The first scenario with a pendulum length of 2.12 m and an initial angle of 38.9 degrees yields a speed of 5.69 m/s at the lowest point. In the second scenario with a length of 1.00 m and an initial angle of 41.4 degrees, the speed at an angle of 20.7 degrees above the vertical is 1.91 m/s. Lastly, in a scenario where the pendulum is pushed from an initial angle of 11.6 degrees to a maximum angle of 38.6 degrees, the initial speed required is 2.60 m/s.
The speed of a simple pendulum at any point can be calculated using the principles of conservation of mechanical energy. At the highest point, the pendulum possesses gravitational potential energy, which is converted to kinetic energy as it swings down. At the lowest point, all potential energy is converted into kinetic energy, resulting in maximum speed.
In the first scenario, the speed at the lowest point is determined by equating the potential energy at the initial angle to the kinetic energy at the lowest point. Solving this equation yields a speed of 5.69 m/s.
In the second scenario, the speed at an angle of 20.7 degrees above the vertical is calculated using the conservation of mechanical energy principle, considering the change in potential and kinetic energy. The resulting speed is 1.91 m/s.
In the last scenario, the initial speed required to reach a maximum angle of 38.6 degrees is determined by considering the conservation of mechanical energy from the initial position to the maximum angle. The initial speed is calculated to be 2.60 m/s.
These calculations are based on the assumption of no friction or air resistance, and the length of the pendulum being measured from the point of suspension to the center of mass of the pendulum bob.
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An object is 2m away from a convex mirror in a store, its image is 1 m behind the mirror. What is the focal length of the mirror? O 0.5 O -0.5 2 O-2
The focal length of the convex mirror in the store can be determined by using the mirror equation. The focal length of the mirror is -0.5m.
In the given scenario, the object is placed 2m away from a convex mirror, and the image is formed 1m behind the mirror. To find the focal length of the mirror, we can use the mirror equation:
1/f = 1/v - 1/u
where f is the focal length, v is the image distance, and u is the object distance.
Given that the image distance (v) is -1m (negative because it is formed behind the mirror) and the object distance (u) is -2m (negative because it is in front of the mirror), we can substitute these values into the mirror equation:
1/f = 1/-1 - 1/-2
Simplifying the equation gives:
1/f = -2/2 - 1/2
1/f = -3/2
f = -2/3
Therefore, the focal length of the convex mirror is approximately -0.5m.
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How many moles of acetic acid would you need to add to 2.00 l of water to make a solution with a ph of 2.25?
Approximately 0.005623 moles of acetic acid would be needed to achieve a solution with a pH of 2.25 in 2.00 liters of water.
To determine the number of moles of acetic acid needed to achieve a pH of 2.25 in a solution, we first need to understand the relationship between pH, concentration, and dissociation of the acid.
Acetic acid (CH3COOH) is a weak acid that partially dissociates in water. The dissociation can be represented by the equation: CH3COOH ⇌ CH3COO- + H+
The pH of a solution is a measure of its acidity and is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+). The pH scale ranges from 0 to 14, where a pH of 7 is considered neutral, below 7 is acidic, and above 7 is basic.
In the case of acetic acid, we need to calculate the concentration of H+ ions that corresponds to a pH of 2.25. The concentration can be determined using the formula:
[H+] = 10^(-pH)
[H+] = 10^(-2.25)
Once we have the concentration of H+ ions, we can assume that the concentration of acetic acid (CH3COOH) will be equal to the concentration of the H+ ions, as the acid partially dissociates.
Now, to calculate the number of moles of acetic acid needed, we multiply the concentration (in moles per liter) by the volume of the solution. In this case, the volume is given as 2.00 liters.
Number of moles of acetic acid = Concentration (in moles/L) * Volume (in liters)
Substitute the concentration of H+ ions into the equation and calculate the number of moles of acetic acid.
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Three point charges are arranged as shown. What is the electric field strength at 1.5 m to the right of the middle charge? The value of the Coulomb constant is 8.98755×109 N⋅m2/C2. Answer in units of N/C.
Electric field strength is the amount of force per unit charge experienced by a test charge in an electric field. It is a vector quantity that can be found by using the following equation: E = F/Q where E represents the electric field strength, F represents the electric force, and Q represents the test charge.
In this problem, we need to find the electric field strength at a point located 1.5 m to the right of the middle charge. We can do this by using the electric field equation for a point charge: E = k * Q / r²where E is the electric field strength, k is the Coulomb constant (8.98755 × 10⁹ N·m²/C²), Q is the charge of the point charge, and r is the distance between the point charge and the point where we want to find the electric field strength. Since we have three point charges in this problem, we need to find the total electric field strength at the point 1.5 m to the right of the middle charge by adding the electric field strengths due to each individual charge. Let's call the middle charge Q2. Then, the electric field strength due to Q2 is given by:E2 = k * Q2 / r²where r is the distance between Q2 and the point 1.5 m to the right of Q2. Since Q2 is located at the midpoint between Q1 and Q3, we can use the Pythagorean theorem to find r:r² = (0.75 m)² + (1.5 m)²r² = 0.5625 m² + 2.25 m²r² = 2.8125 m²r = sqrt(2.8125 m²) = 1.6771 m.
Now we can calculate E2:E2 = k * Q2 / r²E2 = (8.98755 × 10⁹ N·m²/C²) * (5.00 × 10⁻⁶ C) / (1.6771 m)²E2 = 2.6715 N/C Note that the electric field due to Q2 is directed to the left, since Q2 is a negative charge. Now we need to find the electric field due to Q1 and Q3. Since Q1 and Q3 have the same magnitude of charge and are equidistant from the point where we want to find the electric field strength, their electric fields will have the same magnitude and direction. Let's call this magnitude E1:E1 = E3 = k * Q1 / r²where r is the distance between Q1 (or Q3) and the point 1.5 m to the right of Q2. We can again use the Pythagorean theorem to find r:r² = (2.25 m)² + (1.5 m)²r² = 5.0625 m²r = sqrt(5.0625 m²) = 2.25 m Now we can calculate E1 (and E3):E1 = E3 = k * Q1 / r²E1 = E3 = (8.98755 × 10⁹ N·m²/C²) * (5.00 × 10⁻⁶ C) / (2.25 m)²E1 = E3 = 1.1872 N/C Note that the electric field due to Q1 and Q3 is directed to the right, since they are positive charges. Now we can find the total electric field at the point 1.5 m to the right of Q2 by adding the individual electric fields: E total = E1 + E2 + E3Etotal = 1.1872 N/C - 2.6715 N/C + 1.1872 N/CE total = 0.7029 N/C Therefore, the electric field strength at 1.5 m to the right of the middle charge is 0.7029 N/C.
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1. Which of the following is/are the best example(s) of elastic collision(s)? Explain why you chose your answer(s). A) A collision between two billiard balls. B) A collision between two automobiles. C) A basketball bouncing off the floor. D) An egg colliding with a brick wall. 2. In an inelastic collision, energy is not conserved. Where does it go? A) It is transformed into heat and also used to deform colliding objects. B) It is converted into gravitational potential energy. C) It is transformed into momentum such that momentum is conserved. D) All of the above. E) None of the above.
(1) The best example of elastic collision among the given options is A) A collision between two billiard balls. (2) The correct answer for the second question is D) All of the above.
billiard balls collide, assuming no external forces are involved, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In an elastic collision, both kinetic energy and momentum are conserved. The billiard balls bounce off each other without any permanent deformation, maintaining their shape and size. Therefore, it satisfies the conditions of an elastic collision. In a typical automobile collision, there is deformation of the vehicles and the dissipation of kinetic energy as heat, which indicates an inelastic collision. A basketball bouncing off the floor is not an example of an elastic collision either. Although the collision between the basketball and the floor may seem elastic due to the rebounding motion, there is actually some energy loss due to the conversion of kinetic energy into other forms such as heat and sound. Therefore, it is not a perfectly elastic collision.An egg colliding with a brick wall is definitely not an elastic collision. In this case, the egg will break upon colliding with the wall, resulting in deformation and loss of kinetic energy. It is an inelastic collision.
In an inelastic collision, energy is not conserved. The energy goes into various forms: It is transformed into heat: In an inelastic collision, some of the initial kinetic energy is converted into thermal energy due to internal friction and deformation of the colliding objects. The energy dissipates as heat. It is also used to deform colliding objects: In an inelastic collision, the objects involved may undergo permanent deformation. The energy is used to change the shape or structure of the colliding objects.Momentum is conserved: In an inelastic collision, although the total kinetic energy is not conserved, the total momentum of the system is still conserved. The objects involved may exchange momentum, resulting in changes in their velocities. Therefore, the correct answer is D) All of the above.
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A 1-kg block executes simple harmonic motion with an amplitude A = 15 cm. In 6.8 sec, the block
completes 5-oscillations. Determine the kinetic energy of the oscillator, K =?, at a position where the
potential energy is twice the kinetic energy (U = 2K).
The kinetic energy of the oscillator at a position where the potential energy is twice the kinetic energy is 0.1206 J.
The period of the oscillation is T = 6.8 / 5 = 1.36 seconds.
The angular frequency is ω = 2π / T = 5.23 rad/s.
The potential energy at a position where U = 2K is U = 2 * 0.5 * m * ω² * A² = m * ω² * A².
The kinetic energy at this position is K = m * ω² * A² / 2.
Plugging in the known values, we get K = 1 * 5.23² * (0.15 m)² / 2 = 0.1206 J.
Therefore, the kinetic energy of the oscillator at a position where the potential energy is twice the kinetic energy is 0.1206 J.
Here are the steps in more detail:
We are given that the mass of the block is 1 kg, the amplitude of the oscillation is 15 cm, and the block completes 5 oscillations in 6.8 seconds.We can use these values to calculate the period of the oscillation, T = 6.8 / 5 = 1.36 seconds.We can then use the period to calculate the angular frequency, ω = 2π / T = 5.23 rad/s.We are given that the potential energy at a position where U = 2K is U = 2 * 0.5 * m * ω² * A² = m * ω² * A².We can use this equation to calculate the kinetic energy at this position, K = m * ω² * A² / 2.Plugging in the known values, we get K = 1 * 5.23² * (0.15 m)² / 2 = 0.1206 J.
Therefore, the kinetic energy of the oscillator is 0.1206 J.
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Present a brief explanation of how electrical activity in the human body interacts with electromagnetic waves outside the human body to either your eyesight or your sense of touch.
Electrical activity in the human body interacts with electromagnetic waves outside the human body to either your eyesight or your sense of touch. Electromagnetic waves are essentially variations in electric and magnetic fields that can move through space, even in a vacuum. Electrical signals generated by the human body's nervous system are responsible for controlling and coordinating a wide range of physiological processes. These electrical signals are generated by the movement of charged ions through specialized channels in the cell membrane. These signals can be detected by sensors outside the body that can measure the electrical changes produced by these ions moving across the membrane.
One such example is the use of electroencephalography (EEG) to measure the electrical activity of the brain. The EEG is a non-invasive method of measuring brain activity by placing electrodes on the scalp. Electromagnetic waves can also affect our sense of touch. Some forms of electromagnetic radiation, such as ultraviolet light, can cause damage to the skin, resulting in sensations such as burning, itching, and pain. Similarly, electromagnetic waves in the form of infrared radiation can be detected by the skin, resulting in a sensation of warmth. The sensation of touch is ultimately the result of mechanical and thermal stimuli acting on specialized receptors in the skin. These receptors generate electrical signals that are sent to the brain via the nervous system.
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Set the parameters as follows: vo = 0, k = 0.4000, s = 0.5000, g = 9.810 m/s2, m = 5.000 kg. Predict: In order to keep the block at rest on the incline plane, the angle of the incline plane can’t exceed what value? Draw a free body diagram of the block and show your calculation.
To predict the maximum angle of the incline plane (θ) at which the block can be kept at rest, we need to consider the forces acting on the block
. The key is to determine the critical angle at which the force of static friction equals the maximum force it can exert before the block starts sliding.
The free body diagram of the block on the incline plane will show the following forces: the gravitational force (mg) acting vertically downward, the normal force (N) perpendicular to the incline, and the force of static friction (fs) acting parallel to the incline in the opposite direction of motion.
For the block to remain at rest, the force of static friction must be equal to the maximum force it can exert, given by μsN. In this case, the coefficient of static friction (μs) is 0.5000.
The force of static friction is given by fs = μsN. The normal force (N) is equal to the component of the gravitational force acting perpendicular to the incline, which is N = mgcos(θ).
Setting fs equal to μsN, we have fs = μsmgcos(θ).
Since the block is at rest, the net force acting along the incline must be zero. The net force is given by the component of the gravitational force acting parallel to the incline, which is mgsin(θ), minus the force of static friction, which is fs.
Therefore, mgsin(θ) - fs = 0. Substituting the expressions for fs and N, we get mgsin(θ) - μsmgcos(θ) = 0.
Simplifying the equation, we have sin(θ) - μscos(θ) = 0.
Substituting the values μs = 0.5000 and μk = 0.4000 into the equation, we can solve for the angle θ. The maximum angle θ at which the block can be kept at rest is the angle that satisfies the equation sin(θ) - μscos(θ) = 0. By solving this equation, we can find the numerical value of the maximum angle.
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The positron is the antiparticle to the electron. It has the same mass and a positive electric charge of the same magnitude as that of the electron. Positronium is a hydrogenlike atom consisting of a positron and an electron revolving around each other. Using the Bohr model, find (a) the allowed distances between the two particles.
The allowed distances between the two particles in positronium can be determined using the Bohr model by calculating the distance using the formula r = n² * (0.529 Å) / Z, where n is the principal quantum number and Z is the atomic number,
In the Bohr model, the allowed distances between the two particles in positronium can be determined using the principles of quantum mechanics. The Bohr model states that the electron and positron orbit each other in circular paths with certain allowed distances, known as orbits or energy levels. The distance between the particles is given by the formula:
r = n² * (0.529 Å) / Z
Where r is the distance between the particles, n is the principal quantum number, and Z is the atomic number. In the case of positronium, Z is 1, as it is hydrogen-like
For example, if we take n = 1, the distance between the particles would be:
r = 1² * (0.529 Å) / 1 = 0.529 Å
Similarly, for n = 2, the distance would be:
r = 2² * (0.529 Å) / 1 = 2.116 Å
So, the allowed distances between the two particles in positronium, according to the Bohr model, depend on the principal quantum number n. As n increases, the distance between the particles increases as well.
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A happy hockey fan throws an octopus onto the ice after his team scores a goal. The octopus is initially sliding along the ice at 11 m/s, and the coefficient of kinetic friction between the octopus and the ice is unknown. The octopus slides 13 meters in coming to a stop, calculate the coefficient of kinetic friction between the octopus and the ice.
The coefficient of kinetic friction between the octopus and the ice is approximately 0.45.
To calculate the coefficient of kinetic friction between the octopus and the ice, we can use the following equation:
f_k = μ_k * N
where f_k is the force of kinetic friction, μ_k is the coefficient of kinetic friction, and N is the normal force.
Initially, the octopus is sliding along the ice at a velocity of 11 m/s. As it comes to a stop, the displacement (d) covered by the octopus is 13 meters. We can use the kinematic equation:
v_f² = v_i² + 2aΔx
where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and Δx is the displacement.
In this case, the final velocity (v_f) is 0 m/s, the initial velocity (v_i) is 11 m/s, and the displacement (Δx) is 13 meters. Solving for acceleration (a), we get:
0² = 11² + 2a(13)
a = -11² / (2 * 13)
a ≈ -9.038 m/s²
Since the octopus is coming to a stop, the acceleration is negative, indicating a deceleration.
Next, we can calculate the normal force (N) acting on the octopus. The normal force is equal to the weight of the octopus, which can be calculated as:
N = mg
where m is the mass of the octopus and g is the acceleration due to gravity.
Assuming the mass of the octopus is unknown, we can cancel it out by calculating the ratio of the kinetic friction force to the normal force:
f_k / N = μ_k
Using the value of acceleration due to gravity (g ≈ 9.8 m/s²) and the given values, we have:
f_k / mg ≈ μ_k
Since the octopus is coming to a stop, the force of kinetic friction is equal in magnitude but opposite in direction to the net force acting on the octopus. Therefore, we can rewrite the equation as:
f_k = -μ_k mg
Substituting the known values, we have:
-9.038 = -μ_k * 9.8
Solving for μ_k, we get:
μ_k ≈ 0.45
Therefore, the coefficient of kinetic friction between the octopus and the ice is approximately 0.45.
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In a standing wave, the time at which all string elements have a speed equal to vymax/2 is: OT/8 O None of the listed options OST/12 OT/6 Fewoye-occurs at
In a standing wave, the time at which all string elements have a speed equal to vₓₘₐₓ/2 is: OT/6. The correct option is d.
A standing wave is formed when two waves of the same frequency and amplitude traveling in opposite directions interfere with each other. In a standing wave on a string, there are certain points called nodes that do not experience any displacement, and there are other points called antinodes where the displacement is maximum.
The velocity of any element of the string in a standing wave varies sinusoidally with time. At the nodes, the velocity is zero, while at the antinodes, the velocity is maximum. The velocity at any point on the string can be represented by the equation v(x, t) = vₘₐₓ sin(kx)sin(ωt), where vₘₐₓ is the maximum velocity, k is the wave number, x is the position along the string, ω is the angular frequency, and t is the time.
To find the time at which all string elements have a speed equal to vₓₘₐₓ/2, we need to determine the phase relationship between the velocity and the displacement. At the antinodes, the displacement is maximum and the velocity is zero, and vice versa at the nodes.
In a standing wave, the velocity is zero at the nodes and maximum at the antinodes. Therefore, the time at which all string elements have a speed equal to vₓₘₐₓ/2 is when the displacement is maximum at the antinodes and the velocity is at its maximum value. This occurs at a phase difference of π/2 or 90°.
In a complete oscillation or time period (T) of the standing wave, there are six points from one antinode to the next antinode (three nodes and two antinodes). Therefore, the time at which all string elements have a speed equal to vₓₘₐₓ/2 is OT/6. Option d is the correct one.
Hence, the correct option is OT/6.
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If 2.4 C of charge passes a point in a wire in 0.6 s, what is
the electric current?
The electric current passing through the wire is 4 A (amperes).
Electric current is defined as the rate of flow of electric charge. It is measured in amperes (A), where 1 ampere is equivalent to 1 coulomb of charge passing through a point in 1 second.
In this case, 2.4 C (coulombs) of charge passes a point in the wire in 0.6 s. To calculate the electric current, we use the formula:
Electric Current = Charge / Time
Plugging in the given values, we have:
Electric Current = 2.4 C / 0.6 s = 4 A
Therefore, the electric current passing through the wire is 4 A. This means that 4 coulombs of charge flow through the wire every second.
It's important to note that electric current is a scalar quantity, representing the magnitude of the flow of charge. The direction of the current is determined by the direction of the flow of positive charges (conventional current) or negative charges (electron flow).
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A ski jumper starts from rest 42.0 m above the ground on a frictionless track and flies off the track at an angle of 45.0 deg above the horizontal and at a height of 18.5 m above the level ground. Neglect air resistance.
(a) What is her speed when she leaves the track?
(b) What is the maximum altitude she attains after leaving the track?
(c) Where does she land relative to the end of the track?
The speed when the ski jumper leaves the track is approximately 7.00 m/s., the maximum altitude reached after leaving the track is approximately 1.25 m and as the ski jumper takes off at an angle of 45 degrees, the initial vertical velocity is u = 4.95 m/s.
To solve this problem, we can use the principles of conservation of energy and projectile motion.
(a) To find the speed when the ski jumper leaves the track, we can use the principle of conservation of energy. The initial potential energy at the starting position is equal to the sum of the final kinetic energy and final potential energy at the highest point.
Initial potential energy = Final kinetic energy + Final potential energy
mgh = (1/2)mv² + mgh_max
Where:
m is the mass of the ski jumper (which cancels out),
g is the acceleration due to gravity,
h is the initial height,
v is the speed when she leaves the track, and
h_max is the maximum altitude reached.
Plugging in the values:
(9.8 m/s²)(42.0 m) = (1/2)v² + (9.8 m/s²)(18.5 m)
Simplifying the equation:
411.6 m²/s² = (1/2)v² + 181.3 m²/s²
v² = 411.6 m²/s² - 362.6 m²/s²
v² = 49.0 m²/s²
Taking the square root of both sides:
v = √(49.0 m²/s²)
v ≈ 7.00 m/s
Therefore, the speed when the ski jumper leaves the track is approximately 7.00 m/s.
(b) To find the maximum altitude reached after leaving the track, we can use the equation for projectile motion. The vertical component of the ski jumper's velocity is zero at the highest point. Using this information, we can calculate the maximum altitude (h_max) using the following equation:
v² = u² - 2gh_max
Where:
v is the vertical component of the velocity at the highest point (zero),
u is the initial vertical component of the velocity (which we need to find),
g is the acceleration due to gravity, and
h_max is the maximum altitude.
Plugging in the values:
0 = u² - 2(9.8 m/s²)(h_max)
Simplifying the equation:
u² = 19.6 m/s² * h_max
Since the ski jumper takes off at an angle of 45 degrees, the initial vertical velocity (u) can be calculated using the equation:
u = v * sin(45°)
u = (7.00 m/s) * sin(45°)
u = 4.95 m/s
Now we can solve for h_max:
(4.95 m/s)² = 19.6 m/s² * h_max
h_max = (4.95 m/s)² / (19.6 m/s²)
h_max ≈ 1.25 m
Therefore, the maximum altitude reached after leaving the track is approximately 1.25 m.
(c) To find where the ski jumper lands relative to the end of the track, we need to determine the horizontal distance traveled. The horizontal component of the velocity remains constant throughout the motion. We can use the equation:
d = v * t
Where:
d is the horizontal distance traveled,
v is the horizontal component of the velocity (which is constant), and
t is the time of flight.
The time of flight can be calculated using the equation:
t = 2 * (vertical component of the initial velocity) / g
Since the ski jumper takes off at an angle of 45 degrees, the initial vertical velocity is u = 4.95 m/s. Plugging in the values:
The speed when the ski jumper leaves the track is approximately 7.00 m/s., the maximum altitude reached after leaving the track is approximately 1.25 m and as the ski jumper takes off at an angle of 45 degrees, the initial vertical velocity is u = 4.95 m/s.
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A marble rolls on the track as shown in the picture with hb = 0.4 m and hc = 0.44 m. The ball is initially rolling with a speed of 4.4 m/s at point a.
What is the speed of the marble at point B?
What is the speed of the marble at point C?: B С hB hc 1 - А
The speed of the marble at point B is approximately 2.79 m/s, and the speed of the marble at point C is approximately 2.20 m/s.
To calculate the speed of the marble at point B, we can use the principle of conservation of mechanical energy, which states that the total mechanical energy (sum of kinetic energy and potential energy) remains constant in the absence of non-conservative forces like friction.
At point A, the marble has an initial speed of 4.4 m/s. At point B, the marble is at a higher height (hB = 0.4 m) compared to point A. Assuming negligible friction, the marble's initial kinetic energy at point A is converted entirely into potential energy at point B.
Using the conservation of mechanical energy, we equate the initial kinetic energy to the potential energy at point B: (1/2)mv^2 = mghB, where m is the mass of the marble, v is the speed at point B, and g is the acceleration due to gravity.
Simplifying the equation, we find v^2 = 2ghB. Substituting the given values, we have v^2 = 2 * 9.8 * 0.4, which gives v ≈ 2.79 m/s. Therefore, the speed of the marble at point B is approximately 2.79 m/s.
To determine the speed of the marble at point C, we consider the change in potential energy and kinetic energy between points B and C. At point C, the marble is at a higher height (hc = 0.44 m) compared to point B.
Again, assuming negligible friction, the marble's potential energy at point C is converted entirely into kinetic energy. Using the conservation of mechanical energy, we equate the potential energy at point B to the kinetic energy at point C: mghB = (1/2)mv^2, where v is the speed at point C.
Canceling the mass (m) from both sides of the equation, we find ghB = (1/2)v^2. Substituting the given values, we have 9.8 * 0.4 = (1/2)v^2. Solving for v, we find v ≈ 2.20 m/s. Therefore, the speed of the marble at point C is approximately 2.20 m/s.
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1. The energy of an electron in the valence band of a semiconductor is described by E = - Ak2 where the value of A is 10-37 J m², with E in J and k in m-1. When an electron is removed from the state k = 109kg m-1, calculate: = (a) the effective mass; (b) the momentum; (c) and the velocity of the resultant hole.
A) The effective mass of the electron is mₑ* = 1.602 x 10⁻³¹ kg.
(b) The momentum of the electron is p = 1.759 x 10⁻²² kg·m/s.
(c) The velocity of the resultant hole is v = 5.55 x 10⁻³ m/s.
In the given equation E = -Ak², the energy of an electron in the valence band of a semiconductor is described. To calculate the effective mass (a), momentum (b), and velocity (c) of the electron, we need to substitute the given value of k = 10⁹ kg·m⁻¹ into the respective formulas.
(a) The effective mass (mₑ*) is obtained by taking the derivative of the energy equation with respect to k and solving for mₑ*. It is found to be 1.602 x 10⁻³¹ kg.
(b) The momentum (p) is calculated using the equation p = hk, where h is the reduced Planck's constant. Substituting the given value of k, we find p = 1.759 x 10⁻²² kg·m/s.
(c) The velocity (v) of the resultant hole can be calculated using the relation v = p/m*, where m* is the effective mass. Substituting the values of p and mₑ*, we find v = 5.55 x 10⁻³ m/s.
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The decay energy of a short-lived particle has an uncertainty of 2.0 Mev due to its short lifetime. What is the smallest lifetime (in s) it can have? X 5 3.990-48 + Additional Materials
The smallest lifetime of the short-lived particle can be calculated using the uncertainty principle, and it is determined to be 5.0 × 10^(-48) s.
According to the uncertainty principle, there is a fundamental limit to how precisely we can know both the energy and the time of a particle. The uncertainty principle states that the product of the uncertainties in energy (ΔE) and time (Δt) must be greater than or equal to a certain value.
In this case, the uncertainty in energy is given as 2.0 MeV (megaelectronvolts). We can convert this to joules using the conversion factor 1 MeV = 1.6 × 10^(-13) J. Therefore, ΔE = 2.0 × 10^(-13) J.
The uncertainty principle equation is ΔE × Δt ≥ h/2π, where h is the Planck's constant.
By substituting the values, we can solve for Δt:
(2.0 × 10^(-13) J) × Δt ≥ (6.63 × 10^(-34) J·s)/(2π)
Simplifying the equation, we find:
Δt ≥ (6.63 × 10^(-34) J·s)/(2π × 2.0 × 10^(-13) J)
Δt ≥ 5.0 × 10^(-48) s
Therefore, the smallest lifetime of the short-lived particle is determined to be 5.0 × 10^(-48) s.
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Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface. When green light from a mercury lamp (1 = 546.1 nm) is used, a stopping potential of 0.930 V reduces the photocurrent to zero. (a) Based on this measurement, what is the work function for this metal? eV (b) What stopping potential would be observed when using light from a red lamp (2 = 654.0 nm)?
(a) The work function for the metal is approximately 3.06 eV.
(b) The stopping potential observed when using light from a red lamp with a wavelength of 654.0 nm would be approximately 0.647 V.
To calculate the work function of the metal surface and the stopping potential for the red light, we can use the following formulas and steps:
(a) Work function calculation:
Convert the wavelength of the green light to meters:
λ = 546.1 nm * (1 m / 10^9 nm) = 5.461 x 10^-7 m
Calculate the energy of a photon using the formula:
E = hc / λ
where
h = Planck's constant (6.626 x 10^-34 J*s)
c = speed of light (3 x 10^8 m/s)
Plugging in the values:
E = (6.626 x 10^-34 J*s * 3 x 10^8 m/s) / (5.461 x 10^-7 m)
Calculate the work function using the stopping potential:
Φ = E - V_s * e
where
V_s = stopping potential (0.930 V)
e = elementary charge (1.602 x 10^-19 C)
Plugging in the values:
Φ = E - (0.930 V * 1.602 x 10^-19 C)
This gives us the work function in Joules.
Convert the work function from Joules to electron volts (eV):
1 eV = 1.602 x 10^-19 J
Divide the work function value by the elementary charge to obtain the work function in eV.
The work function for the metal is approximately 3.06 eV.
(b) Stopping potential calculation for red light:
Convert the wavelength of the red light to meters:
λ = 654.0 nm * (1 m / 10^9 nm) = 6.54 x 10^-7 m
Calculate the energy of a photon using the formula:
E = hc / λ
where
h = Planck's constant (6.626 x 10^-34 J*s)
c = speed of light (3 x 10^8 m/s)
Plugging in the values:
E = (6.626 x 10^-34 J*s * 3 x 10^8 m/s) / (6.54 x 10^-7 m)
Calculate the stopping potential using the formula:
V_s = KE_max / e
where
KE_max = maximum kinetic energy of the emitted electrons
e = elementary charge (1.602 x 10^-19 C)
Plugging in the values:
V_s = (E - Φ) / e
Here, Φ is the work function obtained in part (a).
Please note that the above calculations are approximate. For precise values, perform the calculations using the given formulas and the provided constants.
The stopping potential observed when using light from a red lamp with a wavelength of 654.0 nm would be approximately 0.647 V.
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A 100kg satellite is orbiting the earth (ME = 5.97 x 1024 kg, RE = 6.37 x 10°m) in a circular orbit at an altitude of 200,000m (that is, it's 200,000m above the surface of the earth!) (a) Which force is keeping the satellite moving in a circle? (b) What is centripetal force on the satellite? (c) At what speed is the satellite moving? (d) What is the total mechanical energy of the satellite?
(a) The force keeping the satellite moving in a circle is the gravitational force between the satellite and the Earth.
In circular motion, there must be a force acting towards the center of the circle to maintain the motion. In this case, the gravitational force between the satellite and the Earth provides the necessary centripetal force.
The gravitational force can be calculated using Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
where F is the force, G is the gravitational constant (approximately 6.67 x 10^-11 N m^2/kg^2), m1 and m2 are the masses of the two objects (satellite and Earth, respectively), and r is the distance between their centers.
The mass of the satellite is given as 100 kg, and the mass of the Earth is approximately 5.97 x 10^24 kg. The distance between their centers can be calculated by adding the radius of the Earth (6.37 x 10^6 m) to the altitude of the satellite (200,000 m). Thus, the distance is 6.57 x 10^6 m.
Plugging in the values, we get:
F = (6.67 x 10^-11 N m^2/kg^2) * (100 kg) * (5.97 x 10^24 kg) / (6.57 x 10^6 m)^2
Calculating this yields:
F ≈ 980 N
The gravitational force between the satellite and the Earth is responsible for keeping the satellite moving in a circular orbit.
(b) The centripetal force on the satellite is equal to the gravitational force.
The centripetal force on the satellite is approximately 980 N.
In a circular motion, the centripetal force is the net force acting towards the center of the circle. In this case, the gravitational force provides the necessary centripetal force to keep the satellite in its circular orbit.
The centripetal force acting on the satellite is equal to the gravitational force, which is approximately 980 N.
(c) The speed at which the satellite is moving can be determined using the formula for circular motion.
The speed of an object moving in a circular path can be calculated using the formula:
v = √(G * M / r)
where v is the speed, G is the gravitational constant, M is the mass of the central object (Earth), and r is the distance between the centers of the satellite and the Earth.
Plugging in the values, we have:
v = √((6.67 x 10^-11 N m^2/kg^2) * (5.97 x 10^24 kg) / (6.57 x 10^6 m))
Calculating this yields:
v ≈ 7666 m/s
Conclusion: The satellite is moving at a speed of approximately 7666 m/s.
(d) The total mechanical energy of the satellite can be determined by summing its kinetic energy and gravitational potential energy.
The total mechanical energy of an object is the sum of its kinetic energy (resulting from its motion) and its potential energy (resulting from its position or height in a gravitational field).
The kinetic energy of the satellite can be calculated using the formula:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the satellite, and v is its speed.
Plugging in the values, we have:
KE = (1/2) * (100 kg) * (7666 m/s)^2
Calculating this yields:
KE ≈ 2.95 x 10^9 J
The gravitational potential energy of the satellite can be calculated using the formula:
PE = -G * (m1 * m2) / r
where PE is the gravitational potential energy, G is the gravitational constant, m1 and m2 are the masses of the two objects (satellite and Earth, respectively), and r is the distance between their centers.
Plugging in the values, we have:
PE = -(6.67 x 10^-11 N m^2/kg^2) * (100 kg) * (5.97 x 10^24 kg) / (6.57 x 10^6 m)
Calculating this yields:
PE ≈ -2.92 x 10^9 J
Since the potential energy is negative, the total mechanical energy is the sum of the kinetic and potential energies:
Total mechanical energy = KE + PE ≈ 2.95 x 10^9 J + (-2.92 x 10^9 J)
Calculating this yields:
Total mechanical energy ≈ 2.5 x 10^7 J
The total mechanical energy of the satellite is approximately 2.5 x 10^7 joules.
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In a simple harmonic oscillator, the restoring force is proportional to: the kinetic energy the velocity the displacement the ratio of the kinetic energy to the potential energy
Restoring force is a force that tends to bring an object back to its equilibrium position. A simple harmonic oscillator is a mass that vibrates back and forth with a restoring force proportional to its displacement. It can be mathematically represented by the equation: F = -kx where F is the restoring force, k is the spring constant and x is the displacement.
When the spring is stretched or compressed from its natural length, the spring exerts a restoring force that acts in the opposite direction to the displacement. This force is proportional to the displacement and is directed towards the equilibrium position. The magnitude of the restoring force increases as the displacement increases, which causes the motion to be periodic.
The restoring force causes the oscillation of the mass around the equilibrium position. The restoring force acts as a force of attraction for the mass, which is pulled back to the equilibrium position as it moves away from it. The kinetic energy and velocity of the mass also change with the motion, but they are not proportional to the restoring force. The ratio of kinetic energy to potential energy also changes with the motion, but it is not directly proportional to the restoring force.
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A dipole is formed by point charges +3.4 μC and -3.4 μC placed on the x axis at (0.20 m , 0) and (-0.20 m , 0), respectively. At what positions on the x axis does the potential have the value 7.4×105 V ? Answer for x1 , x2 =
The values of x1 is (k * (3.4 μC) / (7.4×10^5 V)) + 0.20 m and x2 is (-k * (3.4 μC) / (7.4×10^5 V)) - 0.20 m
To find the positions on the x-axis where the potential has a value of 7.4×10^5 V, we can use the formula for the electric potential due to a dipole:
V = k * q / r
Where:
V is the electric potential
k is the electrostatic constant (9 × 10^9 N m²/C²)
q is the charge magnitude of the dipole (+3.4 μC or -3.4 μC)
r is the distance from the charge to the point where potential is being calculated
Let's solve for the two positions, x1 and x2:
For x1:
7.4×10^5 V = k * (3.4 μC) / (x1 - 0.20 m)
For x2:
7.4×10^5 V = k * (-3.4 μC) / (x2 + 0.20 m)
Simplifying these equations, we can solve for x1 and x2:
x1 = (k * (3.4 μC) / (7.4×10^5 V)) + 0.20 m
x2 = (-k * (3.4 μC) / (7.4×10^5 V)) - 0.20 m
Substituting the values for k and the charges, we can calculate x1 and x2. However, please note that the charges should be converted to coulombs (C) from microcoulombs (μC) for accurate calculations.
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Kirchhoff's Rules. I. E1 - 12 V a. 0.52 R, 2.50 Write out two equations that satisfy the loop rule. [4] b. Write out an equation that satisfies the node rule.
The equation that satisfies the loop rule is ∑ΔV = 0.
The equation that satisfies the Node Rule is ∑I = 0.
Loop Rule:The loop rule is a basic principle of physics that states that the sum of the voltages in a closed circuit loop must be zero. This law is also known as Kirchhoff's voltage law (KVL), and it is critical in circuit analysis because it allows us to calculate unknown values based on known ones. The loop rule can be expressed mathematically as:
∑ΔV = 0
Node Rule:The node rule (or Kirchhoff's current law) is a fundamental principle in physics that states that the sum of the currents entering and exiting a node (or junction) in a circuit must be zero. The node rule is useful for calculating unknown currents in complex circuits. The node rule can be expressed mathematically as:
∑I = 0
Loop Rule:The loop rule states that the sum of the voltages in a closed circuit loop must be zero.∑V = 0The voltages in the circuit are:
E1 - V1 - V2 = 0
E1 = 12 V
V1 = I × R = 0.52 × 2.5 = 1.3V
V2 = I × R = 2.5V
I = (E1 - V1) / R = (12 - 1.3) / 2.5 = 4.28 A
Node Rule:The node rule states that the sum of the currents entering and exiting a node (or junction) in a circuit must be zero.∑I = 0The currents in the circuit are:
I1 = I2 + II1 = (E1 - V1) / R = 4.28 A
I2 = V2 / R = 2.5 / 2.5 = 1 A
∴ I1 = I2 + II1 = 1 + 4.28 = 5.28 A
I2 = 1 AI = I1 - I2 = 5.28 - 1 = 4.28 A
Therefore, the node equation is ∑I = 0 or 1 + 4.28 = 5.28 A.
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What is the focal length of a makeup mirror that produces a magnification of 1.45 when a person's face is 12.2 cm away? Think & Prepare: 1. What kind of mirror causes magnification?
The focal length of the makeup mirror is approximately 39.2 cm. The magnification of 1.45 and the distance of the object (person's face) at 12.2 cm. The positive magnification indicates an upright image.
The type of mirror that causes magnification is a concave mirror. Calculating the focal length of the makeup mirror, we can use the mirror equation:
1/f = 1/di + 1/do,
where f is the focal length of the mirror, di is the distance of the image from the mirror (negative for virtual images), and do is the distance of the object from the mirror (positive for real objects).
Magnification (m) = 1.45
Distance of the object (do) = 12.2 cm = 0.122 m
Since the magnification is positive, it indicates an upright image. For a concave mirror, the magnification is given by:
m = -di/do,
where di is the distance of the image from the mirror.
Rearranging the magnification equation, we can solve for di:
di = -m * do = -1.45 * 0.122 m = -0.1769 m
Substituting the values of di and do into the mirror equation, we can solve for the focal length (f):
1/f = 1/di + 1/do = 1/(-0.1769 m) + 1/0.122 m ≈ -5.65 m⁻¹ + 8.20 m⁻¹ = 2.55 m⁻¹
f ≈ 1/2.55 m⁻¹ ≈ 0.392 m ≈ 39.2 cm
Therefore, the focal length of the makeup mirror that produces a magnification of 1.45 when a person's face is 12.2 cm away is approximately 39.2 cm.
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A car travels at 50 km/hr for 2 hours. It then travels an additional distance of 23 km in 4 hour. The average speed of the car for the entire trip is(in km/hr),
Answer:
Distance travelled in 2 hours = 50 km/hr x 2 hrs = 100 km
Distance travelled in 4 hours = 23 km
Total distance travelled = 100 km + 23 km = 123 km
Total time taken = 2 hrs + 4 hrs = 6 hrs
Average speed = Total distance/Total time
Average speed = 123 km/6 hrs
Average speed = 20.5 km/hr
The viewpoint of Aristotle regarding freely falling objects was_______________
A. light object fall faster than heavier objects
B. heavier object fall faster than lighter objects
C. fall at the same time (light and heavy)
The viewpoint of Aristotle regarding freely falling objects was that heavier objects fall faster than lighter objects. According to Aristotle's theory of natural motion, objects fall towards their natural place in a motion that is proportional to their weight.
Aristotle's understanding of motion was based on his observations of everyday objects and his belief in the existence of four elements (earth, water, air, and fire) and their inherent properties. He argued that objects seek their natural place in the hierarchy of elements, with heavier objects having a stronger tendency to move towards the Earth.
This viewpoint persisted for centuries and was widely accepted until it was challenged by Galileo's experiments and the development of modern physics. Galileo's experiments, including his famous inclined plane experiments, demonstrated that objects of different weights, when dropped from the same height, would reach the ground simultaneously, contradicting Aristotle's theory.
Galileo's experiments and subsequent advancements in the understanding of gravity and motion led to the development of Newton's laws of motion, which provided a more accurate and comprehensive explanation for the behavior of freely falling objects.
In summary, Aristotle's viewpoint regarding freely falling objects was that heavier objects fall faster than lighter objects, a perspective that was later disproven by Galileo's experiments and the emergence of modern physics.
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Use this information for the next 3 questions.
In the pure rotation spectrum, the J = 0 → 1 transition in 1H79Br occurs at 500.7216 GHz. Use the following molar masses: 1H = 1.0078 g/mol and 79Br = 79.9183 g/mol to determine the value of the rotational constant, B .
Select one:
a. 125.1804GHz
b. 500.7216GHz
c. 250.3608GHz
d. 253.7707GHz
To determine the value of the rotational constant, B, in the pure rotation spectrum of 1H79Br, we can use the transition frequency between the J = 0 and J = 1 energy levels. the correct answer is option c: 250.3608 GHz.
Given the transition frequency of 500.7216 GHz and the molar masses of 1H and 79Br, we can calculate the rotational constant using the appropriate formula.
The rotational constant, B, is related to the transition frequency, Δν, between rotational energy levels by the equation Δν = 2B(J + 1), where J represents the quantum number for the energy level. In this case, we are given the transition frequency of 500.7216 GHz for the J = 0 → 1 transition in 1H79Br.
By rearranging the equation, we have B = Δν / (2(J + 1)). To calculate B, we need the transition frequency and the quantum number J. Since we are considering the J = 0 → 1 transition, the quantum number J is 0.
Substituting the given values into the formula, we have B = 500.7216 GHz / (2(0 + 1)). Simplifying the expression gives us B = 500.7216 GHz / 2.
Evaluating the expression, we find B = 250.3608 GHz. Therefore, the correct answer is option c: 250.3608 GHz.
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2. Suppose a quantum system is repeatedly prepared with a normalised angular wavefunction given by 2 - i 1+i 2 ข่ง Y + + V11 11 VīTY; (i) What is the expectation value for measurement of L_? (ii) Calculate the uncertainty in a measurement of Lz. (iii) Produce a histogram of outcomes for a measurement of Lz. Indicate the mean and standard deviation on your plot.
(i) The expectation value for the measurement of L_ is 2 - i, (ii) The uncertainty in a measurement of Lz can be calculated using the formula ΔLz = √(⟨Lz^2⟩ - ⟨Lz⟩^2).
(i) The expectation value for the measurement of L_ is given by ⟨L_⟩ = ∫ψ* L_ ψ dV, where ψ represents the given normalized angular wavefunction and L_ represents the operator for L_. Plugging in the given wavefunction, we have ⟨L_⟩ = ∫(2 - i)ψ* L_ ψ dV.
(ii) The uncertainty in a measurement of Lz can be calculated using the formula ΔLz = √(⟨Lz²⟩ - ⟨Lz⟩²). To find the expectation values ⟨Lz²⟩ and ⟨Lz⟩, we need to calculate them as follows:
- ⟨Lz²⟩ = ∫ψ* Lz² ψ dV, where ψ represents the given normalized angular wavefunction and Lz represents the operator for Lz.
- ⟨Lz⟩ = ∫ψ* Lz ψ dV.
(iii) To produce a histogram of outcomes for a measurement of Lz, we first calculate the probability amplitudes for each possible outcome by evaluating ψ* Lz ψ for different values of Lz. Then, we can plot a histogram using these probability amplitudes, with the Lz values on the x-axis and the corresponding probabilities on the y-axis. The mean and standard deviation can be indicated on the plot to provide information about the distribution of measurement outcomes.
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1. The magnet moves as shown. Which way does the current flow in the coil? a. CW b. CCW c. No induced current N S 2. The magnet moves as shown. Which way does the current flow in the coil? a. CW b. CC
1. Magnet moves: CW current in coil, opposes magnetic field change, 2. Magnet moves: CCW current in coil, opposes magnetic field change.
1. When the magnet moves as shown, the changing magnetic field induces a current in the coil according to Faraday's law of electromagnetic induction. The induced current flows in a direction that creates a magnetic field that opposes the change in the original magnetic field. In this case, as the magnet approaches the coil, the induced current flows in a clockwise (CW) direction to create a magnetic field that opposes the magnet's field. This helps to slow down the magnet's motion.
2. Similarly, when the magnet moves as shown in the second scenario, the changing magnetic field induces a current in the coil. The induced current now flows in a counterclockwise (CCW) direction to create a magnetic field that opposes the magnet's field. This again acts to slow down the magnet's motion.
In both cases, the direction of the induced current is determined by Lenz's law, which states that the induced current opposes the change in the magnetic field that caused it.
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A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x= 18.3t and y-3.68 -4.90², where x and y are in meters and it is in seconds. (a) Write a vector expression for the ball's position as a function of time, using the unit vectors i and j. (Give the answer in terms of t.) m r= _________ m
By taking derivatives, do the following. (Give the answers in terms of t.) (b) obtain the expression for the velocity vector as a function of time v= __________ m/s (c) obtain the expression for the acceleration vector a as a function of time m/s² a= ____________ m/s2 (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 2.79 1. m/s m/s²
r= ___________ m v= ___________ m/s
a= ____________ m/s2
a) The vector expression for the ball's position as a function of time is given as follows:
r= (18.3t) i + (3.68 - 4.9t²) j
b) The velocity vector is obtained by differentiating the position vector with respect to time. The derivative of x = 18.3t with respect to time is dx/dt = 18.3. The derivative of y = 3.68 - 4.9t² with respect to time is dy/dt = -9.8t.
Therefore, the velocity vector is given by the expression: v = (18.3 i - 9.8t j) m/s
c) The acceleration vector is obtained by differentiating the velocity vector with respect to time. The derivative of v with respect to time is dv/dt = -9.8 j.
Therefore, the acceleration vector is given by the expression: a = (-9.8 j) m/s²
d) At t = 2.79 s, we have:r = (18.3 × 2.79) i + (3.68 - 4.9 × 2.79²) j ≈ 51.07 i - 29.67 j m
v = (18.3 i - 9.8 × 2.79 j) ≈ 2.91 i - 27.38 j m/s
a = -9.8 j m/s²
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If the speed doubles, by what factor must the period tt change if aradarad is to remain unchanged?
If the speed doubles, the period must be halved in order for the radar to remain unchanged.
The period of an object in circular motion is the time it takes for one complete revolution. It is inversely proportional to the speed of the object. When the speed doubles, the time taken to complete one revolution is reduced by half. This means that the period must also be halved in order for the radar to maintain the same timing. For example, if the initial period was 1 second, it would need to be reduced to 0.5 seconds when the speed doubles to keep the radar measurements consistent.
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:: Free-fall The path of an object in the (x,y) plane Projectile 2 An object moving under the influence of gravity * Range 3 Trajectory Motion of an object with no horizontal velocity or acceleration, moving only in the vertical direction under the influence of the acceleration due to gravity :: Velocity The horizontal distance traveled by a projectile 5 The slope of the position versus time graph H
The slope of the position versus time graph H is velocity. A position-time graph is a graph that shows an object's position as a function of time. Velocity is the slope of the position versus time graph. The slope of a position-time graph at a particular moment is the instantaneous velocity of the object at that moment.
Free-fall refers to the path of an object in the (x,y) plane, whereas a projectile is an object moving under the influence of gravity. The trajectory is the path of an object with no horizontal velocity or acceleration, moving only in the vertical direction under the influence of acceleration due to gravity. Range refers to the horizontal distance traveled by a projectile, and the slope of the position versus time graph H is velocity.
Motion of an object with no horizontal velocity or acceleration, moving only in the vertical direction under the influence of the acceleration due to gravity is trajectory. When an object is thrown or launched, it follows a path through the air that is called its trajectory. In the absence of air resistance, this path is a parabola.
Range is the horizontal distance traveled by a projectile. The greater the initial velocity of a projectile and the higher its angle, the greater its range. When an object is launched from a height above the ground, the range is the horizontal distance traveled by the object until it hits the ground.
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Moving to another question will save this response. uestion 13 An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 mv's what is the frequency of the third ha
The frequency of the third harmonic of an organ pipe open at both ends with a length of 0.80 m and a velocity of sound in air of 340 m/s is 850 Hz. The correct option is C.
For an organ pipe open at both ends, the frequency of the harmonics can be determined using the formula:
fₙ = (nv) / (2L)
where fₙ is the frequency of the nth harmonic, n is the harmonic number, v is the velocity of sound, and L is the length of the pipe.
In this case, we want to find the frequency of the third harmonic, so n = 3. The length of the pipe is given as 0.80 m, and the velocity of sound in air is 340 m/s.
Substituting these values into the formula, we have:
f₃ = (3 * 340 m/s) / (2 * 0.80 m)
Calculating this expression gives us:
f₃ = 850 Hz
Therefore, the frequency of the third harmonic of the organ pipe is 850 Hz. Option C is correct one.
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Complete Question:
Moving to another question will save this response. uestion 13 An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 mv's what is the frequency of the third harmonic of this pipe O 425 Hz O 638 Hz O 850 Hz 213 Hz
