/16538247
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Categorize the following exercises as being isometric or isotonic.
Pushing constantly against a concrete wall
Running up a hill
Swimming freestyle
Pedaling a bicycle on a flat surface
Holding a bench-press bar in the same position
Doing a plank exercise (holding a push-up position)
Balancing on tiptoes
Doing bicep curls
Isometric pushes against a wall made of concrete, Isotonic running up a hill. isotonic freestyle swimming, bicycle pedalling on a level surface: isotonic.
Static muscle contractions, in which the length of the muscle does not change during the workout, are called isometric exercises. This indicates that during the activity, there is no discernible movement or alteration in joint angle. Instead, the muscles are tense against a constant force or maintained still for a certain period of time. Exercises that are isometric include pushing against a wall, keeping a plank position, and tightening a hand grasp. Exercises that are isometric can help to increase joint stability and balance as well as muscular strength and endurance. They can also be incorporated into normal workout routines for general health and strength training. They are frequently used in physical therapy to aid patients in recovering from injuries or surgery.
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A piece of metal weighing 187.6 g is placed in a graduated cylinder containing 225.2 mL of water. The combined volume of solid and liquid is 250.3 mL. What is the density, in grams per milliliter, of the metal?
The density of the metal in grams per milliliter is 7.87 g/mL.
Given data:The weight of metal, W = 187.6 g,Volume of water, V₁ = 225.2 mL.
The combined volume of solid and liquid, V₂ = 250.3 mL
Volume of the metal can be calculated as:Volume of metal = V₂ - V₁= 250.3 - 225.2= 25.1 mL
The density of the metal can be calculated as:Density = Weight of metal / Volume of metal
Density = W / V= 187.6 g / 25.1 mL= 7.87 g/mL
Thus, the density, in grams per milliliter, of the metal is therefore calculated and found to be 7.87 g/mL.
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help me
plss asap!!!
Answer:B
Explanation:The ray above makes a 90 degree angle. The ray below makes a 60 degree angle.
Ceteris paribus, which of these events would cause both the equilibrium interest rate and the equilibrium quantity of investment to fall?
• A dcrease investor confidance
• A decrease in cosmetic income and wealth • A strengh of time preference
• A decrease in capital productivity
Ceteris paribus, a decrease in capital productivity is the event that would cause both the equilibrium interest rate and the equilibrium quantity of investment to fall. The correct answer is option C.
Ceteris paribus is a Latin expression that means "all other things being equal." Ceteris paribus is a model in which economists use to analyze the effect of one independent variable on a dependent variable while keeping all other independent variables constant. This implies that only one variable is allowed to change while all other variables are held constant at their current level or position.
Therefore, Ceteris paribus, an increase in investor confidence, an increase in cosmetic income and wealth, and a strength of time preference will not cause both the equilibrium interest rate and the equilibrium quantity of investment to fall. However, a decrease in capital productivity is an event that would cause both the equilibrium interest rate and the equilibrium quantity of investment to fall.
When capital productivity is low, firms are unable to produce goods and services efficiently, and as a result, the demand for investment falls. When the demand for investment falls, the equilibrium quantity of investment will also decrease, leading to a decrease in the equilibrium interest rate.
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A 1,600 kg car is moving at 22 m/s. How much work was done to accelerate it to this speed?
O 7.7 x 105 J
O 3.5 x 104 J
○ 3.9 × 105 J
O 1.5 x 106 J
!!! Urgent
The closest answer among the options given is 3.9 x 105 J. . An object can accelerate by increasing its speed, changing its direction, or both.
What is Acceleration?
Acceleration is the rate of change of velocity of an object over time. It is a vector quantity, meaning it has both magnitude and direction, and is expressed in units of meters per second squared (m/s^2) or feet per second squared (ft/s^2)
The work done to accelerate the car can be calculated using the kinetic energy formula:
K = 1/2 mv^2
Substituting the given values, we get:
K = 1/2 (1600 kg) (22 m/s)^2
K = 677,600 J
Therefore, the work done to accelerate the car to this speed is 677,600 J.
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the cardinals kick a 0.43 kg football for a 3-point field goal. if the ball is kicked at 24 m/s at an angle of 53-degrees, how far will it go before landing back on level ground?
The distance which the football which cover before landing back on the ground level will be about 56.4 meters.
What is the distance of football?
The mass of football, m = 0.43 kg, Initial velocity of football (v) = 24 m/s, Angle of inclination(θ) = 53°
From the given data, we know that the vertical component of the initial velocity is given by, vsin(θ) and the horizontal component of initial velocity is given by, vcos(θ). So, the time taken by the football to reach the maximum height is given by,
t = (vsin(θ))/g
Here, g = 9.8 m/s²
Now, the maximum height attained by the football is given by,h = (vsin(θ))²/(2g).
Therefore, the time of flight or the total time which is taken by the football to land on the ground level is given by,
T = 2t
Now, the horizontal distance travelled by the ball is given by, d = (vcos(θ))T
Substituting the given values in the above formulas, we get:
t = (24sin(53°))/9.8 = 1.71 s
h = (24sin(53°))²/(2×9.8) = 23.4m
T = 2×1.71 = 3.42 s
d = (24×cos(53°))×3.42 = 56.4 m
Therefore, the football will go 56.4 m before it is landing back on the level ground.
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for the given input voltage amplitude (200 mvpp), what is the maximum gain that this amplifier will be able to produce? show your calculation below.
The maximum gain of an amplifier that produces an output voltage amplitude of 50 Vpp with an input voltage amplitude of 200 mVpp is 25. The formula to calculate gain is output voltage amplitude divided by input voltage amplitude.
In this case, we are given an input voltage of 200 mVpp, so the maximum gain of this amplifier can be calculated as follows:
Gain = Output Voltage/Input Voltage = Output Voltage/200 mVpp
Therefore, the maximum gain of this amplifier is equal to the output voltage. In other words, the maximum gain of this amplifier is equal to the voltage output of the amplifier.
To calculate the output voltage of the amplifier, we need to know the supply voltage and the resistance of the load. Assuming the supply voltage is 5V and the load resistance is 10k ohms, the output voltage can be calculated as follows:
Output Voltage = Supply Voltage * Load Resistance / (Load Resistance + Output Resistance) = 5V * 10k ohms / (10k ohms + 10k ohms) = 5V
Therefore, the maximum gain of this amplifier is 5V/200 mVpp = 25.
To summarize, the maximum gain of this amplifier is 25, calculated by dividing the output voltage by the input voltage. The output voltage can be calculated by knowing the supply voltage and load resistance.
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a clean nickel surface is exposed to light with a wavelength of 241 nm n m . the photoelectric work function for nickel is 5.10 ev e v . for related problem-solving tips and strategies, you may want to view a video tutor solution of a photoelectric-effect experiment. part a what is the maximum speed of the photoelectrons emitted from this surface?
The maximum speed of the photoelectrons emitted from the clean nickel surface is 6.70 × 10⁵ m/s.
Calculate the energy of a photon.E = hc/λwhere, h = Planck’s constant = 6.626 × 10⁻³⁴ Js, c = speed of light = 3 × 10⁸ m/sE = 6.626 × 10⁻³⁴ × 3 × 10⁸/241 × 10⁻⁹E = 8.21 × 10⁻¹⁸ J
Calculate the kinetic energy of the photoelectrons.
K.E. = E – W₀K.E. = 8.21 × 10⁻¹⁸ J – 5.10 × 1.6 × 10⁻¹⁹ J = 7.09 × 10⁻¹⁹ J
K.E. = 1/2 mv² where, m = mass of photoelectron, v = velocity of photoelectron, and K.E. = kinetic energy of photoelectronv = √(2K.E./m) = √[(2 × 7.09 × 10⁻¹⁹ J)/(9.1 × 10⁻³¹ kg)]v = 6.70 × 10⁵ m/s or 0.224c
So, the maximum speed of the photoelectrons emitted from this surface is 6.70 × 10⁵ m/s.
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how was the heliocentric theory developed by copernicus different from the greek theory of geocentrism?
The geocentric model says that the earth is at the center of the cosmos or universe, and the planets, the sun and the moon, and the stars circles around it. The early heliocentric models consider the sun as the center, and the planets revolve around the sun.
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 the conservation of energy in position A?
[tex]E=\frac{1}{2} mv^{2} \\E=mgh\\E=\frac{1}{2} kx^{2} \\E=\frac{1}{2} k2kx^{2}[/tex]
Answer:
Explanation:
The energy conservation is equal to half of the product of the spring constant and the square of displacement of the spring, so option C is correct.
calculate the magnitude of the gravitational field of the sun at the location of earth, in meters per square second.
The magnitude of the gravitational field of the Sun at the location of Earth is approximately 9.81 m/s2.
This value is derived from Newton's Law of Universal Gravitation, which states that the gravitational force (F) between two objects is equal to the product of the two objects' masses (m1 and m2) multiplied by the gravitational constant (G) divided by the square of the distance between the two objects (r2):
F = G * m1 * m2 / r2
The mass of the Sun is 1.989 × 1030 kg, and the average distance between Earth and the Sun is 1.496 × 1011 meters. Therefore, plugging those values into the equation gives us:
F = 6.67 × 10-11 * 1.989 × 1030 * 5.972 × 1024 / (1.496 × 1011)2
F = 9.81 m/s2
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discuss whether the values found in parts (a) and (b) seem consistent with the fact that tidal effects with earth have caused the moon to rotate with one side always facing earth.
Yes, the values found in parts (a) and (b) are consistent with the fact that tidal effects with earth have caused the moon to rotate with one side always facing earth.
This is because part (a) states that the moon rotates on its axis in the same amount of time it takes to complete one orbit around the Earth, which is a phenomenon known as tidal locking. Part (b) further indicates that the same side of the moon always faces the Earth, further supporting the notion that tidal effects have caused the moon to rotate with one side always facing Earth.
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A 2 kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. The
object's speed after falling for 3 sis 75 m/s. Air resistance is considered to be negligible, Calculate the weight of the 2 kg object on the planet of unknown mass.
2N
B
25 N
50N
D
75 N
The magnitude of the force between two point charges 1. 0 m
apart is 9 x 10°n. If the distance between them is doubled,
what does the force become?
Force will become 2.25 x 10^N. because, According to Coulomb's Law, the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Thus, if the distance between two point charges is doubled, the force between them will decrease by a factor of 4. This is because the inverse square relationship means that the force decreases rapidly with distance. Therefore, if the force between two point charges is 9 x 10^N when they are 1 meter apart, when the distance is doubled to 2 meters, the force will become 9 x 10^N / 4 = 2.25 x 10^N.
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the end result of a theory that is not verified is
Unproven theories ultimately cannot be regarded as scientific facts or principles and are not generally recognised by the scientific community.
A well-supported explanation of a natural occurrence in science that has passed rigorous examination and is backed by empirical data is referred to as a theory. A hypothesis, however, cannot be regarded as a scientific fact or principle if it is not backed up by empirical data or if it has not undergone extensive testing and verification. The scientific community frequently rejects unproven notions with scant empirical backing and may even label them as pseudoscientific or non-scientific. This is so that scientific theories and findings may be evaluated and verified frequently. Science does this by using evidence-based reasoning and critical thinking. Unproven theories are therefore eventually not regarded as being a part of the corpus of scientific knowledge.
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an object is moving to the right in a straight line. the net force acting on the object is also directed to the right, but the magnitude of the force is decreasing with time. the object will
The object will decelerate over time, as the net force acting on it decreases. This is because the net force is the vector sum of all forces acting on the object.
What is the effect on object?When an object is moving to the right in a straight line, and the net force acting on the object is also directed to the right, it means that there is no opposing force to halt its motion.
Therefore, the object will continue to move to the right in a straight line with constant speed since there is no change in the magnitude of the net force.
However, when the net force is directed to the right and is decreasing with time, the object's motion will be altered. The magnitude of the force is decreasing with time, so there will be less force acting on the object.
The force acting on the object is decreasing with time; thus, the object's acceleration will be less than before. As a result, the velocity of the object will decrease with time. Since there is no force opposing the motion, the object will continue to move to the right but with decreasing speed due to the decrease in net force acting on it.
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Find the fourier series of f(x)=x
for 0<=x<=2
The function f(x) = x, where 0 x 2, has the following Fourier series: Given that f(x) has an odd period of 2, we may express its Fourier series as follows: F(x) = A0 + [n=1 to] Ancos (n/x) plus bnsin (n/x).
Since f(x) is an odd function, a0 = 0. We may apply the following formulae to determine the Fourier coefficients: a = (2/1) f(x)cos(nx/1)[0 to 1] dx Bn = (2/1) f(x)sin(nx/1)[0 to 1] dx We may determine the coefficients using the following formulas: an is equal to (2/1) [0 to 1] x*cos(nx/1) dx. Bn is equal to (2/1), [0 to 1]x*sin(nx/1)dx. By integrating in pieces, we obtain: a = (2/π^2) [(1-(-1)^n)/(n^2)] bn = (2/π) [(1-(-1)^n)/(n)] The Fourier series of f(x) = x, where 0 x its Fourier series as follows: F(x) = A0 + [n=1 to] Ancos (n/x) plus bnsin (n/x).2, is as follows: f(x) = Σ(n=1 to ∞) [(2/) (1-(-1)n)/(n))*sin(nx/1)].
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Solve the circuit shown in the figure above, also explain how you did it
Answer:
Explanation:
Using Kirchhoff's laws, we can solve for the current i:
At the node where the 2Ω and 4Ω resistors meet, the current is split into two branches, i and i1. Applying Kirchhoff's current law (KCL), we have:
i + i1 = 12/2 = 6 A
At the loop with the 2Ω, 4Ω, and 5Ω resistors, applying Kirchhoff's voltage law (KVL), we have:
-20 + 2i + 4i1 + 5i1 = 0
-20 + 6i1 + 2i = 0
6i1 + 2i = 20
3i1 + i = 10
We can solve this system of equations by substitution, which gives:
i = 2 A
Therefore, the current through the 2Ω resistor is 2 A. The answer is (A) 2 A.
polar stratospheric clouds are high-altitude clouds made of
Polar stratospheric clouds are high-altitude clouds made of tiny ice crystals that form in the lower stratosphere at very cold temperatures. They exhibit vivid iridescent colors and are associated with ozone depletion.
High-altitude clouds comprised of microscopic ice crystals are referred to as polar stratospheric clouds, nacreous clouds, or mother-of-pearl clouds. At heights of around 15,000 to 25,000 meters and extremely low temperatures of minus 80 to minus 85 degrees Celsius, they occur in the lower stratosphere. The refraction of sunlight as it passes through the ice crystals gives these clouds their distinctive dazzling and vibrant iridescent colors. Polar stratospheric clouds, which are linked to the ozone layer's thinning, are most frequently seen during the winter in polar locations like the Arctic and Antarctic.
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uestion 8: Electron Two-Slit Interference Proctor A beam of electrons with velocity 15.0 m/s pass through two slits separated by 0.500 mm. We place a detector on a distant screen. At which angle measured from the horizontal can we be sure we never detect an electron. (a) The electron could be detected anywhere. O (b) 0.00 rad O (C) 0.0485 rad O (d) 0.195 rad (e) 0.0971 rad Save 5 points available for this attempt
The angle measured from the horizontal is where we can be sure that we never detect an electron is 0.0485 rad.
The correct answer is C.
To find the angle at which we can be sure to never detect an electron, we will use the equation:
θ = λ/d
Where:
θ = angle at which we can be sure to never detect an electron
λ = de Broglie wavelength of the electron = h/p
where h = Planck's constant and p = momentum of electron (m*v)
We know that the velocity of the electron is 15.0 m/s. To find the momentum, we can use the mass of an electron (9.11 x 10⁻³¹ kg).
p = m*v = (9.11 x 10⁻³¹ kg) * (15.0 m/s)
p = 1.37 x 10⁻²⁹ kg m/s
Now we can find the de Broglie wavelength:
λ = h/p
λ = (6.626 x 10⁻³⁴ J s) / (1.37 x 10⁻²⁹ kg m/s)
λ = 4.83 x 10⁻⁵ m
Now we can substitute this into the equation for θ:
θ = λ/d
θ = (4.83 x 10⁻⁵ m)/(0.500 x 10⁻³ m)
θ = 0.0966 rad
However, this is the angle at which we would see destructive interference. To be sure we never detect an electron, we want the angle to be half of this, or:
θ = 0.0966/2
θ = 0.0483 rad, which rounds to 0.0485 rad (to 3 significant figures).
Therefore, the correct answer is (c) 0.0485 rad.
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What is the electromagnetic force?A. a force that governs how elements break down naturallyB. a force that holds atomic nuclei togetherC. a force that attracts objects with mass towards each otherD. a force that acts on charged particles
Option D. The electromagnetic force is a force that acts on charged particles.
The electromagnetic force is a fundamental force of nature that acts on charged particles. It is one of the four fundamental forces of nature, the other three being the strong nuclear force, the weak nuclear force, and gravity. The electromagnetic force is responsible for all electromagnetic phenomena, including electricity, magnetism, and electromagnetic radiation. Charge is the property of matter that is responsible for the electromagnetic force.
All particles that have a charge, including electrons and protons, interact with the electromagnetic force. The electromagnetic force is mediated by the electromagnetic field, which is created by charged particles. When charged particles move, they create electromagnetic waves, which can travel through space at the speed of light.
The electromagnetic force is responsible for a wide range of phenomena, including the structure of atoms, the behavior of magnets, and the behavior of light. It is a very strong force, much stronger than the weak nuclear force and gravity, but weaker than the strong nuclear force. The electromagnetic force is responsible for the repulsion between like charges and the attraction between opposite charges. It is also responsible for the behavior of magnetic materials, such as iron, which can be magnetized by an external magnetic field.
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Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 12 ounces. a. The process standard deviation is 0.14, and the process control is set at plus or minus 2.4 standard deviations. Units with weights less than 11.664 or greater than 12.336 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)? In a production run of 1000 parts, how many defects would be found to the nearest whole number)? b. Through process design improvements, the process standard deviation can be reduced to 0.12. Assume the process control remains the same, with weights less than 11.664 or greater than 12.336 ounces being classified as defects. What is the probability of a defect (to 4 decimals)? In a production run of 1000 parts, how many defects would be found to the nearest whole number)?
a. To calculate the probability of a defect, we need to find the area under the normal distribution curve that falls outside the control limits of 11.664 and 12.336 ounces. We can calculate the z-scores for these limits as follows:
[tex]z_1 = (11.664 - 12) / 0.14 = -2.4[/tex]
[tex]z_2 = (12.336 - 12) / 0.14 = 2.4[/tex]
Using a standard normal distribution table or calculator, we can find that the probability of a defect is approximately 0.0115 (to 4 decimals).
To find the expected number of defects in a production run of 1000 parts, we can use the formula for the binomial distribution:
[tex]P(X = k) = C(n, k) \times p^k \times (1-p)^{(n-k)}[/tex]
where P(X = k) is the probability of exactly k defects in a run of n parts, p is the probability of a single defect (0.0115 in this case), and C(n, k) is the binomial coefficient (the number of ways to choose k defects from n parts).
For k = 0, 1, 2, ..., we can calculate the probabilities and add them up to find the expected number of defects:
E(X) = sum(k=0 to n) [ P(X = k) ] = n * p
Substituting n = 1000 and p = 0.0115, we get:
[tex]E(X) = 1000 \times 0.0115 = 11.5[/tex]
So we can expect to find approximately 12 defects (to the nearest whole number) in a production run of 1000 parts.
b. With a reduced process standard deviation of 0.12, the z-scores for the control limits remain the same as in part a:
[tex]z_1 = (11.664 - 12) / 0.12 = -2.8[/tex]
[tex]z_2 = (12.336 - 12) / 0.12 = 2.8[/tex]
Using a standard normal distribution table or calculator, we can find that the probability of a defect is approximately 0.0004 (to 4 decimals).
To find the expected number of defects in a production run of 1000 parts, we can use the same formula as in part a:
[tex]E(X) = n \times p[/tex]
Substituting n = 1000 and p = 0.0004, we get:
[tex]E(X) = 1000 \times 0.0004 = 0.4[/tex]
So we can expect to find approximately 0 defects (to the nearest whole number) in a production run of 1000 parts.
However, it's important to note that this assumes the process is operating exactly at the mean weight of 12 ounces and there is no other source of variation. In practice, there may still be some small amount of variation that could result in a few defects.
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