The problem statement seems to describe a scenario where charged particles undergo curved motion due to acceleration through a potential difference in the presence of a perpendicular magnetic field. The magnitude of the magnetic field and the period of rotation are sought.
To determine the magnitude of the magnetic field, more information is needed, such as the mass and charge of the particles involved, as well as the radius of the curved path. With these details, one could apply the equation F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field strength.
The period of rotation can be calculated using the formula T = 2πr/v, where T represents the period, r is the radius of the path, and v is the velocity of the particles.
Without specific values or additional information, it is not possible to provide precise answers to the magnitude of the magnetic field or the period of rotation in this scenario.
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Example 1: A wire of length 2 m carries a current of 10 A. Calculate the magnitude of force on the wire when placed in a constant field of 0.5 Tesla a) perpendicular to the wire, b) at an angle of 80° to the wire (10-N, 9.85-N)
a) The magnitude of the force on the wire when the field is perpendicular to the wire is 10 N. b) The magnitude of the force on the wire when the field is at an angle of 80° to the wire is 9.85 N.
When a current-carrying wire is placed in a magnetic field, it experiences a force known as the magnetic force. The magnitude of this force can be calculated using the equation F = BILsinθ, where F is the force, B is the magnetic field strength, I is the current, L is the length of the wire, and θ is the angle between the wire and the magnetic field.
In part a, the field is perpendicular to the wire (θ = 90°). Plugging in the values of B = 0.5 T, I = 10 A, and L = 2 m into the formula, we get F = (0.5 T) * (10 A) * (2 m) * sin(90°) = 10 N. Therefore, the magnitude of the force on the wire when the field is perpendicular to the wire is 10 N.
In part b, the field makes an angle of 80° with the wire. Using the same formula and plugging in the values, we get F = (0.5 T) * (10 A) * (2 m) * sin(80°) ≈ 9.85 N. Thus, the magnitude of the force on the wire when the field is at an angle of 80° to the wire is approximately 9.85 N.
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Two charged point-like objects are located on the x-axis. The point-like object with charge q₁ = 3.40 μC is located at x₁ = 1.25 cm and the point-like object with charge q₂ = -2.16 μC is located at x₂ = -1.80 cm. (a) Determine the total electric potential (in V) at the origin. (b) Determine the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
(a) The total electric potential at the origin (x = 0) due to two charged point-like objects is approximately -2.41 × 10^4 V.
(b) The total electric potential at the point (0, 1.50 cm) due to the same objects is also approximately -2.41 × 10^4 V.
To determine the total electric potential at a point due to two charged objects, we need to calculate the electric potential contributed by each object and then sum them up.
The electric potential at a point due to a charged point-like object is given by the equation:
V = k * q / r
Where:
V is the electric potential,
k is the electrostatic constant (8.99 × 10^9 N m^2/C^2),
q is the charge of the object, and
r is the distance between the object and the point where we are calculating the potential.
Let's calculate the total electric potential at the origin (x = 0) first:
(a) Total electric potential at the origin:
For the object with charge q₁ = 3.40 μC located at x₁ = 1.25 cm (0.0125 m):
r₁ = x₁ - x = 0.0125 m - 0 m = 0.0125 m
Using the formula: V₁ = k * q₁ / r₁
V₁ = (8.99 × 10^9 N m^2/C^2) * (3.40 × 10^(-6) C) / (0.0125 m)
V₁ = 2.447 × 10^5 V
For the object with charge q₂ = -2.16 μC located at x₂ = -1.80 cm (-0.0180 m):
r₂ = x - x₂ = 0 m - (-0.0180 m) = 0.0180 m
Using the formula: V₂ = k * q₂ / r₂
V₂ = (8.99 × 10^9 N m^2/C^2) * (-2.16 × 10^(-6) C) / (0.0180 m)
V₂ = -2.688 × 10^5 V
The total electric potential at the origin is the sum of the potentials contributed by each object:
Total V = V₁ + V₂
Total V = 2.447 × 10^5 V + (-2.688 × 10^5 V)
Total V = -0.241 × 10^5 V
Therefore, the total electric potential at the origin is -0.241 × 10^5 V or -2.41 × 10^4 V.
(b) Total electric potential at (0, 1.50 cm):
For this point, we need to calculate the potential due to each object and then add them up.
For the object with charge q₁ = 3.40 μC located at x₁ = 1.25 cm (0.0125 m):
r₁ = sqrt((x₁ - x)² + (y₁ - y)²) = sqrt((0.0125 m - 0 m)² + (0.015 m - 0.015 m)²) = 0.0125 m
Using the formula: V₁ = k * q₁ / r₁
V₁ = (8.99 × 10^9 N m^2/C^2) * (3.40 × 10^(-6) C) / (0.0125 m)
V₁ = 2.447 × 10^5 V
For the object with charge q₂ = -2.16 μC located at x₂ = -1.80 cm (-0.0180 m):
r₂ = sqrt((x₂ - x)² + (y₂ - y)²) = sqrt((-0.0180 m - 0 m)² + (0.015 m - 0
.015 m)²) = 0.0180 m
Using the formula: V₂ = k * q₂ / r₂
V₂ = (8.99 × 10^9 N m^2/C^2) * (-2.16 × 10^(-6) C) / (0.0180 m)
V₂ = -2.688 × 10^5 V
The total electric potential at (0, 1.50 cm) is the sum of the potentials contributed by each object:
Total V = V₁ + V₂
Total V = 2.447 × 10^5 V + (-2.688 × 10^5 V)
Total V = -0.241 × 10^5 V
Therefore, the total electric potential at (0, 1.50 cm) is -0.241 × 10^5 V or -2.41 × 10^4 V.
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In the figure an electron is shot at an initial speed of Vo = 5.29* 10% m/s, at angle 8. = 43.1 ° from an x axis. It moves through a uniform electric field Ē = (4.45 N/C)). A screen for detecting electrons is positioned parallel to the yaxis, at distance x = 2.55 m. What is they component of the electron's velocity (sign included) when the electron hits the screen? 个 E Detecting screen VO 80 X Number Units
The component of the electron's velocity when it hits the screen is -2.17 * 10^6 m/s. The electric field applies a force on the electron, which will change its velocity.
To find the component of the electron's velocity, we need to consider the motion of the electron in the presence of the electric field.
We can break down the initial velocity of the electron into its x and y components:
Vx = Vo * cos(θ) = 5.29 * 10^6 m/s * cos(43.1°) = 3.82 * 10^6 m/s
Vy = Vo * sin(θ) = 5.29 * 10^6 m/s * sin(43.1°) = 3.41 * 10^6 m/s
Since the electric field only acts in the x-direction, it will affect the x-component of the electron's velocity. The force experienced by the electron due to the electric field is given by:
F = q * E
where q is the charge of the electron (1.6 * 10^-19 C) and E is the electric field (4.45 N/C).
The force on the electron in the x-direction is:
Fx = q * Ex = (1.6 * 10^-19 C) * (4.45 N/C) = 7.12 * 10^-19 N
Using Newton's second law (F = ma), we can find the acceleration of the electron in the x-direction:
Fx = m * ax
where m is the mass of the electron (9.11 * 10^-31 kg).
ax = Fx / m = (7.12 * 10^-19 N) / (9.11 * 10^-31 kg) = 7.81 * 10^11 m/s^2
Since the electron moves in a uniform electric field, the acceleration is constant. Therefore, we can use the kinematic equation to find the final velocity in the x-direction:
Vxf = Vx + ax * t
where t is the time taken for the electron to reach the screen.
To find the time, we can use the equation of motion in the y-direction:
y = Vy * t + (1/2) * ay * t^2
where y is the distance to the screen (2.55 m) and ay is the acceleration in the y-direction (due to gravity, which we assume is negligible).
Since the screen is parallel to the y-axis, the electron will not move in the y-direction. Therefore, we can set ay = 0 and solve for t:
2.55 m = (3.41 * 10^6 m/s) * t
t = 7.48 * 10^-7 s
Now, we can calculate the final velocity in the x-direction:
Vxf = (3.82 * 10^6 m/s) + (7.81 * 10^11 m/s^2) * (7.48 * 10^-7 s)
Vxf = 5.82 * 10^6 m/s
The component of the electron's velocity when it hits the screen is -5.82 * 10^6 m/s (negative sign indicates the velocity is in the opposite direction of the initial velocity).
Note: In the question, the value given for the electric field (4.45 N/C) is written with double parentheses. It is assumed that it should be written as (4.45 N/C).
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A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is 39,000 Hz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.025 times the speed of sound in air. What frequency does the bat hear reflected off the wall? (Speed of sound in air is 343 m/s)
The bat, moving at 0.025 times the speed of sound in air, will hear a frequency of 38,887.5 Hz reflected off the wall.
The frequency of sound heard by an observer is affected by the relative motion between the source of the sound and the observer. In this case, the bat is moving towards the wall, which causes a Doppler shift in the frequency of the reflected sound.
The formula for the Doppler shift in frequency is given by:
f' = f(v + vr) / (v + vs)
where f' is the observed frequency, f is the emitted frequency, v is the speed of sound, vr is the velocity of the observer (bat), and vs is the velocity of the source (wall).
Given that f = 39,000 Hz, v = 343 m/s, vr = 0.025 * v, and vs = 0 (since the wall is stationary), we can substitute these values into the formula:
f' = 39,000 * (343 + 0.025 * 343) / (343 + 0)
f' = 39,000 * 1.025 / 1
f' = 39,877.5 Hz
Therefore, the bat will hear a frequency of approximately 38,887.5 Hz (rounded to the nearest hundredth) reflected off the wall.
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Orbits and gravity (6 questions) Note that while the sections were not specifically assigned in the textbook, chapter 3 may prove useful in answering the following questions. 1) Why do astronauts appear to "float" in the space station? 2) Run NAAP labs app, click lab 5. Planetary orbits, click Planetary Orbit Simulator, a simulation window shows up, click Kepler's 3 rd Law on lower left of the app. Set parameters for Earth, click OK. Answer question a): a) What is the orbital period for Earth? You can find it from the graph plotted in the simulation. Double the value of semimajor axis and observe the change of orbital period on the graph. Answer question b):
Astronauts appear to float in the space station because they are in a state of free fall, continuously falling toward the Earth while also moving forward at the same speed that they would be moving if they were standing on the ground.
This causes them to appear weightless, or as if they are floating. The orbital period for Earth is 1 year, which is equal to 365.25 days. When the value of semimajor axis is doubled, the orbital period of the Earth will increase four times. This is because according to Kepler's third law of planetary motion,
the square of the orbital period of a planet is directly proportional to the cube of the semimajor axis of its orbit. Therefore, if the value of the semimajor axis is doubled, the cube of that value will increase by a factor of eight, which means that the square of the orbital period of the planet will increase by a factor of four.
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A wave travel move in a rope of 50 m and 140 g of mass, with a tension of 700 N. this wave move in direction +x, and the fontion is: y(x,t)=Acos(kx±ωt) Its period is 5.0×10 −4
s and amplitade is a thousond of wove lenght a) find the speed of propagation of the wave b) find the frequency ond wave lenght. c) Calcolate the power soplied at the road for generate this sinusoidal wave d) find the tronsverse speed in the position x=0,04 m and the time
The transverse speed at x = 0.04 m and t = T/4 is 377 m/s for the given wave.
The given parameters are:Tension, T = 700 N Mass, m = 140 g = 0.14 kgLength, L = 50 mPeriod, T =[tex]5.0 * 10^-4 s[/tex]Amplitude, A = λ/1000 = (1/1000)L
A wave is an oscillation or disturbance that moves through a medium or space while only moving energy. Mechanical waves and electromagnetic waves are the two basic forms of waves. A medium (such as water or air) is necessary for the propagation of mechanical waves, such as sound waves or water waves. In contrast, electromagnetic waves like light or radio waves can pass through a vacuum since they don't need a medium to do so. Wavelength, frequency, amplitude, and speed are some of the distinguishing characteristics of waves. They are crucial in many disciplines, including physics, engineering, and communication. They display phenomena like reflection, refraction, diffraction, and interference.
For y(x, t) =[tex]A cos(kx - ωt)T = 2π/ω = 5.0 × 10⁻⁴ s[/tex][tex]2π/T = 2π/(5.0 × 10⁻⁴) rad/s[/tex]
∴ ω = [tex]2π/T = 2π/(5.0 × 10⁻⁴)[/tex]rad/s
The wave speed, v =[tex]√(T/m)[/tex]
For the given values, v = √(700/0.14) m/s= 2652.68 m/s
(a) The speed of propagation of wave is 2652.68 m/s(b) Frequency, f = 1/T = 1/([tex]5.0 * 10^-4[/tex]) Hz
Wavelength, λ = v/f= vT= ([tex]2652.68 * 5.0 * 10^-4[/tex]) m= 1.32634 m
(c) Power supplied = T/2 × v= (700/2) × 2652.68= 929347.8 W(d) At x = 0.04 m and t = T/4, y = A = [tex]λ/1000[/tex]= (1/1000)L= (1/1000) × 50 m= 0.05 m
For y =[tex]A cos(kx - ωt) = A cos(π/2)[/tex] = A
The wave is a transverse wave, hence the transverse speed is given by:[tex]v' = ωk= 2πf × 2π/λ= 4π²f/λ= 4π² × (1/T)/(λ)= 4π² × 2 × 10⁴/(1.32634)[/tex]= 376.99 m/s≈ 377 m/s
Therefore, the transverse speed at x = 0.04 m and t = T/4 is 377 m/s.
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A motor is lifting up a box of mass m=400 kg with a constant speed of v=0.5 m/s. What is the power in W delivered by the motor in lifting the box?
The power delivered by the motor in lifting the box is 200 W.
Power is defined as the rate at which work is done or energy is transferred. In this case, the motor is lifting the box with a constant speed, which means there is no change in potential energy. Therefore, the work done by the motor is equal to the weight of the box (mg) multiplied by the distance it is lifted (v) per unit time. Since the speed is constant, the distance lifted per unit time is equal to the speed (v).
Thus, the power can be calculated as the product of the weight and the speed, which gives a power of 200 W.
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A 0.200-kg particle has a speed of 2.10 m/s at point circled A and kinetic energy of 8.00 J at point circled B.
(a) What is its kinetic energy at circled A? J
(b) What is its speed at circled B? m/s
(c) What is the net work done on the particle by external forces as it moves from circled A to circled B? J
We find that the kinetic energy at circled A is approximately 0.441 J. We find that the speed at circled B is approximately 8.00 m/s. The net work done on the particle as it moves from circled A to circled B is approximately 7.56 J.
a) To find the kinetic energy at circled A, we can use the formula for kinetic energy:[tex]KE = (1/2)mv^2[/tex], where KE is the kinetic energy, m is the mass, and v is the velocity.
Given that the particle has a mass of 0.200 kg and a speed of 2.10 m/s at circled A, we can substitute these values into the formula: [tex]KE = (1/2)(0.200 kg)(2.10 m/s)^2[/tex]. Calculating this, we find that the kinetic energy at circled A is approximately 0.441 J.
b) To find the speed at circled B, we can rearrange the formula for kinetic energy to solve for velocity: [tex]v = \sqrt{(2KE/m)}[/tex].
Given that the particle has a kinetic energy of 8.00 J at circled B and a mass of 0.200 kg, we can substitute these values into the formula: v = √(2(8.00 J) / 0.200 kg). Calculating this, we find that the speed at circled B is approximately 8.00 m/s.
c) The net work done on the particle by external forces can be calculated using the work-energy principle, which states that the net work done is equal to the change in kinetic energy: Net work = KE(B) - KE(A).
Given that the kinetic energy at circled A is 0.441 J and the kinetic energy at circled B is 8.00 J, we can calculate the net work done: Net work = 8.00 J - 0.441 J. The net work done on the particle as it moves from circled A to circled B is approximately 7.56 J.
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What is the rest energy E0 in MeV, the rest mass m in MeV/c2, the momentum p in MeV/c2, kinetic energy K in MeV and relativistic total energy E of a particle with mass (m =1.3367 x 10−27 kg) moving at a speed of v = 0.90c?
The rest energy E0 of a particle with mass m can be calculated using Einstein's famous equation, E0 = mc^2. For a particle with a rest mass of 1.3367 x 10^−27 kg, the rest energy E0 is given by E0 = (1.3367 x 10^−27 kg) x (299,792,458 m/s)^2 ≈ 1.2017 x 10^−10 MeV.
When the particle is moving at a speed of v = 0.90c (where c is the speed of light), we need to consider its relativistic effects. The momentum p of the particle can be calculated as p = mv/sqrt(1 - (v/c)^2), where m is the rest mass. Plugging in the values, we get p ≈ (1.3367 x 10^−27 kg) x (0.90 x 299,792,458 m/s)/sqrt(1 - (0.90)^2) ≈ 4.492 x 10^−19 MeV/c^2.
The kinetic energy K of the particle can be calculated as K = sqrt((pc)^2 + (mc^2)^2) - mc^2. Substituting the values, we get K ≈ sqrt((4.492 x 10^−19 MeV/c^2)^2 + (1.3367 x 10^−27 kg x (299,792,458 m/s)^2)^2) - (1.3367 x 10^−27 kg) x (299,792,458 m/s)^2 ≈ 1.292 x 10^−14 MeV.
The relativistic total energy E of the particle is the sum of the rest energy E0 and the kinetic energy K, so E ≈ E0 + K ≈ 1.2017 x 10^−10 MeV + 1.292 x 10^−14 MeV ≈ 1.2017 x 10^−10 MeV. Thus, the rest energy E0 is approximately 1.2017 x 10^−10 MeV, and the relativistic total energy E is also approximately 1.2017 x 10^−10 MeV.
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A guitar string has length of 0.22 m. The sound of the string has a frequency of 243 Hz when it is oscillating with three antinodes. What is the velocity of the travelling wave in the string? Give your answer to 1 decimal place.
The velocity of the traveling wave in the string is 17.8 m/s. The frequency of the wave is 243 Hz, and the length of the string is 0.22 m. The number of antinodes is 3.
The wavelength of the wave is equal to the length of the string divided by the number of antinodes, which is 0.073 m. The velocity of the wave is equal to the frequency multiplied by the wavelength, which is 17.8 m/s.
* The frequency of a wave is the number of times the wave oscillates per unit time. The length of the string is the distance between two consecutive antinodes. The number of antinodes is the number of times the wave crosses the equilibrium position in a single oscillation.
* The wavelength of a wave is the distance between two consecutive crests or troughs of the wave. The velocity of a wave is the distance traveled by the wave in a unit of time.
* In this problem, the frequency of the wave is given as 243 Hz. The length of the string is given as 0.22 m. The number of antinodes is given as 3.
* The wavelength of the wave can be calculated using the following formula:
```
wavelength = length / number of antinodes
```
Plugging in the known values, we get:
```
wavelength = 0.22 m / 3 = 0.073 m
```
The velocity of the wave can be calculated using the following formula:
```
velocity = frequency * wavelength
```
Plugging in the known values, we get:
```
velocity = 243 Hz * 0.073 m = 17.8 m/s
```
Therefore, the velocity of the traveling wave in the string is 17.8 m/s.
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An echo sounder (SONAR) directly measures the density of seawater the direction of movement of an ocean current the depth of the ocean the salinity of seawater
An echo sounder (SONAR) directly measures the depth of the ocean. An echo sounder (SONAR) is a device that is used to determine the depth of water beneath a vessel.
It sends an acoustic signal to the seabed or any objects in the water and then records the time it takes for the signal to return. It then calculates the distance to the seabed or object, which is equal to twice the time it takes for the signal to travel from the echo sounder to the seabed and back to the echo sounder.
The depth of the water is then measured. The other options are incorrect because:It does not directly measure the density of seawater It does not directly measure the direction of movement of an ocean currentIt does not directly measure the salinity of seawater.
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A grinding wheel is a uniform cylinder with a radius of 8.50 cm and a mass of 0.580 kg ▾ Part A Calculate moment of inertia about its center. Express your answer to three significant figures and include the appropriate units. → ? I: Value Units The unit you entered is not recognized. Review a list of acceptable units. No credit lost. Try again. Submit Previous Answers Request Answer ▾ Part B Calculate the applied torque needed to accelerate it from rest to 1750 rpm in 6.40 s. Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 64.0 s Express your answer to three significant figures and include the appropriate units. ? µA T= Value Units
The moment of inertia of the grinding wheel about its center is approximately 0.102 kg·m².
To calculate the moment of inertia (I) of the grinding wheel, we can use the formula for the moment of inertia of a solid cylinder: I = (1/2) * m * r², where m is the mass of the cylinder and r is its radius. Substituting the given values, we have I = (1/2) * 0.580 kg * (0.085 m)², which simplifies to I ≈ 0.102 kg·m².
To calculate the applied torque (T) needed to accelerate the wheel, we can use the formula: T = I * α, where α is the angular acceleration. First, we need to find the angular acceleration. The initial angular velocity (ω₀) is 0 rpm, and the final angular velocity (ω) is 1750 rpm. Converting these values to radians per second, we have ω₀ = 0 rad/s and ω = (1750 rpm) * (2π rad/60 s) = 183.33 rad/s. The angular acceleration (α) can be calculated as α = (ω - ω₀) / t, where t is the time taken. Substituting the given values, we get α = (183.33 rad/s) / (6.40 s) = 28.64 rad/s².
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At a certain place on the surface of the earth, the sunlight has an intensity of about 1.8 x 103 W/m2. What is the total electromagnetic energy from this sunlight in 3.3 m3 of space?
The electromagnetic energy from sunlight in 3.3 m3 of space is 106.92 J. This is calculated by multiplying the intensity of sunlight by the volume of space, and then by the energy density of sunlight.
The intensity of sunlight is about 1.8 x 103 W/m2. This means that every square meter of space receives 1.8 x 103 watts of sunlight energy. The volume of space is 3.3 m3. This means that there are 3.3 x 106 cubic meters of space.
The total electromagnetic energy from sunlight in 3.3 m3 of space is calculated by multiplying the intensity of sunlight by the volume of space, and then by the energy density of sunlight. This gives us a total energy of 106.92 J.
Total energy = intensity * volume * energy density
= 1.8 x 103 W/m2 * 3.3 m3 * 1.8 x 106 J/m3
= 106.92 J
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Aluminium-26, 2613A1, is an unstable proton-rich nucleus and decays into the stable nuclide Magnesium-2612Mg. a) Name two processes by which the above decay may occur b) Write down balanced equations to describe each of these two processes and name all the particles involved
Answer:
Explanation:
a) The decay of aluminum-26 (26/13Al) into magnesium-26 (26/12Mg) can occur through two processes: beta decay and positron emission.
b) Balanced equations for each process:
1) Beta decay:
26/13Al → 26/12Mg + 0/-1e
In beta decay, a neutron in the aluminum-26 nucleus is converted into a proton, releasing a beta particle (an electron) and an antineutrino. The resulting nucleus is magnesium-26.
2) Positron emission:
26/13Al → 26/12Mg + 0/+1e
In positron emission, a proton in the aluminum-26 nucleus is converted into a neutron, emitting a positron and a neutrino. The resulting nucleus is magnesium-26.
Particles involved:
- Aluminum-26 (26/13Al): The unstable nucleus decaying into magnesium-26.
- Magnesium-26 (26/12Mg): The stable nucleus resulting from the decay.
- Beta particle (0/-1e): An electron emitted during beta decay.
- Antineutrino (V): A neutral particle emitted during beta decay.
- Positron (0/+1e): A positively charged electron emitted during positron emission.
- Neutrino (ν): A neutral particle emitted during positron emission.
Note: The notation used in the equations indicates the mass number as the upper value and the atomic number as the lower value.
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A ball is launched from the ground level with an angle = 30° with the horizontal. The initial speed is vo= 20 m/s. What is the maximal height that the ball reaches? a 5.1 m b 7.8 m c 4.3 m d 9.5 m
The maximal height reached by the ball is approximately 5.1 m, which corresponds to option (a).
The maximal height reached by the ball can be calculated using the kinematic equations of motion. Since the ball is launched at an angle of 30° with the horizontal, we need to consider the vertical motion.
The vertical component of the initial velocity can be calculated as vo * sin(θ), where vo is the initial speed and θ is the launch angle. In this case, vo = 20 m/s and θ = 30°, so the vertical component of the initial velocity is 20 m/s * sin(30°) = 10 m/s.
To find the maximal height, we can use the equation:
h = (v₀y²) / (2g),
where h is the height, v₀y is the vertical component of the initial velocity, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values, we have:
h = (10 m/s)² / (2 * 9.8 m/s²) = 5.1 m.
Therefore, the maximal height reached by the ball is approximately 5.1 m, which corresponds to option (a).
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The maximal height that the ball reaches when launched from the ground level with an angle of 30° and an initial speed of 20 m/s is approximately 9.5 m (option d).
By analyzing the projectile motion, we can determine the time it takes for the ball to reach its maximum height and then calculate the height using the kinematic equations.
When the ball is launched at an angle of 30° with the horizontal, we can break down its motion into horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component follows a parabolic trajectory.
To find the maximal height, we need to determine the time it takes for the ball to reach its highest point. The time of flight for a projectile can be calculated using the equation t = 2vo sin(θ) / g, where vo is the initial speed, θ is the launch angle, and g is the acceleration due to gravity.
Substituting the given values, we have t = (2 * 20 m/s * sin(30°)) / 9.8 m/s² ≈ 2.04 s.
Next, we can calculate the maximal height using the equation h = vo^2 * sin^2(θ) / (2g), where h is the height.
Substituting the given values, we have h = (20 m/s)^2 * sin^2(30°) / (2 * 9.8 m/s²) ≈ 9.5 m.
Therefore, the maximal height that the ball reaches is approximately 9.5 m (option d).
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When volcanoes are moving away from the hotspot (decrease in amount of melt coming in), they erupt lavas that are still basalt but of yet another variety. What is this variety? Maximum number of characters (including HTML tags added by text editor): 32,000
When volcanoes are moving away from the hotspot (decrease in amount of melt coming in), they erupt lavas that are still basalt but of yet another variety.
The variety of basaltic lava that is erupted when volcanoes are moving away from the hotspot is called tholeiitic basalt. Tholeiitic basalt contains a relatively low concentration of sodium and is characterized by the presence of clinopyroxene and orthopyroxene, as well as plagioclase feldspar in its mineral assemblage.
The variety of basaltic lava that is erupted when volcanoes are moving away from Tholeiitic basalt contains a relatively low concentration of sodium and is characterized by the presence of clinopyroxene and orthopyroxene the hotspot is called tholeiitic basalt.
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2 1 point Consider an ac network with an ideal transformer (with turns ratio of 2:1) shown below where ZL is the load impedance. Then the average power delivered to the load is given by 2:1 Z₂ 3+ j4 V₂ V₂ Z₁ 3-j = 15/0 ideal + V₁ D I
The given equation represents the relationship between the voltages (V₁ and V₂) and impedances (Z₁ and Z₂) in the AC network with an ideal transformer. By plugging in the appropriate values and performing the necessary calculations, the equation can be used to determine the average power delivered to the load in the network.
What does the given equation represent in the AC network with an ideal transformer and how can it be used to calculate the average power delivered to the load?The paragraph describes an AC network with an ideal transformer, where ZL represents the load impedance. The average power delivered to the load can be determined using the given equation.
In the equation, V₁ and V₂ represent the voltages on the primary and secondary sides of the transformer, respectively. Z₁ and Z₂ represent the impedances on the primary and secondary sides of the transformer, respectively. The term "15/0" is not clear in the context and may require further clarification.
To calculate the average power delivered to the load, one needs to determine the voltage and impedance values in the equation and perform the necessary calculations. The result will provide the average power delivered to the load.
It is important to note that further information and context may be required to fully understand and interpret the equation and its implications in the given AC network with an ideal transformer.
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An electron has de Broglie wavelength 2.84×10−10 mm .
1)
Determine the electron's kinetic energy in electron volts.
Express your answer in electron volts to three significant figures.
The electron's kinetic energy is approximately 150.1 eV(electron volts).
To determine the electron's kinetic energy in electron volts (eV), we can use the de Broglie wavelength formula:
λ = h / p
where λ is the wavelength, h is the Planck's constant (approximately 6.626 × 10^-34 J·s), and p is the momentum of the electron.
Rearranging the formula to solve for p, we have:
p = h / λ
Substituting the given wavelength:
p = (6.626 × 10^-34 J·s) / (2.84 × 10^-10 m)
p ≈ 2.329 × 10^-24 kg·m/s
The kinetic energy (KE) of an electron is given by:
KE = (p^2) / (2m)
where m is the mass of the electron (approximately 9.109 × 10^-31 kg).
Substituting the values:
KE = [(2.329 × 10^-24 kg·m/s)^2] / (2 × 9.109 × 10^-31 kg)
KE ≈ 2.401 × 10^-17 J
To convert this energy into electron volts, we divide it by the elementary charge (e) which is approximately 1.602 × 10^-19 C:
KE (eV) = (2.401 × 10^-17 J) / (1.602 × 10^-19 C)
KE (eV) ≈ 150.1 eV (rounded to three significant figures)
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A playground merry-go-round with an axis at the center (radius R = 1.9 m and rotational inertia | = 1.8 x 103 kgm2) is initially rotating at angular velocity wo1 rad/s (counter-clockwise). A girl of mass m = 47 kg is running at speed vo= 3.6 m/s in a direction tangent to the disk of the merry-go-round, intending to jump on to the point labeled with the green "x" (she is indicated by the purple circle in the figure). What is the magnitude of the angular velocity of the merry-go-round after the girl has jumped on at the edge? [Hint: treat the girl as a point mass located at a distance R from the axis]. Your answer should be in rad/s, but enter only the numerical part in the box.
The magnitude of the angular velocity of the merry-go-round after the girl has jumped on at the edge is 0.34 rad/s.
We can apply the principle of conservation of angular momentum. Initially, the merry-go-round is rotating at angular velocity wo1. The angular momentum of the system is given by L_initial = I_1 * wo1, where I_1 is the rotational inertia of the merry-go-round.
When the girl jumps onto the edge of the merry-go-round, her angular momentum is conserved as well. The angular momentum of the system after the girl jumps is L_final = (I_1 + m * R^2) * wf, where m is the mass of the girl and R is the radius of the merry-go-round.
Since angular momentum is conserved, we can set L_initial equal to L_final:
I_1 * wo1 = (I_1 + m * R^2) * wf
for wf, the final angular velocity, we find:
wf = (I_1 * wo1) / (I_1 + m * R^2)
Substituting the given values, we get:
wf = (1.8 x 10^3 kgm^2 * wo1) / (1.8 x 10^3 kgm^2 + 47 kg * (1.9 m)^2)
The expression, the magnitude of the angular velocity of the merry-go-round after the girl has jumped on at the edge is approximately 0.34 rad/s.
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A 3.2 kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy (not gravitational potential energy) of this spring-block system is 1.8 J. What is the elastic potential energy of the system when the 3.2kg block is replaced by a 5 kg block?
Hint: Rest position of spring is described by x = 0. Elastic potential energy is .
A) 5.00 J
B) 4.39 J
C) 4.68 J
D) 5.22 J
The elastic potential energy of a spring-block system is given by the formula: Elastic potential energy = (1/2) k x²
where k is the spring constant and x is the displacement from the rest position of the spring.
In this case, the elastic potential energy of the system with the 3.2 kg block is given as 1.8 J.
To find the elastic potential energy when the 3.2 kg block is replaced by a 5 kg block, we need to calculate the new elastic potential energy using the formula above.
Since the rest position of the spring is described by x = 0, the displacement remains the same when the block is replaced. Therefore, the new elastic potential energy will only depend on the change in mass.
The ratio of the new elastic potential energy to the original elastic potential energy is given by the square of the ratio of the masses:
(Elastic potential energy with new block) / (Elastic potential energy with 3.2 kg block) = (mass of new block / mass of 3.2 kg block)²
Let's calculate the ratio of the masses and find the new elastic potential energy:
(mass of new block / mass of 3.2 kg block) = (5 kg / 3.2 kg) = 1.5625
(Elastic potential energy with new block) = (Elastic potential energy with 3.2 kg block) * (ratio of masses)²
= 1.8 J * (1.5625)²
= 1.8 J * 2.4414
= 4.3945 J
Therefore, the elastic potential energy of the system when the 3.2 kg block is replaced by a 5 kg block is approximately 4.39 J.
The correct answer is B) 4.39 J.
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N A 100 turn circular coil of radius 5.0 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.60 T. If the coil carries a current of 30 mA, find the magnitude of the maximum possible tordue exerted on the coil. A) 0.034 N.m B) 0.089 N.m C) 0.010 N.m D) 0.014 N.m E) 0.045 N.m
The maximum possible torque exerted on the coil is 0.089 N⋅m. This is calculated using the formula: τ = NIAB sinθ
where:
* N is the number of turns in the coil (100)
* I is the current in the coil (30 mA = 0.030 A)
* A is the area of the coil (πr^2 = 3.14 × (0.05 m)^2 = 7.85 × 10^-4 m^2)
* B is the strength of the magnetic field (0.60 T)
* θ is the angle between the magnetic field and the plane of the coil (90 degrees)
The torque on a current-carrying loop in a magnetic field is given by the formula:
```
τ = NIAB sinθ
```
where:
* N is the number of turns in the coil
* I is the current in the coil
* A is the area of the coil
* B is the strength of the magnetic field
* θ is the angle between the magnetic field and the plane of the coil
In this case, we have:
* N = 100
* I = 0.030 A
* A = 7.85 × 10^-4 m^2
* B = 0.60 T
* θ = 90 degrees
Substituting these values into the formula, we get:
```
τ = 100 × 0.030 A × 7.85 × 10^-4 m^2 × 0.60 T × sin(90 degrees)
= 0.089 N⋅m
```
Therefore, the maximum possible torque exerted on the coil is 0.089 N⋅m.
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Two metal objects have charges of 70pC and −70pC, resulting in a potential difference (voltage) of 20 V between them. How does the capacitance C change if we double the charge on each object? C doubles. C is cut in half. C does not change. Insufficient information. 1) The magnetic force experienced by a charged particle always points (10pts) B. in the opposite direction as the magnetic field. D. None are correct. A. in the same direction as the magnetic field. C. Either A or B is correct depending on the charge of the particle experiencing the force.
(D) None are correct. The magnetic force does not always point in the opposite direction as the magnetic field.
The capacitance C does not change if we double the charge on each object. The capacitance of a capacitor is determined by its physical characteristics, such as the geometry of its plates and the dielectric material between them, and it does not depend on the magnitude of the charges on the plates.
The magnetic force experienced by a charged particle depends on the charge of the particle, its velocity, and the magnetic field. The direction of the magnetic force is given by the right-hand rule: if you point your thumb in the direction of the particle's velocity and your fingers in the direction of the magnetic field, then the force will be perpendicular to both, following the direction of your palm.
Therefore, the correct answer is (D) None are correct. The magnetic force does not always point in the opposite direction as the magnetic field.
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Give the names of at least two types of pyroclasts (magma that is fragmented into discrete particles).
The two types of pyroclasts are the ash and volcanic bombs.
How do we explain?Ash refers to the fine-grained volcanic material that is produced during explosive volcanic eruptions and consists of tiny, glassy fragments that are less than 2 millimeters in diameter.
Volcanic bombs are described as larger fragments of magma that are ejected during volcanic eruptions.
Volcanic bombs are usually molten or partially molten when airborne and solidify into various shapes as they cool ranging in size from a few centimeters to several meters in diameter and are often aerodynamically shaped due to their movement through the air.
In conclusion, Ash can travel long distances through the air and can have significant impacts on the environment, including air quality and visibility.
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A guitar string is 90 cm long and has a mass of 3.8 g The distance from the bridge to the support post is L = 62 cm, and the string is under a tension of 540 N RAKE 100 DE Part A What are the frequencies of the fundamental and first two overtones? Express your answers using two significant figures. Enter your answers in ascending order separated by commas
The frequency of a vibrating string can be determined using the equation: f = (1/2L) * sqrt(T/μ) where f is the frequency, L is the length of the string, T is the tension in the string, and μ is the linear mass density of the string.
First, we need to calculate the linear mass density μ of the string. The mass of the string is given as 3.8 g, and the length of the string is 90 cm. Therefore μ = (mass/length) = (3.8 g) / (90 cm) = 0.0422 g/cm Next, we can calculate the frequencies of the fundamental and first two overtones.
For the fundamental frequency (n = 1):
f1 = (1/2L) * sqrt(T/μ)
For the first overtone (n = 2):
f2 = (2/2L) * sqrt(T/μ) = (1/L) * sqrt(T/μ)
Plugging in the given values:
f1 = (1/2 * 90 cm) * sqrt(540 N / 0.0422 g/cm) ≈ 220 Hz
f2 = (1/90 cm) * sqrt(540 N / 0.0422 g/cm) ≈ 440 Hz
Therefore, the frequencies of the fundamental and first two overtones are approximately 220 Hz, 440 Hz, and 660 Hz (assuming we consider the second overtone).
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A block of mass m1 = 3.02 kg on a frictionless plane inclined at angle θ = 26.2° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.66 kg hanging vertically (see the figure). (a) What is the acceleration of the hanging block (choose the positive direction down)? (b) What is the tension in the cord?
The acceleration of the hanging block is 0.62 m/s^2. The tension in the cord is 26.1 N.
The system of two blocks is accelerating because the hanging block is pulling down on the cord, which is pulling up on the block on the inclined plane. The acceleration of the system is equal to the acceleration of the hanging block, which can be calculated using Newton's second law:
```
a = (m2 - m1)g / (m1 + m2)
```
where m1 is the mass of the block on the inclined plane, m2 is the mass of the hanging block, and g is the acceleration due to gravity.
Substituting the known values, we get:
```
a = (2.66 kg - 3.02 kg) * 9.8 m/s^2 / (2.66 kg + 3.02 kg) = 0.62 m/s^2
```
The tension in the cord can be calculated using the following equation:
```
T = m2g
```
where T is the tension in the cord and m2 is the mass of the hanging block.
Substituting the known value, we get:
```
T = 2.66 kg * 9.8 m/s^2 = 26.1 N
```
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To make a bounce pass, a player throws a 0.65-kg basketball toward the floor. The ball hits the floor with a speed of 5.7 m/s at an angle of 62 ∘ to the vertical. If the ball rebounds with the same speed and angle, what was the magnitude of the impulse delivered to it by the floor?
Impulse = |0.65 kg * √(v_initial_horizontal² + (-v_initial_vertical)²) - 0.65 kg * √(v_initial_horizontal² + v_initial_vertical²)| Evaluating this expression will give us the magnitude of the impulse delivered to the ball by the floor.
To find the magnitude of the impulse delivered to the basketball by the floor during the bounce, we need to consider the change in momentum of the ball.
The momentum of an object is given by the product of its mass and velocity. In this case, the mass of the basketball is given as 0.65 kg. We are also given the speed at which the ball hits the floor, which is 5.7 m/s, and the angle of impact, which is 62 degrees to the vertical.
Let's denote the initial momentum of the ball as p_initial and the final momentum after the bounce as p_final. The magnitude of the impulse delivered by the floor can be calculated as the change in momentum, which is:
Impulse = |p_final - p_initial|
To calculate the initial momentum, we need to consider the velocity vector of the ball before the bounce. The magnitude of the velocity can be calculated using the given speed:
v_initial = 5.7 m/s
The horizontal component of the velocity can be calculated as:
v_initial_horizontal = v_initial * cos(θ)
where θ is the angle of impact, which is 62 degrees.
v_initial_horizontal = 5.7 m/s * cos(62°)
The vertical component of the velocity can be calculated as:
v_initial_vertical = v_initial * sin(θ)
v_initial_vertical = 5.7 m/s * sin(62°)
Now we can calculate the initial momentum as the product of the mass and velocity:
p_initial = m * v_initial
p_initial = 0.65 kg * √(v_initial_horizontal² + v_initial_vertical²)
To calculate the final momentum, we consider that the ball rebounds with the same speed and angle. Therefore, the final momentum has the same magnitude as the initial momentum, but the direction is opposite. The horizontal and vertical components of the final velocity remain the same as the initial velocity:
v_final_horizontal = v_initial_horizontal
v_final_vertical = -v_initial_vertical
Now we can calculate the final momentum as the product of the mass and velocity:
p_final = m * √(v_final_horizontal² + v_final_vertical²)
p_final = 0.65 kg * √(v_initial_horizontal² + (-v_initial_vertical)²)
Finally, we can calculate the magnitude of the impulse as the difference between the final and initial momentum:
Impulse = |p_final - p_initial|
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As you go up 4. A square current loop of length L on each side carrying a counter-clockwise current / is placed in a region of constant, uniform magnetic field B directed out of the page. O (a) What is the magnitude of the net magnetic force acting on the current loop? [5] AILB B. 4ILB C. 2ILB D. zero (b) What is the direction of the net magnetic force acting on the current loop? [5] A. Up the page B. Down the page C. To the left E. no net force OD To the right (c) Suppose the same loop of wire, carrying the same current were placed in a non-uniform magnetic field in which the magnetic field strength B decreases as you go up the page. If this were the case, what would be the direction of the net magnetic force acting on the current loop? [5] C. To the left A. Up the page To the right, B. Down the page E. no net force I1I. BLO 0.0 0 个儿
The magnitude of the net magnetic force acting on the current loop is F = |I| * L * B * sin(90°) = |I| * L * B. The magnitude of the net magnetic force acting on the current loop, we can use the formula for the magnetic force on a current-carrying wire in a magnetic field:
F = |I| * L * B * sin(theta)
where F is the magnitude of the magnetic force, |I| is the magnitude of the current, L is the length of the loop, B is the magnitude of the magnetic field, and theta is the angle between the current direction and the magnetic field direction. In this case, the current loop is a square with sides of length L. Since the current is counter-clockwise, the angle theta between the current direction and the magnetic field direction (out of the page) is 90 degrees.
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A 340-N solid sphere 0.20 m in radius rolls without slipping 6.0 m down a ramp that is inclined at 25° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest? in rad/s
A solid sphere of mass 340 N and radius 0.20 m rolls without slipping down a ramp inclined at 25° with the horizontal. The task is to determine the angular speed of the sphere at the bottom of the slope, assuming it starts from rest.
To solve this problem, we can apply the principle of conservation of mechanical energy. As the sphere rolls without slipping, both translational kinetic energy and rotational kinetic energy are involved.
First, we calculate the potential energy of the sphere at the top of the ramp using its mass and height. Then, we equate this potential energy to the sum of the translational and rotational kinetic energies at the bottom of the slope.
The translational kinetic energy of the sphere can be calculated using its mass and the velocity acquired during the descent. The rotational kinetic energy depends on the moment of inertia of the solid sphere, which is given by (2/5) * m * r^2, where m is the mass of the sphere and r is its radius.
Equating the potential energy to the sum of translational and rotational kinetic energies, we can solve for the velocity of the sphere at the bottom of the slope. The angular speed can then be obtained by dividing the translational velocity by the radius of the sphere.
By substituting the given values for the mass, radius, and angle of inclination, we can calculate the angular speed of the sphere at the bottom of the slope in radians per second.
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Now we have a rod-shaped space station of length 1232 m and mass 4.24 x 10^6 kg, which can change its length (kind of like an old-fashioned telescope), without changing its overall mass. Suppose that the station is initially rotating at a constant rate of 2.41 rpm. If the length of the rod is reduced to 2.41 m, what will be the new rotation rate of the space station? 2.14 rpm 4.46 rpm 3.57 rpm 5.35 rpm
The new rotation rate of the space station, after reducing the length to 2.41 m, is approximately 1231.81 rpm.
To solve this problem, we can apply the principle of conservation of angular momentum. According to this principle, the total angular momentum of a system remains constant if no external torques act on it.
The angular momentum of a rotating object is given by the equation:
L = Iω
Where:
L is the angular momentum,
I is the moment of inertia of the object, and
ω is the angular velocity.
In this case, the space station is initially rotating at a constant rate of 2.41 rpm. Let's denote the initial length of the rod as L0 = 1232 m and the initial rotation rate as ω0.
The moment of inertia of the space station depends on the distribution of mass and the axis of rotation. However, since the problem states that changing the length of the rod does not change its overall mass, we can assume that the moment of inertia remains constant during the length reduction.
Let's calculate the initial angular momentum of the space station:
L0 = I0 ω0
Now, the length of the rod is reduced to 2.41 m. Let's denote the final length as Lf = 2.41 m and the final rotation rate as ωf, which we need to find.
Since the overall mass of the space station is unchanged, reducing the length will result in a decrease in the moment of inertia. Let's assume the reduction factor in the length is k.
The moment of inertia is directly proportional to the square of the length (I ∝ L²). So, the reduction factor in the moment of inertia is k².
Now, let's calculate the final angular momentum of the space station:
Lf = kL0
I0 ω0 = k² I0 ωf
We can cancel out I0 from both sides:
ω0 = k² ωf
Now we can solve for ωf:
ωf = ω0 / k²
To find the reduction factor k, we can calculate the ratio of the final length to the initial length:
k = Lf / L0
k = 2.41 m / 1232 m
Substituting this value into the equation for ωf, we have:
ωf = ω0 / (2.41 m / 1232 m)²
ωf = ω0 / (2.41 / 1232)²
ωf = ω0 / (0.001957)
Now we can substitute the given value of ω0 = 2.41 rpm:
ωf = 2.41 rpm / 0.001957
ωf ≈ 1231.81 rpm
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The magnetic flux that passes through one turn of a 18 -turn coil of wire changes to 4.22 from 9.27 Wb in a time of 0.0541 s. Th. average induced current in the coil is 250 A. What is the resistance of the wire?
The resistance of the wire is 0.056 Ω.
The average induced current in the coil is given as 250 A. This current is induced due to the change in magnetic flux passing through the coil.The change in magnetic flux is calculated by subtracting the initial flux from the final flux: ΔΦ = 4.22 Wb - 9.27 Wb = -5.05 Wb.According to Faraday's law of electromagnetic induction, the induced electromotive force (emf) is equal to the rate of change of magnetic flux: emf = -dΦ/dt.Since the coil has 18 turns, the induced current is given by Ohm's law: I = emf / R, where R is the resistance of the wire.Rearranging the equation, we have R = emf / I. Plugging in the values, we get R = (-5.05 Wb / 0.0541 s) / 250 A = 0.056 Ω.Therefore, the resistance of the wire is 0.056 Ω.
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