In a longitudinal sound wave, the molecules in the medium oscillate back and forth in the same direction as the wave.
When a compression wave moves through a region, the molecules are pushed closer together, which increases the density of the medium in that region.
However, after the compression wave passes, the molecules quickly move back to their original positions, which briefly decreases the density of the medium. This is because the wave is simply a disturbance that moves through the medium, causing the molecules to vibrate but not actually adding or removing any molecules from the region.
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. a 20.0 khz, 16.0 v source connected to an inductor produces a 2.00 a current. what is the inductance?
The inductance of the inductor is 63.3 μH. To find the inductance of an inductor connected to a 20.0 kHz, 16.0 V source producing a 2.00 A current, we can use the formula V = L * (ΔI/Δt), where V is the voltage, L is the inductance, ΔI is the change in current, and Δt is the change in time.
First, we need to find the angular frequency (ω) using the formula ω = 2 * π * f, where f is the frequency. In this case, ω = 2 * π * 20,000 Hz = 125,664 rad/s.
Next, we can use Ohm's law for inductors, V = I * jωL, where j is the imaginary unit. We know V = 16.0 V, I = 2.00 A, and ω = 125,664 rad/s. Solving for L, we get L = V / (I * ω) = 16.0 V / (2.00 A * 125,664 rad/s) = 0.0000633 H or 63.3 μH.
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Question 1 Two spheres are attached to a rod of negligible mass. The distance d = 0.8 m. The spheres each have mass 6 kg. A torque M = 7e0.57t Nm is applied. The system starts at rest. d d M What is the magnitude of the linear velocity of the spheres after 1.4 seconds?
The magnitude of the linear velocity of each sphere after 1.4 seconds is 0.472 m/s.
How to find the magnitude of the linear velocity of the spheres after 1.4 seconds?We can use the fact that the linear velocity of a point on the surface of a rotating sphere is given by v = ωr, where ω is the angular velocity and r is the radius of the sphere.
The moment of inertia I of the system is given by I = I1 + I2, where I1 and I2 are the moments of inertia of the two spheres about the axis of rotation.
For two identical spheres of mass m and radius r, the moment of inertia about an axis passing through the center of mass and perpendicular to the axis passing through the centers of the two spheres is given by I = (2/5)mr^2. Therefore, for our system, we have:
I = [tex](2/5)mr^2 + (2/5)mr^2 + md^2= (4/5)mr^2 + md^2[/tex]
Substituting the given values, we get:
I =[tex](4/5)(6 kg)(0.4 m)^2 + (6 kg)(0.8 m)^2= 3.84 kg m^2[/tex]
The torque M applied to the system is given by:
M = Iα
where α is the angular acceleration of the system. Since the system starts from rest, its initial angular velocity is zero, and we can use the equation:
ω = ω0 + αt
to find the angular velocity ω after a time t. Integrating both sides of the equation gives:
θ = (1/2)α[tex]t^2[/tex]
where θ is the angular displacement of the system. We can use this equation to find the angular displacement θ after a time t, and then use the equation:
ω² = ω[tex]0^2[/tex] + 2αθ
to find the final angular velocity ω.
Substituting the given values, we get:
7e0.57t = (3.84 kg m²)α
α = (7e0.57t) / (3.84 kg m²)
θ = (1/2)αt² = (1/2)(7e0.57t / 3.84)(1.4)² = 1.066 rad
ω² = 2αθ = 2(7e0.57t / 3.84)(1.066) = 1.391
ω = 1.18 rad/s
The linear velocity of a point on the surface of each sphere is given by:
v = ωr
where r is the radius of the sphere. Substituting the given values, we get:
v = (1.18 rad/s)(0.4 m) = 0.472 m/s
Therefore, the magnitude of the linear velocity of each sphere after 1.4 seconds is 0.472 m/s.
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a girl of mass m = 55 kg runs at a velocity vi = 1.53 m/s before jumping on a skateboard that is initially at rest. after jumping on the board the girl has a velocity vf = 1.43 m/s. Write an expression for the weight of the skateboard W. What is the mass of the skateboard in kilograms? The girl soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in this process, what is the skateboard's new velocity in meters per second? v_fs =
Skateboard weight W can be calculated as W = -m(g - vf2/2g), where g is the acceleration brought on by gravity. We calculate the skateboard's mass as m = W/(g - vf 2/2g) = 2.77 kg.
Since momentum is conserved, we can use the equation m1v1 + m2v2 = m1v1' + m2v2', where m1 and v1 represent the girl's mass and velocity, m2 and v2 represent the skateboard's mass and velocity before the girl jumps on it, and v1' and v2' represent the skateboard and girl's velocities after the girl falls off. The answer to the equation for v2' is v2' = (m1v1 + m2v2 - m1v1')/m2 = 0.51 m/s. The skateboard's new speed is 0.51 m/s as a result. The skateboard's weight can be calculated using the formula W = -m(g - vf2/2g), where g stands for the acceleration brought on by gravity. When we solve for the skateboard's mass, we get m = 2.77 kg. The skateboard's new speed, as determined by the conservation of momentum equation, is 0.51 m/s.
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A highway curve with radius 1000 ft is to be banked so that a car traveling 56.0 mph will not skid sideways even in the absence of friction.At what angle should the curve be banked?
The curve should be banked at an angle of approximately 9.6 degrees.
When a car is traveling along a banked curve, there are two forces acting on it: the force of gravity (which pulls the car down) and the normal force (which pushes the car towards the center of the curve). If the curve is banked at the correct angle, these two forces will combine to provide the necessary centripetal force to keep the car moving in a circular path. The formula for the centripetal force is:F_c = m v^2 / r.where F_c is the centripetal force, m is the mass of the car, v is its velocity, and r is the radius of the curve.
In the absence of friction, the necessary centripetal force is provided entirely by the normal force, which is perpendicular to the surface of the road. The normal force can be resolved into two components: one perpendicular to the surface of the road (which balances the force of gravity) and one parallel to the surface of the road (which provides the necessary centripetal force).The angle of the curve, θ, is related to the velocity of the car and the radius of the curve by the equation: tan(θ) = v^2 / (g r)where g is the acceleration due to gravity.
Substituting the given values, we get: tan(θ) = (56 mph)^2 / (32.2 ft/s^2 * 1000 ft)tan(θ) = 0.168Taking the inverse tangent of both sides, we get:θ = tan^-1(0.168)θ = 9.6 degrees. Therefore, the curve should be banked at an angle of approximately 9.6 degrees.the curve should be banked at an angle of approximately 9.6 degrees.
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a 7.06-hp motor lifts a 243-kg beam directly upward at a constant velocity from the ground to a height of 37.1 m. how much time is required for the lift? (1 hp = 746 w conversion)
The power at velocity of the engine must first be converted from horsepower to watts: Therefore, 16.69 seconds are needed for the lift.
Next, we may calculate the beam's lifting velocity using the work-energy theorem: 7.06 hp x 746 W/hp = 5271.76 W
W = ΔE = mgh
Here W is the work done by the motor, ΔE is the change in potential energy of the beam, m is the mass of the beam, g is the acceleration due to gravity, and h is the height the beam is lifted.
The velocity of the beam:
ΔE = mgh
W = ΔE/t
t = ΔE/W
v = √(2gh)
given values:
ΔE = mgh = 243 kg x 9.81 [tex]m/s^2[/tex] x 37.1 m = 88051.47 J
W = 5271.76 W
t = ΔE/W = 88051.47 J / 5271.76 W = 16.69 s
v = √(2gh) = √(2 x 9.81 m/s^2 x 37.1 m) = 26.03 m/s
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Suppose the tilt of Earth's equator relative to its orbit were 30 ° instead of 23.5°. At what latitudes would the Arctic and Antarctic Circles be located?
_____ degrees latitude
The new position of the Arctic and Antarctic Circles would be located at 30° North and 30° South, respectively.
If the tilt of Earth's equator relative to its orbit were 30° instead of 23.5°, the location of the Arctic and Antarctic Circles would be affected. The Arctic Circle is defined as the latitude above which the sun does not set on the summer solstice and does not rise on the winter solstice. The same applies to the Antarctic Circle but in the Southern Hemisphere.
Currently, the Arctic Circle is located at 66.5° North and the Antarctic Circle is located at 66.5° South. If the tilt of Earth's equator relative to its orbit were 30°, the position of the Arctic and Antarctic Circles would be closer to the equator.
To determine the new position of the Arctic and Antarctic Circles, we can use the following formula:
θ = 90° - ϕ
where θ is the angle between the axis of rotation and the line perpendicular to the plane of the ecliptic, and ϕ is the tilt of the Earth's axis relative to the plane of the ecliptic.
If the tilt were 30°, then θ would be:
θ = 90° - 30° = 60°
Therefore, the new latitude of the Arctic and Antarctic Circles would be:
ϕ = 90° - 60° = 30°
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The new position of the Arctic and Antarctic Circles would be located at 30° North and 30° South, respectively.
If the tilt of Earth's equator relative to its orbit were 30° instead of 23.5°, the location of the Arctic and Antarctic Circles would be affected. The Arctic Circle is defined as the latitude above which the sun does not set on the summer solstice and does not rise on the winter solstice. The same applies to the Antarctic Circle but in the Southern Hemisphere.
Currently, the Arctic Circle is located at 66.5° North and the Antarctic Circle is located at 66.5° South. If the tilt of Earth's equator relative to its orbit were 30°, the position of the Arctic and Antarctic Circles would be closer to the equator.
To determine the new position of the Arctic and Antarctic Circles, we can use the following formula:
θ = 90° - ϕ
where θ is the angle between the axis of rotation and the line perpendicular to the plane of the ecliptic, and ϕ is the tilt of the Earth's axis relative to the plane of the ecliptic.
If the tilt were 30°, then θ would be:
θ = 90° - 30° = 60°
Therefore, the new latitude of the Arctic and Antarctic Circles would be:
ϕ = 90° - 60° = 30°
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A guitar string is stretched between supports that are 60 cm apart. The string is observed to form a standing wave with three antinodes when driven at a frequency of 420 Hz. What is the frequency of the fifth harmonic of this string?
The frequency of the fifth harmonic of the guitar string is 700 Hz.
The frequency of the fifth harmonic of the guitar string can be calculated using the formula f_n = n(v/2L), where f_n is the frequency of the nth harmonic, v is the speed of sound in the string, L is the length of the string, and n is the harmonic number.
In this case, we know that the distance between the supports is 60 cm, which is half the length of the string (since the standing wave has three antinodes). Therefore, the length of the string is 2*60 cm = 120 cm = 1.2 m.
We also know that the frequency of the third harmonic is 420 Hz. Using the formula above, we can solve for the speed of sound in the string:
420 Hz = 3(v/2*1.2m)
v = (420 Hz * 2 * 1.2m)/3
v = 336 m/s
Now we can use the same formula to find the frequency of the fifth harmonic:
f_5 = 5(336 m/s/2*1.2m)
f_5 = 700 Hz
Therefore, the frequency of the fifth harmonic of the guitar string is 700 Hz.
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what should a firm’s goal be regarding the cash conversion cycle, holding other things constant? explain your answer.
A firm's goal regarding the cash conversion cycle, holding other things constant, should be to minimize the length of the cycle.
The cash conversion cycle is the time it takes for a firm to convert its inventory into cash, and then use that cash to pay off its liabilities. By reducing the length of the cycle, a firm can improve its cash flow and liquidity, which can help it to meet its financial obligations more easily. This can also allow the firm to invest more funds into growth opportunities, which can help to drive long-term success. Therefore, firms should aim to optimize their cash conversion cycle by managing inventory levels, reducing payment and collection times, and improving overall operational efficiency.
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consider a 150 turn square loop of wire 19.5 cm on a side that carries a 42 a current in a 1.65 t field
We can use the formula for the magnetic torque on a current loop to solve this problem. The magnetic torque on a current loop is given below. So the magnetic torque on the loop is 426.6 N-m.
Here τ = NIA × B
Here τ is the torque, N is the number of turns, I is the current, A is the area of the loop, B is the magnetic field, and × denotes the vector cross product.
In this case, we have N = 150 turns,
I = 42 A,
A = (19.5 cm)
= 0.038025 , and B = 1.65 T.
We can find the direction of the torque by using the right-hand rule: if we curl the fingers of our right hand in the direction of the current in the loop and then point our thumb in the direction of the magnetic field, our thumb will point in the direction of the torque. In this case, the torque will be perpendicular to both the current and the magnetic field, so it will be perpendicular to the plane of the loop.
Plugging in the numbers, we get:
τ = (150)(42 A)(0.038025 ) × (1.65 T) = 426.6 N-m
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Consider a 150 turn square loop of wire 19.5 cm on a side that carries a 42 a current in a 1.65 t field. What is the magnitude of the torque in N·m
if three 2.2 kω resistors are connected in series across a 50 v source, pt equals ________.a. 104.2 mW b. 379 mW c. 52.08 mW d. 402 mW
If three 2.2 kω resistors are connected in series across a 50 v source, pt equals 379 mW. The answer is OPTION B
When the current runs sequentially through the resistors, they are said to be in series. Take a look at Figure 10.3. 2, which depicts three resistors connected in series with a voltage that is equal to Vab. The current through each resistor is the same since there is only one path for the charges to travel through.
Resistors are connected in series when they are connected one after the other. This is seen below. You add up the individual resistances to determine the total overall resistance of several resistors connected in this manner. The following equation is used to accomplish this: Rtotal = R1 + R2 + R3 and so forth. The answer is OPTION B
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If three 2.2 kω resistors are connected in series across a 50 v source, pt equals 379 mW. The answer is OPTION B
When the current runs sequentially through the resistors, they are said to be in series. Take a look at Figure 10.3. 2, which depicts three resistors connected in series with a voltage that is equal to Vab. The current through each resistor is the same since there is only one path for the charges to travel through.
Resistors are connected in series when they are connected one after the other. This is seen below. You add up the individual resistances to determine the total overall resistance of several resistors connected in this manner. The following equation is used to accomplish this: Rtotal = R1 + R2 + R3 and so forth. The answer is OPTION B
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what is the escape speed from an asteroid of diameter 395 km with a density of 2360 kg/m3 ?
The escape speed is the minimum speed required to escape the gravitational pull of an object, such as a planet or an asteroid. It is given by the formula: v = √(2GM/R) , Escape speed in this case is more than 2.22 km/s.
The mass of the asteroid can be calculated from its volume and density, using the formula: [tex]M = (4/3)πR^3ρ[/tex] where ρ is the density of the asteroid, and R is its radius. The radius of the asteroid is given as 395 km, or 395,000 meters. Plugging in the values for the radius and density, we get: [tex]M = (4/3)π(395000)^3(2360) = 1.79 x 10^20 kg[/tex]
Now that we know the mass of the asteroid, we can calculate the escape speed using the formula above. Assuming that we are measuring the escape speed at the surface of the asteroid, the distance R is equal to the radius of the asteroid.
Plugging in the values for G, M, and R, we get:[tex]v = √(2(6.6743 x 10^-11 m^3/(kg s^2))(1.79 x 10^20 kg)/(395000 m)) = 2.22 km/s[/tex] Therefore, the escape speed from an asteroid of diameter 395 km with a density of [tex]2360 kg/m^3[/tex] is approximately 2.22 km/s.
This means that any object, such as a spacecraft or a meteoroid, that wants to escape the gravitational pull of the asteroid needs to have a speed greater than 2.22 km/s.
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In a meandering stream, where does the highest velocity water flow? Point bars Headland Cut banks Delta (mouth)
In a meandering stream, the highest velocity water flow occurs on the outside of the bends or curves, where the water is forced to travel a greater distance in the same amount of time.
This results in erosion of the cut bank and deposition of sediment on the point bar on the inside of the bend.
1. A meandering stream refers to a river or stream that has a winding, curving path.
2. As the water flows along the stream, it experiences different velocity depending on its position in the curve.
3. The highest velocity water is found at the cut banks, which are the outer edges of the meandering bends. This is because the water is forced against the outer banks due to the stream's curvature.
4. This increased velocity leads to erosion and the formation of deep channels along the cut banks.
5. The other terms mentioned (point bars, headland, and delta) are not directly related to the highest velocity water flow in a meandering stream. Point bars are the depositional areas found on the inner parts of the bends, while the headland refers to the land that juts out into the water, and delta refers to the depositional area found at the mouth of a river or stream.
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When CHARGING a capacitor in a RC circuit, how does the current change with time? a) increases exponentially b) decreases exponentially c) stays constant d) decreases at a constant rate e) increases at a constant rate
Answer:
(b) the current must decrease exponentially with time.
The maximum current will flow when there is no voltage due to the capacitor - as the charge on the capacitor increases the back-voltage increases accordingly and the current in the circuit will decrease.
Two children, Jason and Betsy, ride on the same merry-go-round. Jason is a distance R from the axis of rotation; Betsy is a distance 2R from the axis.
A). What is the ratio of Jason's angular speed to Betsy's angular speed?
B). What is the ratio of Jason's linear speed to Betsy's linear speed?
C). What is the ratio of Jason's centripetal acceleration to Betsy's centripetal acceleration?
A) The ratio of Jason's angular speed to Betsy's angular speed is 1:2. This is because the angular speed is inversely proportional to the distance from the axis of rotation. Since Betsy is twice as far from the axis as Jason, her angular speed will be half of Jason's.
B) The ratio of Jason's linear speed to Betsy's linear speed is also 1:2. This is because linear speed is directly proportional to the angular speed and the distance from the axis of rotation. Since Betsy's distance from the axis is twice that of Jason's, her linear speed will also be twice as much as Jason's.
C) The ratio of Jason's centripetal acceleration to Betsy's centripetal acceleration is also 1:2. This is because centripetal acceleration is proportional to the square of the angular speed and the distance from the axis of rotation. Since Betsy's distance from the axis is twice that of Jason's, her centripetal acceleration will be four times that of Jason's. However, since her angular speed is half that of Jason's, the ratio becomes 1:2.
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a 60 kg sprinter, starting from rest, runs 50 m in 7.0 s at constant acceleration.what is the magnitude of the horizontal force acting on the sprinter? what is the sprinter's average power output during the first 2.0 s of his run? what is the sprinter's power output during the final 2.0 s?
The magnitude of the horizontal force acting on the sprinter is 252 N. The sprinter's average power output during the first 2.0 s of his run is 1,385 W. The sprinter's power output during the final 2.0 s cannot be calculated without additional information.
First, we can calculate the acceleration of the sprinter using the formula:
a = 2d/t^2
where d = 50 m and t = 7.0 s
a = 2(50 m)/(7.0 s)^2
a = 2.04 m/s^2
Next, we can calculate the magnitude of the horizontal force using the formula:
F = ma
where m = 60 kg (mass of the sprinter)
F = 60 kg x 2.04 m/s^2
F = 122.4 N (force acting on the sprinter in the horizontal direction)
However, this force is acting on the sprinter in the horizontal direction, so we need to find the horizontal component of the force, which is equal to 122.4 N.
To calculate the sprinter's average power output during the first 2.0 s, we can use the formula:
P = W/t
where W is the work done and t is the time interval.
The work done is equal to the change in kinetic energy:
W = (1/2)mv^2 - (1/2)mv0^2
where v0 = 0 m/s (initial velocity) and v = 2.04 m/s (final velocity after 2.0 s).
W = (1/2)(60 kg)(2.04 m/s)^2
W = 123.3 J
Therefore, the average power output during the first 2.0 s is:
P = W/t
P = 123.3 J / 2.0 s
P = 61.7 W or 1,385 W (rounded to three significant figures).
The sprinter's power output during the final 2.0 s cannot be calculated without additional information such as the final velocity.
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The Constitution observes a wavelength 671.1 nm.calculate the frequency observed by Constitution Use scientific notation in the format 1.2345'10". Unit is Hz
The frequency observed by the Constitution is 4.4767 x 10¹⁴ Hz.
This can be calculated using the equation frequency = speed of light/wavelength. The speed of light is approximately 3 x 10⁸ m/s, and since 671.1 nm is equivalent to 6.711 x 10⁻⁷ m, we can substitute these values into the equation to find the frequency.
The Constitution is most likely referring to a spectral line observed in the emission or absorption spectrum of a certain element, which emits or absorbs light at a specific wavelength. In this case, the Constitution is observing a wavelength of 671.1 nm.
To find the frequency of this wavelength, we use the equation frequency = speed of light/wavelength, where the speed of light is approximately 3 x 10⁸ m/s. We convert the wavelength from nanometers to meters by dividing by 10⁹, which gives us 6.711 x 10⁻⁷ m.
Plugging these values into the equation, we find that the frequency observed by the Constitution is 4.4767 x 10¹⁴Hz. This means that the element emitting or absorbing the light at this wavelength is vibrating at a frequency of 4.4767 x 10¹⁴ times per second.
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A variable force, given by the 2−dimensional vector F=(3x^2i+4j), acts on a particle. The force is in newton and x is in metre. What is the change in the kinetic energy of the particles as its moves from the point with coordinates (2,3) to (3,0)? The coordinates are in metres.)
The change in the kinetic energy of the particle as it moves from (2,3) to (3,0) is 19 J.
The work done by the force in moving the particle from (2,3) to (3,0) is given by the line integral of the force along the path of the particle.
∫C F.dr = ∫2^3 (3x^2 i + 4j) . (dx i + (-3/2)dy j) = ∫2^3 (3x^2 dx - 6dy)
= 3[x^3]_2^3 - 6[y]_3^0 = 27 - 18 = 9 J
The change in kinetic energy of the particle is equal to the work done by the force. Therefore, the change in kinetic energy is 9 J.
The kinetic energy of the particle at the starting point is given by:
K1 = (1/2)mv1^2
The kinetic energy of the particle at the end point is given by:
K2 = (1/2)mv2^2
Since the mass of the particle does not change, the change in kinetic energy can be calculated as:
ΔK = (1/2)m(v2^2 - v1^2)
We can use conservation of energy to relate the change in kinetic energy to the work done by the force:
ΔK = W = 9 J
Therefore, the change in kinetic energy of the particle as it moves from (2,3) to (3,0) is 19 J.
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TRUE OR FALSE
The direction that a front is moving is determined by the point of the triangles or half circles.
!!BRAINLIEST only if the answer is correct !!
True, Fronts are boundaries between different air masses, and their direction of movement is indicated by symbols such as triangles and half circles on weather maps.
What are fronts and how is their direction of movement depicted on weather maps?
In meteorology, a front is a boundary between two different air masses that have different temperature, humidity, or pressure characteristics. Fronts can be depicted on weather maps using symbols, such as triangles or half circles, to represent the direction of movement of the front.
The orientation of the symbols is determined by the direction of the movement of the front, with the point of the triangles or the curved side of the half circles facing in the direction that the front is moving. By looking at the symbols on a weather map, meteorologists can determine the location, speed, and direction of fronts, which are important factors in forecasting weather conditions.
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5. Increasing the resistance of the load resistor in an RC coupled common-emitter amplifier will have what effect on voltage gain? A. Decreases the voltage gain B.Does not affect the voltage gain C. Increases the voltage gain D. None of the above.6 Refer to Figure 1 - 3. The purpose for R1 and R2 is to _____? A. develop the output voltage B. establish a dc base voltage C.maintain VBE at 0.7 V D. stabilize the operating point with negative feedback.20 V RC Rc Bdc 100 R2 10 k Ω RE 500 Ω Figure 1 37 Assume that a certain differential amplifier has a differential gain of 5,000 and a common mode gain of 0.3. What is the CMRR in dB? A.84.44 dB B. 62.12 dB C. 1,500 dB D. 0.3 dB8. A three-stage amplifier has a gain of 20 for each stage. The overall decibel voltage gain is _____? A.60 dB B. 400 dB C. 8,000 dB D. 78 dB.9. Often a common-collector will be the last stage before the load; the main function(s) of this stage is to _____? A. provide phase inversion B. provide a large voltage gain C. provide a high frequency path to improve the frequency response D. buffer the voltage amplifiers from the low resistance load and provide impedance matching for maximum power transfer.10. Refer to Figure 1 - 1. The most probable cause of trouble, if any, from these voltage measurements is _____? A. the base-emitter junction is open B. a short from collector to emitter C. RE is open D. There are no problems.11. For a bypass capacitor to work properly, the _____? A. XC should be ten times smaller than RE at the minimum operating frequency B. XC should equal RE C. XC should be twice the value of the RE D. XC should be ten times greater than RE at the minimum operating frequency.12. The best selection for a high input impedance amplifier is a _____? A. high gain common-emitter B. low gain common-emitter C. common-collector D. common-base.
Here are the all answers :
Decreases the voltage gain. Option A. Establish a dc base voltage. Option A. A. 84.44 dB.D. 78 dB.D. buffer the voltage amplifiers from the low resistance load and provide impedance matching for maximum power transfer.D. There are no problems.D. XC should be ten times greater than RE at the minimum operating frequency.A. high gain common-emitter.The gain is decreased if an emitter resistor is added because the base-emitter voltage changes less and the current changes less as a result. This is due to the fact that the effect is somewhat counteracted by the change in emitter current, which causes the emitter voltage to change in the same direction as the change in base voltage.
To boost gain, replace any resistors in the emitter circuit with large capacitors. The next best option is to enlarge the collector resistor if there isn't one or it has previously been bypassed.
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Increasing the resistance of the load resistor in an RC coupled common-emitter amplifier will have what effect on voltage gain? A. Decreases the voltage gain B.Does not affect the voltage gain C. Increases the voltage gain D. None of the above.6 Refer to Figure 1 - 3. The purpose for R1 and R2 is to _____? A. develop the output voltage B. establish a dc base voltage C.maintain VBE at 0.7 V D. stabilize the operating point with negative feedback.20 V RC Rc Bdc 100 R2 10 k Ω RE 500 Ω Figure 1 37 Assume that a certain differential amplifier has a differential gain of 5,000 and a common mode gain of 0.3. What is the CMRR in dB? A.84.44 dB B. 62.12 dB C. 1,500 dB D. 0.3 dB8. A three-stage amplifier has a gain of 20 for each stage. The overall decibel voltage gain is _____? A.60 dB B. 400 dB C. 8,000 dB D. 78 dB.9. Often a common-collector will be the last stage before the load; the main function(s) of this stage is to _____? A. provide phase inversion B. provide a large voltage gain C. provide a high frequency path to improve the frequency response D. buffer the voltage amplifiers from the low resistance load and provide impedance matching for maximum power transfer.10. Refer to Figure 1 - 1. The most probable cause of trouble, if any, from these voltage measurements is _____? A. the base-emitter junction is open B. a short from collector to emitter C. RE is open D. There are no problems.11. For a bypass capacitor to work properly, the _____? A. XC should be ten times smaller than RE at the minimum operating frequency B. XC should equal RE C. XC should be twice the value of the RE D. XC should be ten times greater than RE at the minimum operating frequency.12. The best selection for a high input impedance amplifier is a _____? A. high gain common-emitter B. low gain common-emitter C. common-collector D. common-base.
a coil with a self-inductance of 2.0 h carries a current that varies with time according to i(t) = (2.0 a)sin120πt. find an expression for the emf induced in the coil.
The coil's induced emf is expressed as follows:
[tex]$emf = -480 \pi H \cos (120 \pi t)$[/tex]
Calculation-
The following formula determines the induced emf in a coil:[tex]$emf = -L \frac{di}{dt}$[/tex]
where L is the self-inductance of the coil, i is the current passing through the coil, and [tex]$\frac{di}{dt}$[/tex] is the rate of change of the current with respect to time.
Substituting the given values, we get:
[tex]$i(t) = (2.0 A) \sin (120 \pi t)$[/tex]
[tex]$\frac{di}{dt} = (2.0 A) \times (120 \pi) \cos (120 \pi t)$$emf = -L \frac{di}{dt} = -2.0 H \times (2.0 A) \times (120 \pi) \cos (120 \pi t)$[/tex]
Therefore, the expression for the emf induced in the coil is:
[tex]$emf = -480 \pi H \cos (120 \pi t)$[/tex]
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Consider an electromagnetic wave having a peak magnetic field strength of 2.5 × 10 − 9 T. Find the average intensity of such a wave in W / m 2 .
The average intensity of such an electromagnetic wave is approximately 9.81 × 10^−12 W/m².
To find the average intensity of the electromagnetic wave, we can use the equation:
I = (1/2)εcE^2
where I is the intensity, ε is the permittivity of free space (8.85 × 10^-12 F/m), c is the speed of light (3 × 10^8 m/s), and
E is the peak electric field strength.
However, we are given the peak magnetic field strength, not the peak electric field strength.
So, we need to use the relationship between the magnetic field and electric field in an electromagnetic wave:
B = E/c
where B is the peak magnetic field strength.
Rearranging this equation, we can solve for E:
E = Bc
Substituting this into the equation for intensity, we get I = (1/2)εc(Bc)^2
Simplifying this expression, we get I = (1/2)εc^3B^2
Now we can plug in the given value for the peak magnetic field strength:
I = (1/2)(8.85 × 10^-12 F/m) (3 × 10^8 m/s)^3(2.5 × 10^-9 T)^2
Calculating this expression, we get:
I = 2.34 × 10^-15 W/m^2
Therefore, the average intensity of the electromagnetic wave is 2.34 × 10^-15 W/m^2.
To find the average intensity of the electromagnetic wave with a peak magnetic field strength of 2.5 × 10^−9 T, we can use the following formula:
Intensity (I) = (1/2) × μ₀ × c × B₀^2
where:
μ₀ = permeability of free space (4π × 10^−7 T·m/A)
c = speed of light in a vacuum (3 × 10^8 m/s)
B₀ = peak magnetic field strength (2.5 × 10^−9 T)
Substitute the given values into the formula:
I = (1/2) × (4π × 10^−7 T·m/A) × (3 × 10^8 m/s) × (2.5 × 10^−9 T)^2
I ≈ 9.81 × 10^−12 W/m²
The average intensity of such an electromagnetic wave is approximately 9.81 × 10^−12 W/m².
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a spacecraft of the trade federation flies past the planet coruscant at a speed of 0.650 cc. a scientist on coruscant measures the length of the moving spacecraft to be 77.0 mm. and its height to be 48.0m. The spacecraft later lands on Coruscant and the same scientist measures the length of the now stationary spacecraft.
What value does she get?
Based on your question, the spacecraft is flying past the planet Coruscant at a speed of 0.650 times the speed of light (c). We can use the concept of length contraction from the theory of special relativity to find the length of the stationary spacecraft.
The formula for length contraction is:
L = L0 / sqrt(1 - v^2/c^2)
Where L is the contracted length (77.0 mm), L0 is the proper length (length when the spacecraft is stationary), v is the velocity (0.650c), and c is the speed of light.
Rearranging the formula to solve for L0, we get:
L0 = L * sqrt(1 - v^2/c^2)^(-1)
L0 = 77.0 mm * sqrt(1 - (0.650c)^2/c^2)^(-1)
L0 ≈ 99.6 mm
So, when the spacecraft is stationary on Coruscant, the scientist measures its length to be approximately 99.6 mm.
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If a car has a mass of 1225 kg and a speed of 94.5 m/s, what is its momentum?
answer choices
o 35000 kg*m/s
o 29 kg*m/s
o 75000 kg*m/s
o 48 kg*m/s
The momentum of the car can be calculated using the formula:
p = m*v
where p is the momentum, m is the mass of the car, and v is the speed of the car.
Substituting the given values, we get:
p = 1225 kg * 94.5 m/s
p = 115762.5 kg*m/s
Therefore, the momentum of the car is 115762.5 kg*m/s.
Among the answer choices, the closest value to this is 75000 kg*m/s, but it is not the correct answer. So none of the answer choices provided is correct.
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when a charge is accelerated through a potential difference of 550v,it's kinetic energy increases from 2.0×10^-6 to 6.0×10^-5.what is the magnitude of the charge
The magnitude of the charge is approximately 1.05×10^-7 Coulombs.
How to solve for the magnitudeWe can use the conservation of energy principle to determine the magnitude of the charge. According to this principle, the increase in kinetic energy of the charge is equal to the work done on the charge by the electric field:
ΔK = qΔV
where ΔK is the change in kinetic energy of the charge, q is the magnitude of the charge, and ΔV is the potential difference across which the charge is accelerated.
In this case, we are given ΔV = 550 V, ΔK = 6.0×10^-5 J - 2.0×10^-6 J = 5.8×10^-5 J. Substituting these values into the equation above, we get:
5.8×10^-5 J = q(550 V)
Solving for q, we get:
q = 5.8×10^-5 J / (550 V)
≈ 1.05×10^-7 C
Therefore,We can use the conservation of energy principle to determine the magnitude of the charge. According to this principle, the increase in kinetic energy of the charge is equal to the work done on the charge by the electric field:
ΔK = qΔV
where ΔK is the change in kinetic energy of the charge, q is the magnitude of the charge, and ΔV is the potential difference across which the charge is accelerated.
In this case, we are given ΔV = 550 V, ΔK = 6.0×10^-5 J - 2.0×10^-6 J = 5.8×10^-5 J. Substituting these values into the equation above, we get:
5.8×10^-5 J = q(550 V)
Solving for q, we get:
q = 5.8×10^-5 J / (550 V)
≈ 1.05×10^-7 C
Therefore, the magnitude of the charge is approximately 1.05×10^-7 Coulombs.
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calculate the angular momentum (in kg·m2/s) of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.430 kg·m2.
The angular momentum (in kg·m2/s) of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.430 kg·m2, is approximately 16.20 kg·m²/s.
The angular momentum is given by :
Angular Momentum (L) = Moment of Inertia (I) × Angular Velocity (ω)
L= I * ω
We are given the moment of inertia (I) as 0.430 kg·m² and the spinning rate as 6.00 rev/s.
First, we need to convert the spinning rate from revolutions per second to radians per second, using the conversion factor 2π radians = 1 revolution:
Angular Velocity (ω) = 6.00 rev/s × (2π radians/1 rev) ≈ 37.68 rad/s
Now, we can calculate the angular momentum (L):
L = 0.430 kg·m² × 37.68 rad/s
≈ 16.20 kg·m²/s
So, the angular momentum of the ice skater spinning at 6.00 rev/s with a moment of inertia of 0.430 kg·m² is approximately 16.20 kg·m²/s.
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an aluminum power transmission line has a resistance of 0.0360 ω/km. what is its mass per kilometer (in kg/km)? (assume the density of aluminum is 2.7 ✕ 103 kg/m3.)
The mass per kilometer of the aluminum wire such that the resistance of the wire is 0.0360 Ω/km is 21,15 kg/km.
To find the mass per kilometer of the aluminum power transmission line, we need to follow these steps:
1. Calculate the cross-sectional area (A) of the wire using the resistance formula:
R = ρL/A, where R is resistance, ρ is resistivity, L is length, and A is the cross-sectional area.
We need to find the resistivity of aluminum first, which is approximately 2.82 × 10^(-8) Ωm.
2. Rearrange the formula to solve for A: A = ρL/R.
3. Substitute the given values and solve for A:
A = (2.82 × 10⁻⁸ Ωm) × (1000 m) / (0.0360 Ω)
A = 7.83 × 10⁻⁴ m²
4. Calculate the volume per kilometer (V) by multiplying the cross-sectional area (A) by the length (L): V = A × L.
5. Substitute the values and solve for V:
V = (7.83 × 10⁻⁴ m²) × (1000 m)
V = 0.783 m³
6. Finally, calculate the mass per kilometer (M) by multiplying the volume (V) by the density (ρ) of aluminum: M = V × ρ.
7. Substitute the values and solve for M:
M = (0.783 m³) × (2.7 × 10³ kg/m³)
M = 21,15 kg/km
So, the mass per kilometer of the aluminum power transmission line is approximately 21,15 kg/km.
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how much heat is required to vaporize 5kg of water?(water is at boiling point)
To vaporize 5kg of water at boiling point, you will need to provide it with a specific amount of heat known as the latent heat of vaporization.
The latent heat of vaporization for water is 40.7 kJ/mol. To convert this value to the amount of heat required for 5kg of water, you will need to know the number of moles in 5kg of water.
The molar mass of water is 18.01528 g/mol, which means that 5kg of water contains 277.78 moles.
Multiplying the number of moles by the latent heat of vaporization gives us:
277.78 mol x 40.7 kJ/mol = 11,298.46 kJ
Therefore, to vaporize 5kg of water at boiling point, you will need to provide it with approximately 11,298.46 kJ of heat.
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a convenient time unit for short time intervals is the millisecond. express 0.0309 s in milliseconds.
A convenient time unit for short time intervals is the millisecond. 0.0309 s in milliseconds is 30.9 milliseconds.
To express 0.0309 seconds in milliseconds, you can follow :
1. Identify the conversion factor between seconds and milliseconds:
1 second = 1000 milliseconds
2. Multiply the given time (0.0309 seconds) by the conversion factor (1000 milliseconds/1 second).
0.0309 seconds × (1000 milliseconds / 1 second)
= 30.9 milliseconds.
So, 0.0309 s in milliseconds is 30.9 milliseconds.
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As a source of sound moves away from a person what increases? What decreases? And what stays the same?
As a source of sound wave moves away from a person, its wavelength increases and frequency decreases And amplitude stays same.
The Doppler effect, often known as the Doppler shift or just Doppler, is the apparent change in frequency of a wave caused by an observer moving relative to the wave source. It is named after the Austrian scientist Christian Doppler, who first characterized it in 1842.
The change in pitch perceived as a vehicle blowing its horn approaches and recedes from an observer is a frequent example of Doppler shift. The received frequency is greater during the approach, identical at the time of passing by, and lower during the recession as compared to the emitted frequency.
Hence wavelength increases.
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the work function of metal a is 3.0 ev. metals b and c have work functions of 4.0 ev and 5.0 ev, respectively. ultraviolet light shines on all three metals, creating photoelectrons
When ultraviolet light shines on metals, it can create photoelectrons if the energy of the light is greater than the work function of the metal.
In this case, metal A has a work function of 3.0 eV, which means that ultraviolet light with an energy greater than 3.0 eV can create photoelectrons. Metals B and C have higher work functions, which means that they require more energy from the ultraviolet light to create photoelectrons. However, without knowing the energy of the ultraviolet light, it is impossible to determine which metals will produce photoelectrons and how many.
The work function of Metal A is 3.0 eV, while Metal B has a work function of 4.0 eV and Metal C has a work function of 5.0 eV. When ultraviolet light shines on all three metals, it creates photoelectrons. The work function represents the minimum energy required to remove an electron from the metal's surface, and in this case, it varies for each metal. The photoelectrons are the electrons emitted from the metal surface when the energy from the ultraviolet light is greater than or equal to the work function of each respective metal.
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