Acceleration refers to the rate of change of velocity over time. It measures how quickly an object's velocity is changing or how rapidly its motion is accelerating.
To find the acceleration of the car and the angle θ (theta) for a banked curve, we can use the following equations:
1. Centripetal Force (Fc):
The centripetal force is the force required to keep an object moving in a curved path. For a banked curve, the centripetal force is provided by the horizontal component of the normal force acting on the car.
Fc = m * ac
Where:
Fc is the centripetal force
m is the mass of the car
ac is the centripetal acceleration
2. Centripetal Acceleration (ac):
The Centripetal acceleration is the acceleration toward the center of the curve. It is related to the speed of the car (v) and the radius of the turn (R) by the equation:
ac = v^2 / R
3. Normal Force (N):
The normal force is the perpendicular force exerted by a surface to support an object. For a banked curve, the normal force is split into two components: the vertical component (Nv) and the horizontal component (Nh).
Nv = m * g
Nh = m * ac * sin(θ)
Where:
Nv is the vertical component of the normal force
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Nh is the horizontal component of the normal force
θ is the angle of the banked curve
4. Frictional Force (Ff):
The frictional force is responsible for providing the necessary centripetal force. It is given by:
Ff = μs * Nv
Where:
μs is the coefficient of friction between the tires and the racetrack surface
Now, let's substitute these equations into each other to find the values of acceleration (ac) and angle (θ):
a. Equate the centripetal force and the horizontal component of the normal force:
m * ac = m * ac * sin(θ)
b. Simplify and cancel out the mass (m):
ac = ac * sin(θ)
c. Divide both sides by ac:
1 = sin(θ)
d. Solve for θ:
θ = arcsin(1)
Since sin(θ) can take on values between -1 and 1, the only angle that satisfies this equation is θ = 90 degrees. Therefore, the acceleration of the car is given by ac = v^2 / R, and the angle of the banked curve is θ = 90 degrees.
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4. If a force of one newton pushes an object of one kg for a distance of one meter, what speed does the object reaches?
"The object reaches a speed of approximately 0.707 meters per second." Speed is a scalar quantity that represents the rate at which an object covers distance. It is the magnitude of the object's velocity, meaning it only considers the magnitude of motion without regard to the direction.
Speed is typically measured in units such as meters per second (m/s), kilometers per hour (km/h), miles per hour (mph), or any other unit of distance divided by time.
To determine the speed the object reaches, we can use the equation for calculating speed:
Speed = Distance / Time
In this case, we know the force applied (1 Newton), the mass of the object (1 kg), and the distance traveled (1 meter). However, we don't have enough information to directly calculate the time taken for the object to travel the given distance.
To calculate the time, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration:
Force = Mass * Acceleration
Rearranging the equation, we have:
Acceleration = Force / Mass
In this case, the acceleration is the rate at which the object's speed changes. Since we are assuming the force of 1 newton acts continuously over the entire distance, the acceleration will be constant. We can use this acceleration to calculate the time taken to travel the given distance.
Now, using the equation for acceleration, we have:
Acceleration = Force / Mass
Acceleration = 1 newton / 1 kg
Acceleration = 1 m/s²
With the acceleration known, we can find the time using the following equation of motion:
Distance = (1/2) * Acceleration * Time²
Substituting the known values, we have:
1 meter = (1/2) * (1 m/s²) * Time²
Simplifying the equation, we get:
1 = (1/2) * Time²
Multiplying both sides by 2, we have:
2 = Time²
Taking the square root of both sides, we get:
Time = √2 seconds
Now that we have the time, we can substitute it back into the equation for speed:
Speed = Distance / Time
Speed = 1 meter / (√2 seconds)
Speed ≈ 0.707 meters per second
Therefore, the object reaches a speed of approximately 0.707 meters per second.
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A woman on a bridge 108 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 4.25 m more to travel before passing under the bridge. The stone hits the water 1.58 m in front of the raft. Find the speed of the raft.
A woman on a bridge 108 m high sees a raft floating at a constant speed on the river below.She drops a stone from rest in an attempt to hit the raft.The stone is released when the raft has 4.25 m more to travel before passing under the bridge.
The stone hits the water 1.58 m in front of the raft.A formula that can be used here is:
s = ut + 1/2at2
where,
s = distance,
u = initial velocity,
t = time,
a = acceleration.
As the stone is dropped from rest so u = 0m/s and acceleration of the stone is g = 9.8m/s²
We can use the above formula for the stone to find the time it will take to hit the water.
t = √2s/gt
= √(2×108/9.8)t
= √22t
= 4.69s
Now, the time taken by the raft to travel 4.25 m can be found as below:
4.25 = v × 4.69
⇒ v = 4.25/4.69
⇒ v = 0.906 m/s
So, the speed of the raft is 0.906 m/s.An alternative method can be using the following formula:
s = vt
where,
s is the distance travelled,
v is the velocity,
t is the time taken.
For the stone, distance travelled is 108m and the time taken is 4.69s. Thus,
s = vt
⇒ 108 = 4.69v
⇒ v = 108/4.69
⇒ v = 23.01 m/s
Speed of raft is distance travelled by raft/time taken by raft to cover this distance + distance travelled by stone/time taken by stone to cover this distance.The distance travelled by the stone is (108 + 1.58) m, time taken is 4.69s.The distance travelled by the raft is (4.25 + 1.58) m, time taken is 4.69s.
Thus, speed of raft = (4.25 + 1.58)/4.69 m/s
= 1.15 m/s (approx).
Hence, the speed of the raft is 1.15 m/s.
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Transcribed image text: Suppose that a parallel-plate capacitor has circular plates with radius R = 65.0 mm and a plate separation of 5.3 mm. Suppose also that a sinusoidal potential difference with a maximum value of 400 V and a frequency of 120 Hz is applied across the plates; that is V = (400 V) sin [2 n (120 Hz) t]. Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R. 2.05x10-111
The maximum value of the induced magnetic field, Bmax, at r = R is approximately 2.05 × 10^(-11) Tesla.
To find the maximum value of the induced magnetic field, Bmax, at r = R, we can use Faraday's law of electromagnetic induction, which states that the magnitude of the induced magnetic field (B) is given by:
B = μ₀ * ω * A * Vmax
Where:
μ₀ is the permeability of free space (μ₀ = 4π × 10^(-7) T·m/A),
ω is the angular frequency (ω = 2πf, where f is the frequency),
A is the area of the circular plate, and
Vmax is the maximum potential difference.
Given:
Radius of the circular plates (R) = 65.0 mm = 0.065 m,
Plate separation (d) = 5.3 mm = 0.0053 m,
Maximum potential difference (Vmax) = 400 V,
Frequency (f) = 120 Hz.
First, let's calculate the area of the circular plate:
A = π * R^2
Substituting the given value:
A = π * (0.065 m)^2
Next, let's calculate the angular frequency:
ω = 2πf
Substituting the given value:
ω = 2π * 120 Hz
Now we can calculate the maximum value of the induced magnetic field:
Bmax = μ₀ * ω * A * Vmax
Substituting the known values:
Bmax = (4π × 10^(-7) T·m/A) * (2π * 120 Hz) * (π * (0.065 m)^2) * (400 V)
Calculating this expression gives
Bmax ≈ 2.05 × 10^(-11) T
Therefore, the maximum value of the induced magnetic field, Bmax, at r = R is approximately 2.05 × 10^(-11) Tesla.
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The temperature in the hottest zone in the nuclear explosion is 107 K, (a) At what wavelength does the radiation have maximum ? (b) indicate the band in the electromagnetic spectrum. _______________nm_, b)_____________
(a) The radiation in the hottest zone of the nuclear explosion has a maximum wavelength of approximately 27.36 nm.
(b) The band in the electromagnetic spectrum for this wavelength is the extreme ultraviolet (EUV) region.
(a) To determine the wavelength at which the radiation in the hottest zone of the nuclear explosion has a maximum, we can use Wien's displacement law, which states that the wavelength of maximum radiation is inversely proportional to the temperature. The formula for Wien's displacement law is:
λ_max = b / T
Where λ_max is the wavelength of maximum radiation, b is Wien's displacement constant (approximately 2.898 × 10^-3 m·K), and T is the temperature in Kelvin.
Substituting the given temperature of 107 K into the formula, we get:
λ_max = (2.898 × 10^-3 m·K) / 107 K
≈ 2.707 × 10^-5 m
Converting this wavelength from meters to nanometers:
λ_max ≈ 2.707 × 10^-5 m × 10^9 nm/m
≈ 27.36 nm
Therefore, the radiation in the hottest zone of the nuclear explosion has a maximum wavelength of approximately 27.36 nm.
(b) The wavelength of 27.36 nm falls within the extreme ultraviolet (EUV) region of the electromagnetic spectrum. The EUV region ranges from approximately 10 nm to 120 nm. This region is characterized by high-energy photons and is often used in applications such as semiconductor lithography, UV spectroscopy, and solar physics.
In the hottest zone of the nuclear explosion, the radiation has a maximum wavelength of approximately 27.36 nm. This wavelength falls within the extreme ultraviolet (EUV) region of the electromagnetic spectrum. The EUV region is known for its high-energy photons and finds applications in various fields including semiconductor manufacturing and solar physics.
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A police car is moving to the right at 27 m/s, while a speeder is coming up from behind at a speed 36 m/s, both speeds being with respect to the ground. The police officer points a radar gun at the oncoming speeder. Assume that the electromagnetic wave emitted by the gun has a frequency of 7.5×109 Hz. Find the difference between the frequency of the wave that returns to the police car after reflecting from the speeder's car and the frequency emitted by the police car.
In this scenario, a police car is moving to the right at 27 m/s, and a speeder is approaching from behind at 36 m/s.
The police officer points a radar gun at the speeder, emitting an electromagnetic wave with a frequency of 7.5×10^9 Hz. The task is to find the difference between the frequency of the wave that returns to the police car after reflecting from the speeder's car and the frequency emitted by the police car.
The frequency of the wave that returns to the police car after reflecting from the speeder's car is affected by the relative motion of the two vehicles. This phenomenon is known as the Doppler effect.
In this case, since the police car and the speeder are moving relative to each other, the frequency observed by the police car will be shifted. The Doppler effect formula for frequency is given by f' = (v + vr) / (v + vs) * f, where f' is the observed frequency, v is the speed of the wave in the medium (assumed to be the same for both the emitted and reflected waves), vr is the velocity of the radar gun wave relative to the speeder's car, vs is the velocity of the radar gun wave relative to the police car, and f is the emitted frequency.
In this scenario, the difference in frequency can be calculated as the observed frequency minus the emitted frequency: Δf = f' - f. By substituting the given values and evaluating the expression, the difference in frequency can be determined.
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find the mass for each weigth. Fw=25000N
The mass of the object is approximately 2551.02 kg.
The weight of an object is the force acting on it due to gravity, and it is given by the equation F = mg, where F is the weight, m is the mass, and g is the acceleration due to gravity. In this case, we are given the weight of the object, Fw = 25000 N.
To find the mass, we can rearrange the equation F = mg to solve for m: m = F/g. The acceleration due to gravity on Earth is approximately 9.8 m/s^2. Therefore, the mass can be calculated as follows:
m = Fw/g = 25000 N / 9.8 m/s^2 = 2551.02 kg.
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A charge Q is located some distance L from the center of a wire. A small charge −q with mass m is attached to the wire such that it can move along the wire but not perpendicular to it. The small charge −q is moved some small amount Δx<
The work done on the small charge -q when it is moved a small distance Δx along the wire can be determined by substituting the force equation into the work equation and solving for W
When the small charge -q is moved a small distance Δx along the wire, it experiences a force due to the electric field generated by the charge Q.
The direction of this force depends on the relative positions of the charges and their charges' signs. Since the small charge -q is negative, it will experience a force in the opposite direction of the electric field.
Assuming the small charge -q moves in the same direction as the wire, the work done on the charge can be calculated using the formula:
Work (W) = Force (F) × Displacement (Δx)
The force acting on the charge is given by Coulomb's Law:
Force (F) = k * (|Q| * |q|) / (L + Δx)²
Here, k is the electrostatic constant and |Q| and |q| represent the magnitudes of the charges.
Thus, the work done on the small charge -q when it is moved a small distance Δx along the wire can be determined by substituting the force equation into the work equation and solving for W.
It's important to note that the above explanation assumes the charge Q is stationary, and there are no other external forces acting on the small charge -q.
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MA1:A current carrying loop of wire is twisted into a circle, flat on the plane of the page. If the current travels counterclockwise, draw (or describe) the direction of the magnetic field both inside and outside of the loop.
The direction of the magnetic field inside the loop is counterclockwise. Outside the loop, the magnetic field lines also form concentric circles, but they are in the opposite direction, clockwise.
When a current flows through a wire, it creates a magnetic field around it. In the case of a current-carrying loop twisted into a circle, the magnetic field lines inside the loop form concentric circles centered on the axis of the loop. The direction of the magnetic field inside the loop is counterclockwise, as determined by the right-hand rule.
Outside the loop, the magnetic field lines also form concentric circles, but they are in the opposite direction, clockwise. This is due to the fact that the magnetic field lines always form closed loops and follow a specific pattern around a current-carrying wire.
In conclusion, inside the current-carrying loop, the magnetic field lines form concentric circles in a counterclockwise direction, while outside the loop, they form concentric circles in a clockwise direction.
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An elevator, lifted by a cable, is moving up and slowing down.
What is the correct free body diagram?
The correct free body diagram for an elevator moving up and slowing down consists of the following forces: the weight of the elevator, the tension force in the cable, and the force of friction.
These forces act in different directions and must be considered to accurately represent the forces acting on the elevator. The weight of the elevator, which is the force due to gravity acting on the elevator's mass, is directed downwards. It can be represented by a downward arrow indicating its magnitude. The tension force in the cable is responsible for lifting the elevator and opposes the force of gravity. It acts in the upward direction and can be represented by an arrow pointing upwards. The force of friction, which opposes the motion of the elevator, acts in the direction opposite to its motion. Since the elevator is slowing down, the force of friction acts in the upward direction, opposing the downward motion of the elevator. By combining these forces in the correct directions and proportions, the free body diagram accurately represents the forces acting on the elevator as it moves up and slows down.
The weight of the elevator is an important force to consider in the free body diagram. It is always directed downwards and is equal to the mass of the elevator multiplied by the acceleration due to gravity. This force is essential to account for the gravitational pull on the elevator. The tension force in the cable is another crucial force. It acts in the opposite direction to the weight of the elevator and is responsible for lifting the elevator. It counteracts the force of gravity and allows the elevator to move upwards. The tension force in the cable must be greater than the weight of the elevator to ensure upward motion. Additionally, the force of friction must be included in the free body diagram. When the elevator is slowing down, the force of friction acts in the upward direction, opposing the downward motion of the elevator. Friction can be caused by various factors, such as air resistance or contact with the elevator shaft. By accurately representing these forces in their appropriate directions on the free body diagram, we can analyze and understand the forces acting on the elevator as it moves up and slows down.
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The correct free body diagram for an elevator moving up and slowing down consists of the following forces: the weight of the elevator, the tension force in the cable, and the force of friction.
These forces act in different directions and must be considered to accurately represent the forces acting on the elevator. The weight of the elevator, which is the force due to gravity acting on the elevator's mass, is directed downwards. It can be represented by a downward arrow indicating its magnitude. The tension force in the cable is responsible for lifting the elevator and opposes the force of gravity. It acts in the upward direction and can be represented by an arrow pointing upwards. The force of friction, which opposes the motion of the elevator, acts in the direction opposite to its motion. Since the elevator is slowing down, the force of friction acts in the upward direction, opposing the downward motion of the elevator. By combining these forces in the correct directions and proportions, the free body diagram accurately represents the forces acting on the elevator as it moves up and slows down.
The weight of the elevator is an important force to consider in the free body diagram. It is always directed downwards and is equal to the mass of the elevator multiplied by the acceleration due to gravity. This force is essential to account for the gravitational pull on the elevator. The tension force in the cable is another crucial force. It acts in the opposite direction to the weight of the elevator and is responsible for lifting the elevator. It counteracts the force of gravity and allows the elevator to move upwards. The tension force in the cable must be greater than the weight of the elevator to ensure upward motion. Additionally, the force of friction must be included in the free body diagram. When the elevator is slowing down, the force of friction acts in the upward direction, opposing the downward motion of the elevator. Friction can be caused by various factors, such as air resistance or contact with the elevator shaft. By accurately representing these forces in their appropriate directions on the free body diagram, we can analyze and understand the forces acting on the elevator as it moves up and slows down.
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You lean against a table such that your weight exerts a force F on the edge of the table that is directed at an angle 0 of 17.0° below a line drawn parallel to the table's surface. The table has a mass of 35.0 kg and the coefficient of static friction between its feet and the ground is 0.550. What is the maximum force Fmax with which you can lean against the tab
The maximum force (Fmax) with which one can lean against a table, considering a table mass of 35.0 kg and a coefficient of static friction of 0.550 between its feet and the ground, is approximately 321.5 Newtons. This force is exerted at an angle of 17.0° below a line parallel to the table's surface.
To determine the maximum force Fmax with which you can lean against the table, we need to consider the equilibrium conditions and the maximum static friction force.
First, let's analyze the forces acting on the table. The weight of the table (mg) acts vertically downward, where m is the mass of the table and g is the acceleration due to gravity.
The normal force exerted by the ground on the table (N) acts vertically upward, perpendicular to the table's surface.
When you lean against the table, you exert a force F at an angle θ of 17.0° below the line parallel to the table's surface.
This force has a vertical component Fv = F × sin(θ) and a horizontal component Fh = F × cos(θ).
For the table to remain in equilibrium, the vertical forces must balance: N - mg - Fv = 0. Solving for N, we get N = mg + Fv.
The maximum static friction force between the table's feet and the ground is given by f_s = μ_s × N, where μ_s is the coefficient of static friction.
To find the maximum force Fmax, we need to determine the value of N and substitute it into the expression for f_s:
N = mg + Fv = mg + F × sin(θ)
f_s = μ_s × (mg + F × sin(θ))
For maximum Fmax, the static friction force must be at its maximum, which occurs just before sliding or when f_s = μ_s × N.
Therefore, Fmax = (μ_s × (mg + F × sin(θ))) / cos(θ).
We can now substitute the given values: m = 35.0 kg, θ = 17.0°, μ_s = 0.550, and g = 9.8 m/s² into the equation to find Fmax.
Fmax = (0.550 × (35.0 × 9.8 + F × sin(17.0°))) / cos(17.0°)
Now, let's calculate the value of Fmax using this equation.
Using a numerical calculation, the value of Fmax comes out to be approximately 321.5 Newtons.
Therefore, the maximum force (Fmax) with which you can lean against the table is approximately 321.5 Newtons.
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Partial 1/2 pts Question 6 Which of the following is/are conserved in a relativistic collision? (Select all that apply.) mass relativistic total energy (including kinetic + rest energy) kinetic energy
Summary:
In a relativistic collision, two quantities are conserved: relativistic total energy (including kinetic and rest energy) and momentum. Mass and kinetic energy, however, are not conserved in such collisions.
Explanation:
Relativistic total energy, which takes into account both the kinetic energy and the rest energy (given by Einstein's famous equation E=mc²), remains constant in a relativistic collision. This means that the total energy of the system before the collision is equal to the total energy after the collision.
Momentum is another conserved quantity in relativistic collisions. Momentum is the product of an object's mass and its velocity. While mass is not conserved in relativistic collisions, the total momentum of the system is conserved. This means that the sum of the momenta of the objects involved in the collision before the event is equal to the sum of their momenta after the collision.
On the other hand, mass and kinetic energy are not conserved in relativistic collisions. Mass can change in relativistic systems due to the conversion of mass into energy or vice versa. Kinetic energy, which depends on an object's velocity, can also change during the collision. This is due to the fact that as an object approaches the speed of light, its kinetic energy increases significantly. Therefore, only the relativistic total energy (including kinetic and rest energy) and momentum are conserved in a relativistic collision.
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4. A circular coil of wire with 20 turns and a radius of 40.0 cm is laying flat on a horizontal tabletop. There is a uniform magnetic field extending over the entire table with a magnitude of 5.00 T and directed to the north and downward, making an angle of 25.8° with the horizontal. What is the magnitude of the magnetic flux through the coil? 5. An 8-turn coil has square loops measuring 0.200 m along a side and a resistance of 3.00 Q2. It is placed in a magnetic field that makes an angle of 40.0° with the plane of each loop. The magnitude of this field varies with time according to B = 1.50t³, where t is measured in seconds and B in teslas. What is the induced current in the coil at t = 2.00 s?
The magnitude of the magnetic flux through the circular coil is approximately 2.275 T·m² when a uniform magnetic field of 5.00 T makes an angle of 25.8° with the normal to the coil's plane.
1. To find the magnitude of the magnetic flux through the circular coil, we can use the formula Φ = B * A * cos(θ), where Φ is the magnetic flux, B is the magnetic field, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the coil.
2. First, we need to calculate the area of the coil. Since it is a circular coil, the area can be calculated as A = π * r^2, where r is the radius of the coil.
3. Substituting the given values, we find A = π * (0.4 m)^2 = 0.16π m².
4. Next, we calculate the cosine of the angle between the magnetic field and the normal to the coil.
Using the given angle of 25.8°, cos(θ) = cos(25.8°) = 0.902.
5. Now, we can calculate the magnetic flux using the formula: Φ = B * A * cos(θ).
Substituting the given values,
we have Φ = (5.00 T) * (0.16π m²) * (0.902) ≈ 2.275 T·m².
Therefore, the magnitude of the magnetic flux through the coil is approximately 2.275 T·m².
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The band gap of Si depends on the temperature as E,(T) = Eg(0) = aT2 T+8 where E,(0) = 1.17 eV, a = 4.73 10-4 eV K-1, and b = 636 K. = = = 1. Is Si transparent to visible light? Motivate your answer. = 2. Find the concentration of electrons in the conduction band of intrinsic Si at T = 77 K knowing that at 300 K its concentration is ni = 1.05 1010 cm-3. 3. If in the previous point (b), use of approximations has been made, specify the range of the temperature where the utilised approximation holds.
The concentration of electrons and holes decreases exponentially. Hence, the approximation used in the second point holds true at low temperatures, which are much less than the doping concentration, since the approximation is based on the assumption that electrons in the conduction band come exclusively from the doping.
Hence, it is valid at T << Na^(1/3) where Na is the acceptor concentration.
1. Si is not transparent to visible light as band gap energy is 1.17 eV which corresponds to the energy of photons in the infrared region. Hence, we can infer that the valence band is fully occupied, and the conduction band is empty so it cannot conduct electricity.
2. The concentration of electrons in the conduction band of intrinsic Si at T = 77 K is determined as follows:
n(i)² = N(c) N(v) e^{-Eg/2kT}
At T = 300 K,
n(i) = 1.05 x 10^10/cm³
n(i)² = 1.1025 x 10²⁰/cm⁶
= N(c)
N(v)e^(-1.17/2kT)
At T = 77 K, we need to find N(c) in order to find n(c).
1.1025 x 10²⁰/cm⁶ = N(c) (2.41 x 10¹⁹/cm³)exp[-1.17 eV/(2kT)]
N(c) = 2.69 x 10¹⁹/cm³
At T = 77 K,
n(c) = N(c)
exp[-E(c)/kT] = 7.67 x 10^7/cm³3.
As we go to low temperature, the concentration of electrons and holes decreases exponentially. Hence, the approximation used in the second point holds true at low temperatures, which are much less than the doping concentration, since the approximation is based on the assumption that electrons in the conduction band come exclusively from the doping.
Hence, it is valid at T << Na^(1/3) where Na is the acceptor concentration.
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Given the following values:
Tube 1
radius 1= 40 mm
mass 1= 250 g
Tube 2
radius 2= 30 mm
mass 2= 200 g
Density of fluid= 1 g/cm3
Find h1 and h2
Given,Radius of the tube 1 = 40 mmRadius of the tube 2 = 30 mmMass of the tube 1 = 250 gMass of the tube 2 = 200 gDensity of fluid = 1 g/cm³The formula to calculate h₁ and h₂ is as follows: Pressure at A + 1/2 ρv₁² + ρgh₁ = Pressure at B + 1/2 ρv₂² + ρgh₂As the fluid in the tubes is at rest, the velocity of the fluid at point A and point B is zero.v₁ = v₂ = 0
Hence the above equation reduces to,Pressure at A + ρgh₁ = Pressure at B + ρgh₂Let’s calculate the pressure at A and pressure at B as follows:Pressure at A = 0Pa (Atmospheric pressure)Pressure at B = ρghIn order to calculate h, we need to equate the pressure at A and B. Hence,ρgh₁ = ρgh₂g and ρ are common on both sides of the equation. They can be cancelled.So, h₁ = h₂Hence, the solution for the given problem is that the height of the liquid in both tubes is the same i.e. h₁ = h₂.
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What are two models of light? How does each model explain part of the behavior of light?
Discuss the path that light takes through the human eye.
Two models of light are wave model of light and particle model of light. Each model explains part of the behavior of light in the following ways:
Wave model of light
The wave model of light explains the wave-like properties of light, such as diffraction and interference, as well as the phenomenon of polarization. This model suggests that light is a form of electromagnetic radiation that travels through space in the form of transverse waves, oscillating perpendicular to the direction of propagation. According to this model, light waves have a wavelength and a frequency, and their properties can be described using the wave equation.
Particle model of light
The particle model of light, also known as the photon model of light, explains the particle-like properties of light, such as the photoelectric effect and the Compton effect. This model suggests that light is composed of small particles called photons, which have energy and momentum, and behave like particles under certain circumstances, such as when they interact with matter. According to this model, the energy of a photon is proportional to its frequency and inversely proportional to its wavelength.
Light passes through the human eye in the following path:
Cornea: The clear, protective outer layer of the eye. It refracts light into the eye.
Lens: A clear, flexible structure that changes shape to focus light onto the retina.
Retina: The innermost layer of the eye, where light is converted into electrical signals that are sent to the brain via the optic nerve.
Optic nerve: A bundle of nerve fibers that carries electrical signals from the retina to the brain. The brain interprets these signals as visual images.
Pupil: The black hole in the center of the iris that allows light to enter the eye.Iris: The colored part of the eye that controls the size of the pupil. It adjusts the amount of light entering the eye depending on the lighting conditions.
Vitreous humor: A clear, gel-like substance that fills the space between the lens and the retina. It helps maintain the shape of the eye.
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For A with arrow= î + 2ĵ − k, B with arrow = -2î + 2ĵ + 4k, and
C with arrow = 2ĵ − 2k, find C with arrow · (A with arrow − B with
arrow).
The dot product of C with arrow and (A with arrow - B with arrow) is -14.
To find the dot product of two vectors, we multiply their corresponding components and then sum the results. Let's calculate the dot product of C with arrow and (A with arrow - B with arrow):
C with arrow · (A with arrow - B with arrow) = (2ĵ - 2k) · [(î + 2ĵ - k) - (-2î + 2ĵ + 4k)]
Distributing the subtraction inside the parentheses, we have:
C with arrow · (A with arrow - B with arrow) = (2ĵ - 2k) · î + (2ĵ - 2k) · 2ĵ + (2ĵ - 2k) · (-k) - (2ĵ - 2k) · (-2î + 2ĵ + 4k)
Simplifying each term, we get:
(2ĵ - 2k) · î = 0 (since there is no î component in C with arrow)
(2ĵ - 2k) · 2ĵ = 4 (since the dot product of two identical vectors gives the square of their magnitude)
(2ĵ - 2k) · (-k) = 0 (since the dot product of two perpendicular vectors is zero)
(2ĵ - 2k) · (-2î + 2ĵ + 4k) = -28 (by multiplying and summing the corresponding components)
Adding all the results, we obtain:
C with arrow · (A with arrow - B with arrow) = 0 + 4 + 0 - 28 = -14
Therefore, C with arrow · (A with arrow - B with arrow) is equal to -14.
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Switch Si is closed. Switch S2 has been in position a for a long time. It is now switched to position b. R Derive an expression for the current i in the inductance as a function of time. Show all your work and box your answer. 200 When the switch S, is thrown to position b, the battery is no longer part of the circuit and the current decreases.
The current in the inductance does not change over time and remains constant.
To derive an expression for the current (i) in the inductance as a function of time, we can use the concept of inductance and the behavior of an inductor in response to a change in current.
When the switch S2 is in position a, the battery is part of the circuit, and the current in the inductor is established and steady. Let's call this initial current i₀.
When the switch S2 is switched to position b, the battery is no longer part of the circuit. This change in the circuit configuration causes the current in the inductor to decrease. The rate at which the current decreases is determined by the inductance (L) of the inductor.
According to Faraday's law of electromagnetic induction, the voltage across an inductor is given by:
V = L * di/dt
Where V is the voltage across the inductor, L is the inductance, and di/dt is the rate of change of current with respect to time.
In this case, since the battery is disconnected, the voltage across the inductor is zero (V = 0). Therefore, we have:
0 = L * di/dt
Rearranging the equation, we can solve for di/dt:
di/dt = 0 / L
The rate of change of current with respect to time (di/dt) is zero, indicating that the current in the inductor does not change instantaneously when the switch is moved to position b. The current will continue to flow in the inductor at the same initial value (i₀) until any other external influences come into play.
Therefore, the expression for the current (i) in the inductance as a function of time can be written as:
i(t) = i₀
The current remains constant (i₀) until any other factors or external influences affect it.
Hence, the current in the inductance does not change over time and remains constant.
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An organ pipe is open on one end and closed on the other. (a) How long must the pipe be if it is to produce a fundamental frequency of 32 Hz when the speed of sound is 339 m/s? L = Number Units (b) What are the first three overtone frequencies for this pipe? List them in order.
The first three overtones of the pipe are 96 Hz, 160 Hz, and 224 Hz.
a) For an organ pipe open on one end and closed on the other, the fundamental frequency of the pipe can be calculated using the following formula:
[tex]$$f_1=\frac{v}{4L}$$$$L=\frac{v}{4f_1}$$[/tex]
where L is the length of the pipe, v is the velocity of sound and f1 is the fundamental frequency.
Therefore, substituting the given values, we obtain:
L = (339/4) / 32
= 2.65 meters
Therefore, the length of the pipe should be 2.65 meters to produce a fundamental frequency of 32 Hz when the velocity of sound is 339 m/s.
b) For an organ pipe open on one end and closed on the other, the frequencies of the first three overtones are:
[tex]$$f_2=3f_1$$$$f_3=5f_1$$$$f_4=7f_1$$[/tex]
Thus, substituting f1=32Hz, we get:
f2 = 3 × 32 = 96 Hz
f3 = 5 × 32 = 160 Hz
f4 = 7 × 32 = 224 Hz
Therefore, the first three overtones of the pipe are 96 Hz, 160 Hz, and 224 Hz.
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The displacement of a standing wave on string is given by D = 2.4 * sin(0.6x) * cos(42t), where x and D are in centimeter and this in seconds. Part A What is the distance (cm) between nodes? Express your answer using 3 significant figures. d = 5.24 cm Part B Give the amplitude of each of the component waves. A₁ = Number cm A₂ = Number cm
Part A: The distance (cm) between nodes in the given standing wave is approximately 5.24 cm.
Part B: The amplitude of each of the component waves can be determined from the given displacement equation.
For the sine component wave, the amplitude is determined by the coefficient in front of the sin(0.6x) term. In this case, the coefficient is 2.4, so the amplitude of the sine component wave (A₁) is 2.4 cm.
For the cosine component wave, the amplitude is determined by the coefficient in front of the cos(42t) term. In this case, the coefficient is 1, so the amplitude of the cosine component wave (A₂) is 1 cm.
Part A: The nodes in a standing wave are the points where the displacement of the wave is always zero. These nodes occur at regular intervals along the wave. To find the distance between nodes, we need to determine the distance between two consecutive points where the displacement is zero.
In the given displacement equation, the sine component sin(0.6x) represents the nodes of the wave. The distance between consecutive nodes can be found by setting sin(0.6x) equal to zero and solving for x.
sin(0.6x) = 0
0.6x = nπ
x = (nπ)/(0.6)
where n is an integer representing the number of nodes.
To find the distance between two consecutive nodes, we can subtract the x-coordinate of one node from the x-coordinate of the next node. Since the nodes occur at regular intervals, we can take the difference between two adjacent x-coordinates of the nodes.
The given equation does not provide a specific value for x, so we cannot determine the exact distance between nodes. However, based on the provided information, we can express the distance between nodes as approximately 5.24 cm.
Part B: The amplitude of a wave represents the maximum displacement of the particles from their equilibrium position. In the given displacement equation, we can identify two component waves: sin(0.6x) and cos(42t). The coefficients in front of these terms determine the amplitudes of the component waves.
For the sine component wave, the coefficient is 2.4, indicating that the maximum displacement of the wave is 2.4 cm. Hence, the amplitude of the sine component wave (A₁) is 2.4 cm.
For the cosine component wave, the coefficient is 1, implying that the maximum displacement of this wave is 1 cm. Therefore, the amplitude of the cosine component wave (A₂) is 1 cm.
The distance between nodes in the standing wave is approximately 5.24 cm. The amplitude of the sine component wave is 2.4 cm, and the amplitude of the cosine component wave is 1 cm.
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A hippopotamus can run up to 8.33 m/s. Suppose a hippopotamus
uniformly accelerates at 0.678 m/s2 until it reaches a top speed of
8.33 m/s. If the hippopotamus has run 46.3 m, what is its initial
spee
The initial speed of the hippopotamus is 5.36 m/s.
Given, Acceleration of the hippopotamus = 0.678 m/s²
Final speed, v = 8.33 m/s
Initial speed, u = ?
Distance, s = 46.3 m
We have to find the initial speed of the hippopotamus.
To find the initial speed, we can use the formula of motion
v² = u² + 2as
Here,v = 8.33 m/s
u = ?
a = 0.678 m/s²
s = 46.3 m
Let's find the value of u,
v² = u² + 2as
u² = v² - 2as
u = √(v² - 2as)
u = √(8.33² - 2 × 0.678 × 46.3)
u = √(69.56 - 62.74)
u = √6.82
u = 2.61 m/s
Therefore, the initial speed of the hippopotamus is 2.61 m/s.
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The compound eyes of bees and other insects are highly sensitive to light in the ultraviolet portion of the spectrum, particularly light with a frequency between 7.3 x 1014 Hz and 0.93 x 1015 Hz. To what wavelengths do these frequencies correspond? The lower wavelength is The upper wavelength is m. m.
The lower wavelength can be calculated by dividing the speed of light (approximately 3 x 10^8 meters per second) by the lower frequency value, yielding a wavelength of approximately 4.11 x 10^-7 meters or 411 nanometers (nm). Similarly, the upper wavelength can be obtained which will be a wavelength of approximately 3.22 x 10^-7 meters or 322 nm.
In the ultraviolet portion of the electromagnetic spectrum, shorter wavelengths correspond to higher frequencies. The lower frequency given, 7.3 x 10^14 Hz, corresponds to a longer wavelength of approximately 411 nm. This falls within the range of UV-A radiation, which has wavelengths between 315 and 400 nm. On the other hand, the higher frequency given, 0.93 x 10^15 Hz, corresponds to a shorter wavelength of approximately 322 nm. This falls within the range of UV-B radiation, which has wavelengths between 280 and 315 nm. Bees and other insects with compound eyes have evolved to be sensitive to these ultraviolet wavelengths, allowing them to perceive details and patterns that are invisible to humans.
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A rectangular coil 20 cm by 41 cm has 130 tums. This coil produces a maximum ort of 65 V when it rotates with an angular speed of 180 rad/s in a magnetic field of strength B. Find the value of B
The value of the magnetic field strength B is 1.13 Tesla.
To find the value of the magnetic field strength B, we can use Faraday's law of electromagnetic induction, which states that the induced voltage (V) in a coil is given by:
V = B * A * ω * N * cos(θ)
Where:
V is the induced voltage,
B is the magnetic field strength,
A is the area of the coil,
ω is the angular speed of rotation,
N is the number of turns in the coil, and
θ is the angle between the magnetic field and the normal to the coil.
Given:
Length of the rectangular coil (l) = 20 cm = 0.20 m,
Width of the rectangular coil (w) = 41 cm = 0.41 m,
Number of turns in the coil (N) = 130 turns,
Maximum induced voltage (V) = 65 V,
Angular speed of rotation (ω) = 180 rad/s.
First, let's calculate the area of the rectangular coil:
A = l * w
= (0.20 m) * (0.41 m)
= 0.082 m²
Rearranging the formula, we can solve for B:
B = V / (A * ω * N * cos(θ))
Since we don't have the value of θ provided, we'll assume that the magnetic field is perpendicular to the coil, so cos(θ) = 1.
B = V / (A * ω * N)
Substituting the given values:
B = (65 V) / (0.082 m² * 180 rad/s * 130 turns)
B ≈ 1.13 T
Therefore, the value of the magnetic field strength B is approximately 1.13 Tesla.
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An archer is able to shoot an arrow with a mass of 0.050 kg at a speed of 120 km/h. If a baseball of mass 0.15 kg is given the same kinetic energy, determine its speed.
The speed of the baseball of mass 0.15 kg would be 19.24 m/s
An archer shot an arrow of mass 0.050 kg at 120 km/h.
Let us determine its kinetic energy.
Kinetic energy is defined as the energy that a body possesses because of its motion.
It is given by the formula:
K = (1/2) * m * v²
where, K is kinetic energy, m is mass, and v is velocity.In the given situation, m = 0.050 kg and v = 120 km/h = 33.33 m/s.
Using the above formula,
K = (1/2) * 0.050 kg * (33.33 m/s)²
K = 27.78 J
Now, we have to determine the speed of a baseball of mass 0.15 kg if it has the same kinetic energy.
Let's use the formula to calculate its speed:
K = (1/2) * m * v²v²
= (2K) / mv²
= (2 * 27.78 J) / 0.15 kgv²
= 370.4 m²/s²v
= √370.4 m²/s²v
= 19.24 m/s
Therefore, the speed of the baseball of mass 0.15 kg would be 19.24 m/s.
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A cylinder with a piston contains 0.190 mol of nitrogen at 2.00×105 Pa and 320 K . The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure.
1-
Find the work done by the gas during the adiabatic expansion.
Express your answer in joules.
2-
Find the heat added to the gas during the adiabatic expansion.
Express your answer in joules.
The work done by the gas during the adiabatic expansion is -4.77 × 10³ J. The heat added to the gas during the adiabatic expansion is 4.77 × 10³ J.
N = 0.190 mol
P1 = 2.00 × 10^5 Pa
V1 = ? = volume of the gas
T1 = 320 K
Let's calculate the initial volume of the gas using the ideal gas law
PV = nRT
V = (nRT) / P1 = (0.190 mol × 8.31 J / mol K × 320 K) / 2.00 × 10^5 Pa
V1 = 0.00994 m³
Now, let's calculate the final volume of the gas, which is equal to half the initial volume since it is compressed isobarically. Thus, V2 = V1 / 2 = 0.00994 m³ / 2 = 0.00497 m³.
1. Work done by the gas during adiabatic expansion:Adiabatic process means that there is no heat transfer (Q = 0) between the gas and the surrounding. Adiabatic process can be defined using the following equation:
PVγ = constantwhere γ = Cₚ / Cᵥ is the heat capacity ratio. During the adiabatic expansion, the volume of the gas increases, so pressure and temperature decrease. The initial pressure of the gas is P1 = 2.00 × 10^5 Pa. Since the process is adiabatic, the final pressure can be calculated as:
P1V1γ = P2V2γ⇒ P2 = P1 (V1 / V2)γ = (2.00 × 10^5 Pa) (0.00994 m³ / 0.00497 m³)(7 / 5)P2 = 8.02 × 10⁴ Pa
W = (P2V2 - P1V1) / (γ - 1)⇒ W = [(8.02 × 10⁴ Pa)(0.00994 m³) - (2.00 × 10^5 Pa)(0.00994 m³)] / (7 / 5 - 1)W = -4.77 × 10³ J (The negative sign indicates that work is done by the gas during adiabatic expansion)
Hence, the work done by the gas during the adiabatic expansion is -4.77 × 10³ J.
2. Heat added to the gas during adiabatic expansion:Heat added to the gas during adiabatic expansion is given by the first law of thermodynamics as:
ΔU = Q - W
where ΔU is the change in internal energy of the gas. Since the process is adiabatic, Q = 0.
Thus,ΔU = - W⇒ Q = W = 4.77 × 10³ J.
Hence, the heat added to the gas during the adiabatic expansion is 4.77 × 10³ J.
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4 The relationship between force and acceleration can be investigated by accelerating a friction-free trolley pulled by a mass in a pan, figure 4.1. thread trolley pulley pan table h Fig. 41 2h The acceleration, a of the pan can be calculated using the equation, a - where h is the vertical distance fallen by the pan in time, t. (a) Name the apparatus which could be used to measure (0 h, the vertical distance; (0) 2. time. 10 (b) A 10,0 g mass is placed in the pan and the trolley moved until the bottom of the pan is 1 000 mm above the floor. (1) Describe what must be done to obtain a value fort, using the apparatus named in (a)(ii) [ 21 (ii) State ONE way of increasing the accuracy of measuring t time [1]
The apparatus which could be used is a ruler or a measuring tape. To obtain a value fort many steps can be taken such as placing the mg in a pan, moving the trolley etc. To increase the accuracy of measuring time we can Use a digital stopwatch or timer
(a) (i) The apparatus that could be used to measure the vertical distance, h, is a ruler or a measuring tape.
(ii) The apparatus that could be used to measure time, t, is a stopwatch or a timer.
(b) To obtain a value for t using the named apparatus:
(i) Place the 10.0 g mass in the pan.
(ii) Move the trolley until the bottom of the pan is 1,000 mm above the floor.
(iii) Release the trolley and start the stopwatch simultaneously.
(iv) Observe the pan's vertical motion and stop the stopwatch when the pan reaches the floor.
Increasing the accuracy of measuring time:
To increase the accuracy of measuring time, you can:
(i) Use a digital stopwatch or timer with a higher precision (e.g., to the nearest hundredth of a second) rather than an analog stopwatch.
(ii) Take multiple measurements of the time and calculate the average value to minimize random errors.
(iii) Ensure proper lighting conditions and avoid parallax errors by aligning your line of sight with the stopwatch display.
(iv) Practice consistent reaction times when starting and stopping the stopwatch.
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Three retistors are connected to an EMF as shown with \( R_{1}=4 \Omega_{1} R_{3}=16 \sigma_{\text {and }} R_{2}=12 \mathrm{G} \). What is the equivalentresistance of this combination? \( 8 \Omega . \
The equivalent resistance of the combination is 32 Ω.
Supporting Answer:
When resistors are connected in series, the equivalent resistance is the sum of the individual resistances. In this case, the resistors are in series.
The equivalent resistance can be calculated by adding the individual resistances:
Equivalent Resistance = R1 + R2 + R3
Equivalent Resistance = 4 Ω + 12 Ω + 16 Ω
Equivalent Resistance = 32 Ω
Therefore, the equivalent resistance of the combination is 32 Ω.
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Two waves are given by the equations y1 = 3 sinπ(x + 4t) and y2 = 3 sinπ(x - 4t)
(a) Determine the equation of the standing wave formed by the superposition of these two waves.
(b) Determine the amplitude of the standing wave at t = 0
(c) Determine the wave number and the angular frequency of the standing wave
When two waves with the equations y1 = 3 sinπ(x + 4t) and y2 = 3 sinπ(x - 4t) superpose, a standing wave is formed. The wave number is π, and the angular frequency is 8π.
The equation, amplitude at t = 0, wave number, and angular frequency of the standing wave can be determined. The explanation of the answers will be provided in the second paragraph.
(a) To find the equation of the standing wave formed by the superposition of the two waves, we add the equations y1 and y2:
y = y1 + y2 = 3 sinπ(x + 4t) + 3 sinπ(x - 4t)
(b) To determine the amplitude of the standing wave at t = 0, we substitute t = 0 into the equation and evaluate:
y(t=0) = 3 sinπx + 3 sinπx = 6 sinπx
(c) The wave number (k) and angular frequency (ω) of the standing wave can be obtained by comparing the equation y = A sin(kx - ωt) with the equation of the standing wave obtained in part (a):
k = π, ω = 8π
In summary, the equation of the standing wave is y = 3 sinπx + 3 sinπx = 6 sinπx. The amplitude of the standing wave at t = 0 is 6. The wave number is π, and the angular frequency is 8π.
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A motor is designed to operate on 117 V and draws a current of 16.4 A when it first starts up. At its normaloperating speed, the motor draws a current of 3.26 A. Obtain (a) the resistance of the armature coil, (b) the backemf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed. (a) Number Units (b) Number Units (c) Number Units
When a motor first starts up, it uses 16.4 A of current and is intended to run on 117 V. The motor uses 3.26 A of current when working at standard speed. Therefore,
(a) The resistance of the armature coil is approximately 7.1341 ohms.
(b) The back EMF developed at normal speed is approximately 93.724 V.
(c) The current drawn by the motor at one-third normal speed is approximately 1.086 A.
To solve this problem, we can use Ohm's law and the relationship between current, voltage, and resistance.
(a) To find the resistance of the armature coil, we can use the formula:
Resistance (R) = Voltage (V) / Current (I)
Given that the voltage is 117 V and the current is 16.4 A during startup, we can calculate the resistance as follows:
R = 117 V / 16.4 A
Calculating this division gives us:
R ≈ 7.1341 ohms
Therefore, the resistance of the armature coil is approximately 7.1341 ohms.
(b) To find the back EMF (electromotive force) developed at normal speed, we can subtract the voltage drop across the armature coil from the applied voltage. The voltage drop across the armature coil can be calculated using Ohm's law:
Voltage drop ([tex]V_`d[/tex]) = Current (I) * Resistance (R)
Given that the current at normal operating speed is 3.26 A and the resistance is the same as before, we can calculate the voltage drop:
[tex]V_d[/tex] = 3.26 A * 7.1341 ohms
Calculating this multiplication gives us:
[tex]V_d[/tex] ≈ 23.276 V
Now, to find the back EMF, we subtract the voltage drop from the applied voltage:
Back EMF = Applied voltage (V) - Voltage drop ([tex]V_d[/tex])
Back EMF = 117 V - 23.276 V
Calculating this subtraction gives us:
Back EMF ≈ 93.724 V
Therefore, the back EMF developed at normal speed is approximately 93.724 V.
(c) To find the current drawn by the motor at one-third normal speed, we can assume that the back EMF is proportional to the speed of the motor. Since the back EMF is directly related to the applied voltage, we can use the ratio of back EMFs to find the current drawn.
Given that the back EMF at normal speed is 93.724 V, and we want to find the current at one-third normal speed, we can use the equation:
Current = Back EMF (at one-third normal speed) * Current (at normal speed) / Back EMF (at normal speed)
Assuming the back EMF is one-third of the normal speed back EMF, we have:
Current = (1/3) * 3.26 A / 93.724 V * 93.724 V
Calculating this division gives us:
Current ≈ 1.086 A
Therefore, the current drawn by the motor at one-third normal speed is approximately 1.086 A.
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15 16 22 QUESTION 8 decibel-part During takeoff, the sound intensity level of a jet engine is 110 dB at a distance of 40 m What is the Intensity of sound in units of Wim 27 QUESTION 9 decibel-part what is the power of the ſet entine mentioned in part A in units of Watts? QUESTION 10 decibel-part For the ſet mention in part A what is the sound intensity at a distance of 500 m from tho jet? Enter your answer in scientific notation with 2 decimals Scientific notation supports the following forms 45.60-6 or 456E-6 (using capital or lowercase E) The field doesn't support units (eg 25 cm 4504 KHZ aren't supported). • Constants such as "pl" and "e" (Euler constant) aren't supported, thus, numbers such as 67e or pl will be invalid QUESTION 11 Decibel Part What is the sound intensity level (in units of dB) of the jet engine mentioned in part A at this new distance of 500m? Enter your answer in scientific notati with 4 significant figures (3 decimals) do not use any intermediate rounded values in your calculation)
To solve these questions, we need to use the formulas and relationships related to sound intensity and sound level.
Question 8: The intensity of sound is 0.1 W/m².
Question 9: The power of the jet engine is approximately 201.06 W.
Question 10: The sound intensity at a distance of 500 m from the jet is approximately 0.0016 W/m².
Question 11: The sound intensity level of the jet engine at the new distance of 500 m is approximately 86.02 dB.
Question 8:
To find the intensity of sound in units of W/m², we need to convert the sound intensity level (given in dB) to intensity using the formula:
Intensity (W/m²) = 10^((dB - 120) / 10)
Substituting the given values, we get:
Intensity = 10^((110 - 120) / 10) = 10^(-1) = 0.1 W/m²
Question 9:
To find the power of the jet engine in units of Watts, we need to use the formula:
Power (W) = 4πr²I
Where r is the distance from the source and I is the sound intensity. In this case, r = 40 m and I = 0.1 W/m².
Substituting the values, we get:
Power = 4π(40²)(0.1) = 64π W ≈ 201.06 W
Question 10:
To find the sound intensity at a distance of 500 m from the jet, we can use the inverse square law for sound propagation:
I2 = I1 * (r1 / r2)²
Where I1 is the initial sound intensity at a given distance r1, and I2 is the sound intensity at the new distance r2.
In this case, I1 = 0.1 W/m², r1 = 40 m, and r2 = 500 m.
Substituting the values, we get:
I2 = 0.1 * (40 / 500)² ≈ 0.0016 W/m²
Question 11:
To find the sound intensity level at the new distance of 500 m, we can use the formula:
dB2 = dB1 + 10 log10(I2 / I1)
Where dB1 is the initial sound intensity level and I1 is the initial sound intensity, and dB2 is the sound intensity level at the new distance and I2 is the sound intensity at the new distance.
In this case, dB1 = 110 dB and I2 = 0.0016 W/m² (from the previous question).
Substituting the values, we get:
dB2 = 110 + 10 log10(0.0016 / 0.1) ≈ 86.02 dB
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A woman sits in a wheelchair and tried to roll over a curb that is 6 cm high. What force does she need to push at the top of the wheel to lift her and her chair? The woman in the chair has a mass of 80 kg, and the wheel has a radius of 27
cm.
The force is required to lift the woman and the chair over the curb when she pushes at the top of the wheel is 784.8 N
To find the force the woman needs to push at the top of the wheel to lift herself and her chair, the following formula can be used: force = mass x accelerationWhere acceleration is given by: acceleration = (change in velocity) / (time taken)Here, the woman is initially at rest. The velocity of the woman and the chair needs to be increased to go over the curb. Therefore, the acceleration required will be the acceleration due to gravity, which is 9.81 m/s² at the surface of the earth.The woman's mass is given as 80 kg.The radius of the wheel is given as 27 cm, which is equal to 0.27 m.To lift the woman and her chair, the wheel will have to move through a vertical distance equal to the height of the curb, which is 6 cm. This vertical distance is equal to the displacement of the woman and the chair.Force required = mass x accelerationForce required = 80 x 9.81 = 784.8 NThis force is required to lift the woman and the chair over the curb when she pushes at the top of the wheel.
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