In a month it will be the next full moon
Explanation:
An electron that has a velocity of v = (2×10^6m/s)i + (3×10^6m/s)j. Find the force on the electron due to the magnetic field
The force on an electron due to a magnetic field is given by F = qv × B, where q is the charge on the electron, v is its velocity, and B is the magnitude of the magnetic field.
What is magnetic field ?A magnetic field is an invisible force that can be generated by electric currents and can be used to exert a force on other magnetic materials. It is most commonly found around magnets and electric currents. It is made up of lines of force that are created by the magnetic poles of a magnet and can be used to attract or repel other materials. The strength of the magnetic field is determined by the number of lines of force and the distance between them. Magnetic fields are used in a variety of applications, including navigation, electrical power generation.
Therefore,the force on the electron due to the magnetic field is F = (1.6 × 10∧-19C)(2 × 10∧6m/s)i + (3 × 10∧ × 6m/s)j × B,where B is the magnitude of the magnetic field.
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A 3-kilogram ball is accelerated from rest to a speed of 10 m/sec
The result of multiplying a particle's mass by its velocity is the fluctuation in momentum of a ball, which is 30 kg per second. Since momentum has both a magnitude and a direction, it is a vector quantity.
In the actual world, what is momentum?Almost every action that involves motion has momentum. It is an important tenet of physicsFor instance, if a team is moving forward and trying to stop, it will be difficult.
mv - mu, where u = 0 and v = 10 m/s, equals change in momentum. Note that the ball moved from rest, therefore its initial velocity was zero (u = 0).
Momentum change is equal to mv mu, which is 3*10 - 3*0, or 30.
30 kg/s = change in momentum.
What are examples and momentum?Momentum can be compared to the "power" a moving body has, or the amount of force it can exert on another body. For instance, a baseball that is thrown quickly (high velocity) and has a small bulk (big mass) can have the exact same momentum as a bowling ball that is travelling very slowly (low velocity).
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The electric potential is 300 V at x = 0 cm , is -100 V at 1x = 5 cm, and varies linearly with x. If a positive charge is released from rest at x = 2.5 cm , and is subject only to electric forces, what will the charge do?(a) Move to the right.(b) Move to the left.(c) Stay at 2.5 cm(d) Not enough information to tell.
The charge will move to the left.
The electric potential varies linearly with x, so we can find the equation for the potential as:
V(x) = mx + b,
where m is the slope and b is the y-intercept. We can use the given points to solve for m and b:
300 V = m(0 cm) + b
-100 V = m(5 cm) + b
Subtracting the first equation from the second, we get:
-400 V = 5m
m = -80 V/cm
Substituting this value for m into one of the equations, we get:
300 V = -80 V/cm * (0 cm) + b
b = 300 V
So the equation for the potential is:
V(x) = -80 V/cm * x + 300 V
Now we can use the electric field to determine the direction of the force on the positive charge. The electric field is the negative gradient of the potential:
E(x) = -dV/dx = 80 V/cm
At x = 2.5 cm, the electric field is:
E(2.5 cm) = 80 V/cm
Since the charge is positive, it will experience a force in the direction of the electric field, which is to the left. Therefore, the charge will move to the left.
The answer is (b) Move to the left.
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STT 13.2 Water is slowly poured into the container until the water level has risen into tubes 1, 2 and 3. The water doesn't overflow from any tubes. How do the water depths in the three columns compare to each other?A d1>d2>d3B D1 d3E D1=d2
Water is slowly poured into the container until the water level has risen into tubes 1, 2 and 3. The water doesn't overflow from any tubes. We have to compare the water depths in the three columns.
Since the water level has risen into all three tubes, the pressure at the bottom of all three tubes is the same.
The pressure at the bottom of each tube is given by the height of the water column multiplied by the density of the water and the acceleration due to gravity.
Since the density and acceleration due to gravity are the same for all three tubes, the pressure at the bottom of each tube depends only on the height of the water column. Therefore, the water depths in the three columns must be the same, i.e., D1 = D2 = D3.
So, the correct answer is: D1 = D2 = D3.
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By what factor must one change the weight suspended vertically from a spring coil in order to triple its period of simple harmonic motion?
In order to triple the period of simple harmonic motion of a spring coil, one must change the weight suspended vertically from the spring by a factor of 9.
This can be derived from the formula for the period of simple harmonic motion, which is T = 2π√(m/k), where T is the period, m is the mass of the object attached to the spring, and k is the spring constant.
If we assume that the spring constant remains constant, we can use the formula to determine the change in mass required to triple the period. We know that the period is directly proportional to the square root of the mass, so if we want to triple the period, we need to increase the mass by a factor of 9 (since [tex]3^2[/tex] = 9). This means that the weight suspended from the spring must also be increased by a factor of 9, since weight is proportional to mass.
Therefore, if the weight suspended vertically from a spring coil is x, in order to triple the period of simple harmonic motion, the weight must be changed to 9x. This change in weight will result in a change in mass that will cause the period to triple, as required.
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In order to triple the period of simple harmonic motion of a spring coil, one must change the weight suspended vertically from the spring by a factor of 9.
This can be derived from the formula for the period of simple harmonic motion, which is T = 2π√(m/k), where T is the period, m is the mass of the object attached to the spring, and k is the spring constant.
If we assume that the spring constant remains constant, we can use the formula to determine the change in mass required to triple the period. We know that the period is directly proportional to the square root of the mass, so if we want to triple the period, we need to increase the mass by a factor of 9 (since [tex]3^2[/tex] = 9). This means that the weight suspended from the spring must also be increased by a factor of 9, since weight is proportional to mass.
Therefore, if the weight suspended vertically from a spring coil is x, in order to triple the period of simple harmonic motion, the weight must be changed to 9x. This change in weight will result in a change in mass that will cause the period to triple, as required.
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(E) E = V/d so if V à 2, E à 2 and if d ÷ 5, E à 5 so the net effect is E à 10
Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2,000 N/C. If the voltage is doubled and the distance between the plates is reduced to 1/5 the original distance, the magnitude of the new electric field is
(A) 800 N/C
(B) 1,600 N/C
(C) 2,400 N/C
(D) 5,000 N/C
(E) 20,000 N/C
When two parallel plates are connected with a voltage source, the electric field E = V / d. Thus, option E-20,000 N/C is correct.
The electric field in the two parallel plate conductors depends on the voltage or potential difference (V) and the distance (d) between the two plates. The electric field between the two plates is 2000 N/C. If the voltage is doubled (V = 2V), and the distance is reduced to 1/5 (d = d/5),then
Electric field (E) = V / d
= 2V / (d/5)
= 10 (V/d) = 10 ×2
Electric field (E) = 20,000 N/C
Thus, the ideal solution is E) 20,000 N/C.
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A solid conducting sphere is given a positive charge Q. How is the charge Q distributed in or on the sphere?
(A) It is concentrated at the center of the sphere.
(B) It is uniformly distributed throughout the sphere.
(C) Its density decreases radially outward from the center.
(D) Its density increases radially outward from the center.
(E) It is uniformly distributed on the surface of the sphere only.
The charge Q is uniformly distributed throughout the sphere. The correct answer is (B)
When a solid conducting sphere is given a positive charge Q, the charge will distribute itself evenly throughout the surface of the sphere due to the repulsion of like charges. This is known as the "Faraday's ice pail experiment".
According to the principle of electrostatics, the charge on a conductor always resides on its surface and distributes itself in a way that the electric field inside the conductor is zero. Since the charge on a conductor always resides on its surface, it follows that the charge Q in this case must be uniformly distributed throughout the surface of the sphere.
Option (A) is not true because the charge is not concentrated at the center of the sphere. If the charge was concentrated at the center of the sphere, the electric field would not be zero inside the conductor, which contradicts the principle of electrostatics.
Option (C) and (D) are not true because the density of the charge does not change radially outward from the center. If the density decreased or increased radially outward, the electric field inside the conductor would not be zero, which again contradicts the principle of electrostatics.
Option (E) is not true because the charge is distributed throughout the entire volume of the sphere, not just on its surface. A solid conductor has free charges that can move throughout its entire volume, so the charge will distribute itself throughout the entire volume of the sphere until the electric field inside the conductor is zero.
Therefore, the correct answer is (B) it is uniformly distributed throughout the sphere.
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. Tripling the mass per unit length of a guitar string will result in changing the wave speed in the string by what factor? a. 1.73 b. 1.00 (i.e., no change) c. 3.00 d. 0.58
The wave speed will be reduced by a factor of 0.58, which is the same as dividing the original wave speed by 1.73 Option D is correct.
To answer this question, we need to use the formula for wave speed in a string, which is:
v = √(T/μ)
Where v is the wave speed, T is the tension in the string, and μ is the mass per unit length of the string. We can see that the wave speed is proportional to the square root of μ.
If we triple the mass per unit length of the guitar string, then μ will become 3 times its original value. Plugging this into the wave speed formula, we get:
v = √(T/(3μ))
Simplifying this expression, we get:
v = (1/√3)√(T/μ)
We can see that the wave speed has been divided by the square root of 3. Therefore, the answer is:d) 0.58
The wave speed will be reduced by a factor of 0.58, which is the same as dividing the original wave speed by 1.73 (since 1/√3 ≈ 0.58). This makes sense because a higher mass per unit length means that the string will be less flexible and harder to vibrate, resulting in a slower wave speed. Therefore option D is correct.
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If an object engaging in simple harmonic motion has its amplitude doubled, the maximum acceleration changes by what factor?
If an object engaging in simple harmonic motion has its amplitude doubled then the maximum acceleration changes by factor 2.
Simple harmonic motion is a specific kind of periodic motion of a body that arises from a dynamic equilibrium between an inertial force that is proportional to the body's acceleration away from the static equilibrium position and a restoring force on the moving object that is directly proportional to the magnitude of the object's displacement and acts towards the object's equilibrium position. If friction or any other energy dissipation is not present, it leads to an oscillation that is represented by a sinusoid and that lasts indefinitely.
The differential equation for SHM is given by,
[tex]\frac{d^2x}{dt^2} + \sqrt{\frac{k}{m}} x=0[/tex]
where [tex]\frac{d^2x}{dt^2}[/tex] is the acceleration of SHM.
when x = A
[tex]\frac{d^2x}{dt^2} = \omega^{2} A[/tex]
if A = 2A
[tex](\frac{d^2x}{dt^2})' = \omega^{2} 2A[/tex]
[tex](\frac{d^2x}{dt^2})' = 2(\frac{d^2x}{dt^2})[/tex]
Hence acceleration gets doubled if amplitude gets doubled.
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On the cosmic calendar, which compresses the history of the universe into a single year, about when did early humans first walk on Earth?
-in June
-in September
-in mid-December
-on December 30
-just a few hours before midnight on December 31
According to the cosmic calendar, which compresses the entire history of the universe into a single year, early humans first walked on Earth just a few hours before midnight on December 31.
To put it in perspective, if the cosmic calendar starts on January 1 with the Big Bang and ends on December 31 with the present day, then early humans appeared on the scene around 11:59 pm on December 31. This means that humans have only been around for a tiny fraction of the universe's existence, which spans over 13 billion years.
It's important to note that the exact timing of when early humans first walked on Earth is still up for debate among scientists, as there is limited fossil evidence and the timeline can be affected by various factors such as climate change and evolutionary processes. However, based on current knowledge and research, it is estimated that early humans appeared on Earth around 2-3 million years ago during the Paleolithic era.
Over time, humans evolved and developed various tools and technologies that allowed them to survive and thrive in different environments, leading to the diverse cultures and societies we see today.
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As the wind blows across a field of grain, the top of the plants move back and forth when a wave travels across the field. This wave is
a) Transverse wave. c) Longitudinal wave.
b) Polarized wave. d) Electromagnetic wave.
The wave described in the question is a longitudinal wave. Option C
Longitudinal waves are waves in which the particles of the medium vibrate back and forth in the same direction as the wave travels. This is in contrast to transverse waves, in which the particles of the medium vibrate perpendicular to the direction of the wave.
In the example given, the wind creates a disturbance in the field of grain, causing the plants to move back and forth. This movement creates a wave that travels across the field. The particles of the plants are moving in the same direction as the wave, making it a longitudinal wave.
Polarized waves are waves in which the oscillations occur in a single plane, while electromagnetic waves are waves that consist of oscillating electric and magnetic fields. Neither of these types of waves are applicable to the scenario described in the question.
In summary, the wave described in the question is a longitudinal wave, as the particles of the medium (the plants) vibrate back and forth in the same direction as the wave travels.So, option C is correct.
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A gasoline engine with an efficiency of 30.0% operates between a high temperature T1 and a low temperature T2 = 320 K. If this engine operates with Carnot efficiency, what is the high-side temperature T1?
The efficiency of the Carnot engine is calculated by the formula efficiency, η = 1 - (T₂ / T₁). The efficiency of the Carnot engine η = 30% with low-temperature T₁= 320 k and high-temperature T₂ = 457K.
The Carnot engine is a theoretical model of the thermodynamic cycle proposed by Leonard Carnot. Carnot's theorem states that an engine works between a hot and cold reservoir can have more efficiency than an engine that works between the same reservoirs. The efficiency of Carnot's engine is, η = 1 ₋ (T₂ /T₁ ), T₂ represents high temperature and T₁ represents low temperature. η represents efficiency.
From the given,
η = 1 ₋ (T₂ /T₁ )
T₁ = T₂ / (1 ₋η) = 320 / (1₋0.30)
T₁ = 427 K.
Thus, T₁ is 427K and T₂ is 320 K giving an efficiency of 30%.
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What frequencies may be HEARD due to the nonlinear effects of the ear when two sinusoids of frequency 500 and 600 Hz are directed into the ear at fairly high levels?
When two sinusoids of frequency 500 Hz and 600 Hz are directed into the ear at fairly high levels, the frequencies that may be heard due to the nonlinear effects of the ear include the original frequencies, their sum, and their difference. So, due to the nonlinear effects of the ear, the frequencies that may be heard are 500 Hz, 600 Hz, 1100 Hz, and 100 Hz.
Step 1: Identify the original frequencies:
- Frequency 1: 500 Hz
- Frequency 2: 600 Hz
Step 2: Calculate the sum and difference of the original frequencies:
- Sum: 500 Hz + 600 Hz = 1100 Hz
- Difference: 600 Hz - 500 Hz = 100 Hz
Step 3: List all the frequencies that may be heard:
- 500 Hz (Frequency 1)
- 600 Hz (Frequency 2)
- 1100 Hz (Sum)
- 100 Hz (Difference)
So, due to the nonlinear effects of the ear, the frequencies that may be heard are 500 Hz, 600 Hz, 1100 Hz, and 100 Hz.
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What is the total force on the bottom of a 2.0-m-diameter by 1.0-m-deep round wading pool due to the weight of the air and the weight of the water? (Note the pressure contribution from the atmosphere is 1.0 ´ 105 N/m2, the density of water is 1 000 kg/m3, and g = 9.8 m/s2.)
The total force on the bottom of the wading pool, considering the weight of the air and the weight of the water, is 3.47 × 10⁵ Newtons.
To calculate the total force on the bottom of the wading pool, we need to consider the weight of the air and the weight of the water.
Weight of the air:
The pressure contribution from the atmosphere is given as 1.0 × 10⁵ N/m². Since the pressure acts equally in all directions, we can assume it applies uniformly over the surface of the wading pool.
The force due to the weight of the air can be calculated using the formula:
Force = Pressure × Area
Weight of the air:
Area = π × (1.0 m)²
Area = π m²
Force_air = (1.0 × 10⁵ N/m²) × (π m²)
Force_air = 3.14 × 10⁵ N
Weight of the water:
The weight of the water can be calculated using the formula:
Weight = Mass × Acceleration due to gravity
The mass of the water can be determined using its density and the volume of the pool:
Volume = Area × Depth
Volume = π m² × 1.0 m = π m³
Mass = (1000 kg/m³) × (π m³)
Mass = 1000π kg
Weight_water = (1000π kg) × (9.8 m/s²)
Weight_water ≈ 9800π N
Total force on the bottom of the pool:
Total force = Force_air + Weight_water
Total force ≈ 3.14 × 10⁵ N + 9800π N
Total force ≈ 3.14 × 10⁵ N + 30794 N
Using the approximation π ≈ 3.14:
Total force ≈ 3.47 × 10⁵ N
Therefore, the total force on the bottom of the wading pool is approximately 3.47 × 10⁵ Newtons.
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If you calculated 105 ± 2 Ohms and the manufacturer says it is 100 Ohms with a 5% tolerance, do your data support the manufacturer's claims?
Yes, your data support the manufacturer's claims. You calculated 105 ± 2 Ohms, Ohm's law which means the range is 103 to 107 Ohms. The manufacturer claims 100 Ohms with a 5% tolerance, which means the acceptable range is 95 to 105 Ohms.
According to Ohm's law, the voltage across two places is precisely proportional to the current flowing through a conductor between them. One arrives at the standard mathematical equation that captures this relationship by include the resistance as the constant of proportionality: I = V/R, where V is the voltage applied across the conductor in volts, R is the resistance of the conductor in ohms, and I is the current flowing through the wire in amperes.
domestic conventional fans.Electrified heaters. Around the world, electric heaters are a typical wintertime device.Irons and kettles with electricity. There are numerous resistors within the electric kettle and irons.Electric device design.Since your calculated range falls within the manufacturer's specified tolerance, your data support their claims.
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Two force vectors, F1=(3.0N)x^−(4.0N)y^ and F2=(−6.0N)x^+(4.5N)y^, are applied to a particle.What third force F3 would make the net, or resultant, force on the particle zero?
To make the net force on the particle zero, the third force vector, F3, must be equal and opposite to the vector sum of F1 and F2.
To find the vector sum of F1 and F2, we add their respective x and y components:
Fx = F1x + F2x = (3.0N) - (6.0N) = -3.0N
Fy = F1y + F2y = (-4.0N) + (4.5N) = 0.5N
Therefore, the vector sum of F1 and F2 is F1+F2 = (-3.0N)x^ + (0.5N)y^.
To make the net force zero, the third force vector, F3, must be equal and opposite to F1+F2:
F3 = -(F1+F2)
F3 = -(-3.0N)x^ - (0.5N)y^
F3 = (3.0N)x^ + (0.5N)y^
Therefore, a third force of (3.0N)x^ + (0.5N)y^ would make the net force on the particle zero.
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when a buck-boost transformer has current less than nine amperes, an overcurrent protection device is allowed to be rated at not more than ? of the input current.
When a buck-boost transformer has a current less than nine amperes, an overcurrent protection device is allowed to be rated at not more than 125% of the transformer's primary current.
To calculate the rating of the overcurrent protection device:
1. Determine the primary current of the transformer (let's assume it's less than 9 amperes).
2. Multiply the primary current by 125% (or 1.25) to find the maximum allowed rating for the overcurrent protection device.
For example, if the primary current is 8 amperes, the maximum allowed rating for the overcurrent protection device would be 8 x 1.25 = 10 amperes.
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where does the dme indicator have the greatest error between the ground distance and displayed distance to the vortac?
When the aircraft is directly above or below the VORTAC station, the DME (Distance Measuring Equipment) indicator has the maximum error between the ground distance and the displayed distance to the VORTAC.
What is the most common error in DME?The most extreme case of "slant range error" occurs when an aircraft passes directly over the station; rather than reading zero, the DME displays the airplane's altitude above the station (in nautical miles).
What are DME's flaws?Only when flying directly to or from the ground station will DME ground speed and time-to-station be accurate. Any other direction will result in incorrect ground speed and time-to-station.
Where is the DME indicator's error between ground the greatest?The greatest DME indication error occurs at. - High altitudes near the VORTAC Because the DME measures slant range distance, its greatest error occurs at high altitudes near the VORTAC. Assume you're at 12,000 feet.
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The greatest error occurs at higher altitudes when using DME, as it measures the line-of-sight distance, rather than the actual ground distance to the VORTAC.
The DME (Distance Measuring Equipment) indicator will have the greatest error between the ground distance and the displayed distance to the VORTAC (VHF Omnidirectional Range Tactical Air Navigation) when the aircraft is flying at higher altitudes. This is because the DME measures 'line-of-sight' distance, which forms a hypotenuse of a right triangle, hence higher from the ground level. When at higher altitudes, this slant range error becomes more prominent compared to when flying closer to the ground level, hence often leading to overestimates of the true ground distance to the VORTAC station.
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A 5.0 kg box slides a 10 m distance on ice.If the coefficient of kinetic friction is 0.20, what is the work done by the friction force?
To calculate the work done by the friction force on a 5.0 kg box sliding a 10 m distance on ice with a coefficient of kinetic friction of 0.20, follow these steps:
1. Calculate the normal force (N): Since the box is on a flat surface, the normal force is equal to its weight, which is the mass (m) times gravity (g).
N = m * g
N = 5.0 kg * 9.81 m/s²
N ≈ 49.05 N
2. Calculate the friction force (F_friction): Use the coefficient of kinetic friction (μ_k) and the normal force (N).
F_friction = μ_k * N
F_friction = 0.20 * 49.05 N
F_friction ≈ 9.81 N
3. Calculate the work done by the friction force (W): Use the friction force (F_friction) and the distance (d) the box slides.
W = F_friction * d * cos(θ)
Since the friction force opposes the motion, the angle between the force and the displacement is 180 degrees (π radians), so cos(θ) = -1.
W = 9.81 N * 10 m * -1
W ≈ -98.1 J
The work done by the friction force on the 5.0 kg box sliding a 10 m distance on ice with a coefficient of kinetic friction of 0.20 is approximately -98.1 joules.
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(D) W = âK = QV (mass doesn't have an effect on the kinetic energy, just on the speed in this case)
A particle of charge Q and mass m is accelerated from rest through a potential difference V, attaining a kinetic energy K. What is the kinetic energy of a particle of charge 2Q and mass m/2 that is accelerated from rest through the same potential difference?
(A) ¼ K (B) ½ K (C) K (D) 2K (E) 4K
The kinetic energy of the second particle is 2K.
Charge of the first particle = Q
Charge on the second particle = 2Q
mass of the first particle = m
mass of the second particle = m/2
Potential difference applied = V
Given that, mass doesn't have an effect on the kinetic energy, just on the speed.
The kinetic energy attained by the first particle,
K = charge x potential difference
K = Q x V = QV
So, the accelerating potential, V = K/Q
Since, the second particle is accelerated from rest through the same potential difference, its kinetic energy,
K' = 2Q x V
K' = 2K
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The primary coil of a transformer makes 240 turns around the iron core, and the secondary coil of that transformer makes 80 turns. If the primary voltage is 120-V AC, what is the secondary voltage?
Entry field with correct answer
Zero.
40-V AC.
360-V AC.
160-V AC.
The secondary voltage is 40-V AC. The correct answer is B.
The voltage induced in the secondary coil of a transformer is proportional to the ratio of the number of turns in the secondary coil to the number of turns in the primary coil. This is known as the turn ratio.
In this case, the turns ratio is 80/240, which simplifies to 1/3. So, the voltage induced in the secondary coil is one-third of the voltage applied to the primary coil, or:
V_secondary = (1/3) * V_primary = (1/3) * 120 V = 40 V
The other options are not true because:
Zero: This is not possible because there is a voltage applied to the primary coil.
360-V AC: This is not possible because the turns ratio is 1/3, which means the secondary voltage is one-third of the primary voltage.
160-V AC: This is not correct because the turns ratio is 1/3, not 3/1.
Therefore, the secondary voltage is 40-V AC.
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the electrical force between charges depends only on the charges' magnitude and separation distance. separation distance. magnitude. none of the above choices are correct.
The statement "the electrical force between charges depends only on the charges' magnitude and separation distance" is correct.
An electric force is the interaction of either attractive force or repulsive force between two charged bodies. This force is similar to other forces because it affects and impacts towards a particular object and can be easily demonstrated by Newton’s law of motion. Electric force is one of the forces which is exerted over other bodies.
The force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the separation distance between them, according to Coulomb's Law. Therefore, the electrical force between charges is determined solely by the magnitude of the charges and their separation distance.
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The earth takes 8.62 x 10^4 s to complete one rotation about its axis of rotation. Find the angular velocity of the earth.
The earth takes 8.62 ×10⁴ s to complete one rotation about its axis and hence, the angular velocity of the earth is 7.285×10⁻⁵ rad / s.
When an object rotates or revolves around its axis is called angular velocity. It also defines the angular displacement between two bodies with respect to time. The unit of angular velocity is rad/s.
Angular velocity, ω = Δθ / Δt. Δθ represents the change in angular displacement and Δt represents the time taken for the rotating body. When an object completes one revolution, the angle in radians is 2π and the time taken to complete one revolution is T.
From the given,
Time taken to complete to rotation = 8.62 x 10^4 s
ω = 2π / T
= 2×3.14 / 8.62 x 10^4
= 7.285×10⁻⁵ rad / s.
The angular velocity of the earth is 7.285×10⁻⁵ rad / s.
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is there a point between a 10nC charge and a -20nC charge at which the electric field is zero
The neutral point is located at a distance x from the 10 nC charge and (d+x) from the -20 nC charge.
Yes, there is a point between a 10 nC charge and a -20 nC charge at which the electric field is zero. This point is known as the "neutral point" or the "equipotential point" and it lies on the line that joins the two charges.
To find the position of the neutral point, we can use the principle of superposition of electric fields. According to this principle, the electric field at any point due to a collection of charges is the vector sum of the electric fields due to each individual charge.
Let's assume that the 10 nC charge is located at the origin and the -20 nC charge is located on the x-axis at a distance of d from the origin. The electric field due to the 10 nC charge at any point on the x-axis is given by:
E1 = k*q1/x^2
where k is Coulomb's constant, q1 is the charge on the 10 nC charge, and x is the distance from the 10 nC charge to the point on the x-axis.
Similarly, the electric field due to the -20 nC charge at any point on the x-axis is given by:
E2 = k*q2/(d+x)^2
where q2 is the charge on the -20 nC charge and (d+x) is the distance from the -20 nC charge to the point on the x-axis.
For the neutral point, the electric field due to the 10 nC charge and the electric field due to the -20 nC charge must cancel each other out. In other words, E1 + E2 = 0. Solving this equation for x, we get:
x = d*q2/(q1-q2)
Therefore, the neutral point is located at a distance x from the 10 nC charge and (d+x) from the -20 nC charge.
If q1 and q2 have the same magnitude (in this case, 10 nC), the neutral point will be located at the midpoint between the two charges, which is at a distance of d/2 from each charge. However, in this case, since the charges have opposite signs, the neutral point will be located closer to the negative charge (-20 nC).
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A boy takes a toy top and pulls on a string to make the top spin. The top can be considered a solid disk (I=½MR2) and has a mass of 0.100kg and a radius of 0.0200m. The top starts from rest and ends up spinning at 15.0rev/s after 0.800s. What is the torque applied to the top?
The torque applied to the toy top is approximately 3.75×10^-3 N·m.
We can use the rotational equivalent of Newton's second law of motion, which states that the net torque applied to an object is equal to the moment of inertia times the angular acceleration:
Στ = Iα
where Στ is the net torque, I is the moment of inertia, and α is the angular acceleration.
We can also use the equation for angular acceleration
α = Δω/Δt
where Δω is the change in angular velocity and Δt is the time interval over which the change occurs.
In this problem, the initial angular velocity is 0 rev/s, the final angular velocity is 15.0 rev/s, and the time interval is 0.800 s. Therefore:
Δω = 15.0 rev/s - 0 rev/s = 15.0 rev/s
Δt = 0.800 s
α = Δω/Δt = 15.0 rev/s / 0.800 s = 18.75 rad/s^2
The moment of inertia of a solid disk is I = 1/2MR^2, where M is the mass and R is the radius. Plugging in the given values, we get:
I = 1/2 (0.100 kg) (0.0200 m)^2 = 2.00×10^-5 kg·m^2
Now we can solve for the torque:
Στ = Iα = (2.00×10^-5 kg·m^2) (18.75 rad/s^2) = 3.75×10^-3 N·m
Therefore, the torque applied to the top is 3.75×10^-3 N·m.
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a 6 uf capacitor a 10 uf capacitor and a 16 uf capacitor are connected in parallel what is their equivalent capacitance
The equivalent capacitance is 32 µF.
A capacitor is a passive electronic component that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, an electric field is created between them, causing one plate to become positively charged and the other negatively charged. The capacitor can then store this charge, which can be used later to perform work in an electrical circuit. The amount of charge that a capacitor can store is determined by its capacitance, which is measured in farads (F).
When capacitors are connected in parallel, the equivalent capacitance is calculated by adding up the capacitance values of individual capacitors. Therefore, the equivalent capacitance of a 6 µF capacitor, a 10 µF capacitor, and a 16 µF capacitor connected in parallel is:
Ceq = C1 + C2 + C3 = 6 µF + 10 µF + 16 µF = 32 µF
Hence, the equivalent capacitance is 32 µF.
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.
In an isovolumetric process by an ideal gas, the system's heat gain is equivalent to a change in: a. temperature. b. volume. c. pressure.
In an isovolumetric process, the volume remains constant in the system. The heat gain is equivalent to a change in the internal energy of the system. Thus, option D is correct.
In a thermodynamic system, when the volume remains constant it is called as an isovolumetric process. In an isovolumetric process, the work done by the system is zero. The heat energy is equivalent to the internal energy of the system.
In an isovolumetric process, ΔW = 0 , ΔU = ΔH. ΔU is the internal energy of the system whereas ΔH represents the heat flow in the system.
Hence, the ideal solution is D) internal energy.
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Your question is incomplete, most probably your full question was, n an isovolumetric process by an ideal gas, the system's heat gain is equivalent to a change in a. temperature. b. volume. c. pressure. d. internal energy.
and according to, option d is the answer to the question.
a person who weighs 830 n is riding a 91-n mountain bike. suppose the entire weight of the rider plus bike is supported equally by the two tires. if the gauge pressure in each tire is 6.80 x 105 pa, what is the area of contact between each tire and the ground?
The area of contact between each tire and the ground is approximately 6.77 x 10⁻⁴ m².
To find the area of contact between each tire and the ground, we can use the formula for pressure:
Pressure = Force / Area
We are given the gauge pressure (6.80 x 10⁵ Pa) and the total weight of the rider plus the bike (830 N + 91 N = 921 N). Since the weight is supported equally by the two tires, the force on each tire is half the total weight:
Force on each tire = 921 N / 2 = 460.5 N
Now, we can rearrange the formula to find the area of contact:
Area = Force / Pressure
Substitute the values for force and pressure:
Area = 460.5 N / (6.80 x 10⁵ Pa)
Area ≈ 6.77 x 10⁻⁴ m²
So, between each tire and the ground, the area of contact is approximately 6.77 x 10⁻⁴ square meters.
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How are the arrowheads on the cutting-plane related to the view in section? What other types of lines are associated with sectional views?
The arrowheads on the cutting-plane in a sectional view indicate the direction of sight. In other words, they show the direction in which the observer is looking at the object.
The cutting-plane is an imaginary plane that slices through the object, and the arrowheads point in the direction of the cut. This helps to clarify the relationship between the view and the object.
In addition to the arrowheads, there are other types of lines associated with sectional views. For example, the section line shows the location of the cutting-plane, and it is typically a dashed line. The section hatching or shading is used to distinguish the cut surfaces from the uncut surfaces. The hatch lines are typically at a 45-degree angle and spaced evenly. The outline or contour lines are used to show the shape of the object and to differentiate between the cut and uncut portions.
Overall, the use of arrowheads, section lines, section hatching, and contour lines in a sectional view helps to provide a clear and detailed representation of the object. This type of drawing is particularly useful in engineering and architecture, where precise visualization and communication of designs are essential.
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Which is true about the transmission of the electric fields of light by a polarizing sheet?
a. Any component of the field is transmitted.
b. Only the components perpendicular to the polarizing axis of the sheet are transmitted.
c. Only the components parallel to the polarizing axis of the sheet are transmitted.
The transmission of the electric fields of light by a polarizing sheet C. Only the components parallel to the polarizing axis of the sheet are transmitted.
A polarizing sheet is a material that selectively transmits light waves based on their orientation. The sheet has a specific polarizing axis, which is a direction in the material that allows certain components of the electric fields of light to pass through. When light encounters a polarizing sheet, it consists of both components parallel and perpendicular to the polarizing axis.
The polarizing sheet only transmits the components of the electric fields of light that are parallel to its polarizing axis. This occurs because the sheet's molecules preferentially absorb the light components that are perpendicular to the polarizing axis, effectively blocking them from passing through. On the other hand, the components parallel to the polarizing axis are allowed to continue without being absorbed.
By transmitting only the parallel components, the polarizing sheet effectively polarizes the light, which means the light waves become more aligned in a single direction. This property of polarizing sheets is useful in various applications, such as reducing glare in sunglasses and improving image clarity in optical instruments. Therefore the correct option is C
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