The wind direction, speed, and temperature that a pilot should expect when planning for a flight over EMI at FL 270 are as follows:
Wind direction: 240 degrees True
Wind speed: 25 knots
Temperature: -10 degrees Celsius
EMI is a waypoint in the North Atlantic Track System, located in the middle of the ocean. When planning for a flight over this area, a pilot must take into account the wind and temperature conditions at that altitude (FL 270) to ensure the safety and efficiency of the flight.
These conditions can be obtained from weather forecasts and/or real-time data provided by the aircraft's instruments or other sources. The wind direction, speed, and temperature are all factors that affect the aircraft's performance, fuel consumption, and other operational parameters, and must be carefully considered in the flight planning process.
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the photo at right was taken through a specroscope. what color was the pigment extract used to produce this spectrum? what colo(s) did this extract absorb?
Light is a form of energy. All the properties of light can be explained by Considering the Wave length and lespuscutar theory.
The Wave Theory states that waves are how light moves across space. When Visible light is passed through a prim it is split up into seven colours which corresponds to definite wave length. a phenomenon Called dispersion. The study of interaction between matter and electromagnetic radiation is defined as spectroscopy.
A spectrophotometer is a device which detect the percentage transmittance of light radiation. When light of certain intensity and frequency range is passed through the Sample Thus the instrument Compare the intensity of the transmitted light with that of the incident light.
A spectroscope is a device that divides light into its individual wavelengths to produce a spectrum.
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which stereo microphone technique works only from level differences produced by off-axis attenuation and not from difference in time delay
The stereo microphone technique that works only from level differences produced by off-axis attenuation is called Mid-Side (M-S) stereo recording.
Mid-Side (M-S) stereo recording technique utilizes two microphones, one of which is pointed directly at the sound source and the other pointing 90 degrees to the left or right. The result is two channels of audio which are then combined in the mixing process to create a full stereo image. The two signals have a 3 dB difference in level between them, with the off-axis microphone being the lower of the two. This level difference produces the stereo image and is the only factor used in M-S stereo recording. Time delays are not used in this technique, as the level differences are sufficient to create the stereo image.
M-S stereo recording has a number of benefits, such as being able to adjust the width of the stereo image in the mixing process, and allowing for mono compatibility. It is therefore a great choice for stereo recordings that need to be heard in both mono and stereo, such as live broadcasts or online streams.
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if each charge has two field lines per unit of charge (q), what is the ratio of the total positive (red) charge to the total negative (blue) charge?
The ratio of total positive charge (red) to total negative charge (blue) is 1:1. This is because for each unit of charge (q), there are two field lines, one for the positive charge and one for the negative charge.
What are field lines?Field lines are a visual tool used to represent the direction and strength of an electrical field. The direction of a field line shows the direction of the force that a positive test charge would experience if it were placed at that point in the field. Meanwhile, the density of the field lines indicates the strength of the electric field.
Since each charge has two field lines per unit of charge (q), it means that the total number of field lines is proportional to the total charge. If there are equal numbers of field lines coming from both the positive and negative charges, it means that the ratio of the total positive charge to the total negative charge is 1:1.
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Question about my assignment
Answer: A - Gold Foil Experiment
Explanation:
Answer:
A
Explanation: Rutherford's gold foil experiment.
A load of 100g placed on a spiral spring, A extends it spring by 2cm when the same load is placed on spiral spring, B it extends it by 5cm, which spring has smaller stiffness
Spring A has smaller stiffness.Stiffness is a measure of the spring's resistance to deformation. The stiffer the spring, the more resistant it is to deformation. When a load is applied to a spring, it experiences deformation. Stiffness is a measure of how much force is required to deform the spring by a certain amount.
Springs with higher stiffness require more force to deform them than springs with lower stiffness.A load of 100g placed on a spiral spring, A extends its spring by 2cm, whereas the same load placed on spiral spring, B extends it by 5cm. The stiffness of a spring is inversely proportional to the amount of deformation it experiences. Spring B will be less stiff because it experiences more deformation than Spring A.Spring stiffness is measured in units of force per unit of length. The spring constant k is a measure of stiffness. It is defined as the amount of force required to extend the spring by one unit of length.The spring constant k can be calculated as follows:F = kxWhere F is the force applied, k is the spring constant, and x is the amount of deformation experienced by the spring. We can use this formula to calculate the spring constants for A and B:kA = F/x = 100g/(2/100) = 5000 N/mkB = F/x = 100g/(5/100) = 2000 N/mSpring A has a higher stiffness (5000 N/m) than spring B (2000 N/m) because it requires more force to deform it by the same amount. Hence, spring A has smaller stiffness.
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To be able to calculate the energy of a charged capacitorand to understand the concept of energy associated withan electric field.The energy of a charged capacitor is given byU= QV/2, where Q is the charge of the capacitor andV is the potential difference across the capacitor. Theenergy of a charged capacitor can be described as theenergy associated with the electric field created insidethe capacitor.In this problem, you will derive two more formulas for theenergy of a charged capacitor; you will then use aparallel-plate capacitor as a vehicle for obtaining theformula for the energy density associated with an electricfield. It will be useful to recall the definition ofcapacitance, C = Q/V, and the formula for thecapacitance of a parallel-plate capacitor,Co A/d, where A is the area of each of the platesand d is the plate separation. As usual, eo is thepermittivity of free space.
The energy of a charged capacitor can also be written as [tex]U = \frac {CV^2}{2}[/tex] and [tex]U = \frac {Q^2d}{2\epsilon_o A}[/tex].
To derive two more formulas for the energy of a charged capacitor, we start with the definition of capacitance:
C = Q/V
Solving for Q, we get:
Q = CV
Substituting this expression for Q into the original formula for the energy of a charged capacitor, [tex]U = QV/2[/tex], we get:
[tex]U = (CV)V/2[/tex]
[tex]U = CV^2/2[/tex]
This is one of the additional formulas for the energy of a charged capacitor.
Next, we can use the formula for the capacitance of a parallel-plate capacitor to derive the energy density associated with an electric field. The capacitance of a parallel-plate capacitor is given by:
[tex]C = \epsilon _o A/d[/tex]
where εo is the permittivity of free space, A is the area of each plate, and d is the distance between the plates. Solving this equation for the potential difference, V, we get:
[tex]V = Q/C[/tex]
[tex]V = Q/(\epsilon_o A/d)[/tex]
[tex]V = Qd/(\epsilon_o A)[/tex]
Substituting this expression for V into the formula for the energy of a charged capacitor, [tex]U = QV/2[/tex], we get:
[tex]U = \frac {Q^2d}{2\epsilon_o A}[/tex]
This expression gives us the energy associated with the electric field in the capacitor.
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The trachea is found ______ to the esophagus; connects the larynx to primary bronchi; inferiorly the trachea divides into right and left ______ ______.
The trachea is found anterior to the esophagus; connects the larynx to primary bronchi; inferiorly the trachea divides into right and left main bronchi.
The trachea is around 10-12 cm long and 2-3 cm wide, and it is made up of cartilage rings that support the tube and keep it from collapsing during inhalation. Lining the trachea is ciliated mucosa, which captures and eliminates foreign particles and mucus from the respiratory system. Mucus is also secreted by the mucosa to assist moisten and warm the inspired air. The trachea splits inferiorly into right and left primary bronchi, which divide further into secondary and tertiary bronchi, finally delivering air to the lungs. The trachea is a tube.
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you throw a 0.80kg snowball at 7.0m/s straight down off a 5.0m tall bridge
The kinetic energy of the snowball just before it hits the ground is 19.6 joules.
How to calculate the kinetic energy?To determine the kinetic energy of the snowball just before it hits the ground, we can use the formula for kinetic energy:
KE = 1/2 m v²
where KE is the kinetic energy, m is the mass of the snowball, and v is the velocity of the snowball just before it hits the ground.
In this case, we know that the mass of the snowball is 0.80 kg and the velocity just before it hits the ground is equal to the initial velocity with which it was thrown (7.0 m/s) since air resistance is assumed to be negligible. Therefore, we can substitute these values into the formula:
KE = 1/2 * 0.80 kg * (7.0 m/s)²
KE = 19.6 J
Therefore, the kinetic energy of the snowball just before it hits the ground is 19.6 joules.
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Correct question is:
You throw a 0.80kg snowball at 7.0m/s straight down off a 5.0m tall bridge then what is the kinetic energy of the snowball just before it hits the ground?
a semi-circular gate on an inclined wall is in contact with water. calculate the resultant force of the water on the gate
The resultant force of the water on the semi-circular gate on an inclined wall can be calculated using the equations of hydrostatics.
R = √([tex]F1^2 + F2^2 - 2*F1*F2*cos[/tex])α, where 'R' is the resultant force and 'α' is the angle of the wall.
First, determine the pressure of the water at any given point along the gate. To do this, multiply the density of the water, 'ρ', by the acceleration of gravity, 'g', and then the vertical height of the water relative to the gate, 'h', to get the pressure 'p':
p = ρ*g*h
Second, determine the force acting on the gate. This is done by multiplying the pressure with the area of the gate, 'A':
F = p*A
Finally, find the resultant force, 'R', by adding the forces together and taking into account the angle of the wall:
R = √([tex]F1^2 + F2^2 - 2*F1*F2*cos[/tex])α
where α is the angle of the wall.
By following these steps, you can calculate the resultant force of the water on the semi-circular gate on an inclined wall.
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assuming that the magnetic field is uniform between the pole faces and negligible elsewhere, write an expression for the induced emf in the coil.
The induced EMF (electromotive force) in a coil can be calculated using Faraday's Law of Electromagnetic Induction:
EMF = -N(dΦ/dt)
What is the induced emf?In a uniform magnetic field, the flux through the coil can be calculated as:
Φ = BAcos(θ)
where B is the magnitude of 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.
Assuming that the coil is moving perpendicular to the magnetic field (θ = 0), the rate of change of flux is:
dΦ/dt = BA(d/dt)(cos(0))
= 0
Therefore, the induced EMF in the coil is zero.
However, if the coil is moving at an angle to the magnetic field, or if the magnetic field is changing in time, then the induced EMF will not be zero and can be calculated using the above equations.
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a 70.0-g arrow, fired at a speed of 115 m/s to the left, impacts a tree, which it penetrates to a depth of 12.2 cm before coming to a stop. assuming the force of friction exerted by the tree is constant, what are the magnitude and direction of the friction force acting on the arrow?
The magnitude of the friction force is 0.788 N and it is directed to the right.
The friction force acting on the arrow is equal to the force required to stop the arrow and is directed opposite the direction of motion.
The magnitude of the friction force is equal to the product of the mass of the arrow (70.0 g) and the deceleration of the arrow (11.2 cm/s^2).
When the arrow hits the tree, the friction force of the tree will slow down the arrow's motion. The magnitude of this friction force is equal to the product of the mass of the arrow (70.0 g) and the deceleration of the arrow (11.2 cm/s^2).
The direction of the friction force will be opposite to the direction of the arrow's motion.
Therefore, the magnitude of the friction force is 0.788 N and it is directed to the right. This is because the arrow was fired to the left and the friction force must be equal and opposite in order to bring the arrow to a stop.
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Two crates are dropped from the same height in a vacuum. The mass and force of the crates are shown in the diagram. Which of the following statements is supported by the data?
The momentum of both crates will be the same right before they hit the ground.
The normal force will be the same on both crates when they hit the ground.
Both of the crates are accelerating at the same rate.
Both of the crates are falling at a constant speed.
The statement that is supported by the data is that both of the crates are accelerating at the same rate. That is option C.
What is acceleration?Acceleration is defined as the quantity that measures the rate of change of velocity of a moving object.
According to Newton's second law, the force (F) acting on an object is equal to the mass (m) of object times its acceleration (a).
That is F = ma
For the first crate = a = 686/70 = 9.8 m/s²
Fro the second crate = 490/50 = 9.8 m/s²
Therefore, Both of the crates are accelerating at the same rate.
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kim holds a 2.0 kg air rifle loosely and fires a bullet of mass 1.0 g. the muzzle velocity of the bullet is 150 m/s. calculate the recoil speed of the rifle.
The recoil speed of the rifle is 0.075 m/s in the opposite direction to the direction of the bullet.
To calculate the recoil speed of the rifle, we can use the conservation of momentum principle. According to this principle, the total momentum of the system (bullet + rifle) is conserved before and after the firing of the bullet.
Initially, the total momentum of the system is zero because the rifle and bullet are at rest. After firing the bullet, the total momentum of the system is given by:
m1v1 + m2v2 = 0
where m1 and v1 are the mass and velocity of the bullet, and m2 and v2 are the mass and recoil velocity of the rifle, respectively.
Substituting the given values, we get:
(0.001 kg)(150 m/s) + (2.0 kg)(v2) = 0
Solving for v2, we get:
v2 = -(0.001 kg)(150 m/s) / (2.0 kg)
v2 = -0.075 m/s
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how does radio waves change with respect to their position to the event horizon
Radio waves, like all forms of electromagnetic radiation, are affected by the gravitational field of massive objects such as black holes, which have an event horizon beyond which nothing, not even light, can escape.
What is radio waves?Radio waves are a type of electromagnetic radiation with a wavelength longer than infrared light and shorter than microwaves. Radio waves are used for communication and information transfer, and are commonly used in radio and television broadcasting, cellular communication, and satellite communication.
As radio waves approach a black hole, they are subject to gravitational redshift, a phenomenon whereby the wavelength of the wave is stretched out and its frequency decreases as it moves away from the black hole's gravitational field.
This means that the closer the radio waves get to the event horizon, the more they are redshifted, and the longer their wavelengths become.
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A lamp in child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.250 s, during which it produces an average 0.500 W from an average 3.00 V. A) What energy does it dissipate? B) How much charge moves through the lamp? C) Find capacitance. D) What is the resistance of the lamp?
A) the energy dissipated is 0.125 Joules. B) the charge that moves through the lamp is 0.0278 Coulombs. C) the capacitance is 0.0278 Farads. D) the resistance of the lamp is approximately 8.99 Ohms.
A) To find the energy dissipated, use the formula:
Energy = Power × Time
where Power = 0.500 W and Time = 0.250 s.
Energy = 0.500 W × 0.250 s = 0.125 J
So, the energy dissipated is 0.125 Joules.
B) To find the charge that moves through the lamp, use the formula:
Energy = (1/2) × Charge × Voltage^2
Rearrange the formula to solve for Charge:
Charge = (2 × Energy) / Voltage^2
Charge = (2 × 0.125 J) / (3.00 V)^2 = 0.0278 C
So, the charge that moves through the lamp is 0.0278 Coulombs.
C) To find the capacitance, use the formula:
Energy = (1/2) × Capacitance × Voltage^2
Rearrange the formula to solve for Capacitance:
Capacitance = (2 × Energy) / Voltage^2
Capacitance = (2 × 0.125 J) / (3.00 V)^2 = 0.0278 F
So, the capacitance is 0.0278 Farads.
D) To find the resistance of the lamp, use the RC time constant formula:
Time constant (τ) = Resistance × Capacitance
Since the effective duration of the flash is 0.250 s, we can assume it is approximately equal to the time constant (τ).
Rearrange the formula to solve for Resistance:
Resistance = Time constant / Capacitance
Resistance = 0.250 s / 0.0278 F = 8.99 Ω
So, the resistance of the lamp is approximately 8.99 Ohms.
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a system releases 690 kj of heat and does 110 kj of work on the surroundings. part a what is the change in internal energy of the system?
A system releases 690 kj of heat and does 110 kj of work on the surroundings then part a what i the change in internal energy of the system -800 kJ.
The change in internal energy of the system can be calculated using the formula
ΔU = Q - W,
where ΔU is the change in internal energy, Q is the heat exchanged, and W is the work done.
In this case, the system releases 690 kJ of heat (Q = -690 kJ) and does 110 kJ of work on the surroundings (W = 110 kJ).
So, ΔU = -690 kJ - 110 kJ = -800 kJ.
The change in internal energy of the system is -800 kJ.
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g what is the expected acceleration of the cart if its mass is 220 g and two masses of 250 g each are added to it. a mass of 60 g is placed in the mass hanger of 5 g.
The expected acceleration of the cart when two masses of 250 g are added to it, and a mass of 60 g is placed in the mass hanger of 5 g, given that its mass is 220 g is 1.55 m/s².
What is acceleration?
Acceleration is the change in velocity with respect to time. It can be defined as the rate at which the velocity of a body changes with respect to time. It is denoted by "a".
Mass is the amount of matter in a body or object. It is a scalar quantity, which is denoted by "m".
acceleration: a = (v - u) / t
Where; a = acceleration
v = final velocity
u = initial velocity
t = time taken
The expected acceleration of the cart:
Mass = Mass of cart + Mass of 2 masses
Mass = 220 g + (2 × 250 g)
Mass = 720 g
The total mass hanging on the mass hanger:
Mass on hanger = Mass of hanger + Mass on a hanger
Mass on hanger = 5 g + 60 g
Mass on hanger = 65 g
The net force acting on the system.
Net force = (Mass on hanger + Mass) × gNet force
= (65 g + 720 g) × 9.8 m/s²
Net force = 7.06 N
The expected acceleration of the cart;
a = F / ma
= (7.06 N) / (720 g)
a = (7.06 N) / (0.72 kg)a
= 9.81 m/s² × (7.06 / 0.72)a
= 1.55 m/s²
Therefore, the expected acceleration of the cart when two masses of 250 g are added to it, and a mass of 60 g is placed in the mass hanger of 5 g, given that its mass is 220 g is 1.55 m/s².
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suppose i drop a football from a tall building, and 4 seconds elapse before it hits the ground. neglecting air resistance, roughly how fast is the football moving upon impact?
The football is moving at approximately 59 meters/second (212.3 kilometers/hour) upon impact, neglecting air resistance.
The football is moving at approximately 59 meters/second (212.3 kilometers/hour) upon impact, neglecting air resistance. This can be calculated using the equation s=1/2at^2, where 's' is the displacement, 'a' is the acceleration due to gravity (9.8m/s^2), and 't' is the time of the fall (4 seconds). Therefore, the displacement is s=1/2(9.8m/s^2)(4s)^2, which simplifies to s=78.4m.
Since the displacement is known (78.4m), the velocity can be determined using the equation v^2=u^2+2as, where 'v' is the velocity upon impact, 'u' is the initial velocity (0m/s), and 'a' is the acceleration due to gravity (9.8m/s^2). Therefore, the velocity is v=sqrt(2as), which simplifies to v=sqrt(2(9.8m/s^2)(78.4m)), which simplifies to v=59m/s.
This is the speed of the football upon impact, neglecting air resistance. This assumes that the football is dropped from rest, and experiences a uniform acceleration throughout the fall, which is due to gravity. The acceleration due to gravity is a constant 9.8m/s^2, regardless of the speed of the falling object.
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william herschel tried to locate the center of our galaxy by counting the number of stars in different directions. this did not work because
William Herschel's approach failed due to the fact that some parts of the Milky Way galaxy are denser than others.
This means that the number of stars would be greater in these regions, making it difficult to determine the galaxy's center simply by counting the number of stars in different directions. Herschel's pioneering work, including his discovery of Uranus and his cataloging of hundreds of nebulae, helped pave the way for future astronomers to explore and understand the universe. However, his method for locating the center of the Milky Way was limited by the technology of his time.
In modern times, astronomers have employed a range of techniques to study the galaxy, including measuring the positions and motions of stars, observing the behavior of gas and dust clouds, and using radio and other wavelengths of light to observe the galaxy's structure and composition.
Despite these advances, the center of the Milky Way remains difficult to observe directly due to the presence of dense dust and gas clouds, which block visible light. Nonetheless, astronomers have been able to estimate the location and size of the galaxy's central region through careful analysis of the behavior of stars and other objects orbiting around its center.
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what is the maximum speed with which a 1500 kg car can make a turn around a curve of a radius 20 m. on a level. road without sliding.
The maximum speed with which a 1500 kg car can make a turn around a curve of a radius 20 m. on a level road without sliding is approximately 0.823 m/s.
The maximum speed with which a 1500 kg car can make a turn around a curve of a radius 20 m. on a level road without sliding is determined by the maximum static friction force, Fsmax, that can be created between the car tires and the road surface. The formula for the static friction force is Fsmax = μsFn, where μs is the coefficient of static friction and Fn is the normal force acting on the car. The normal force can be calculated using Newton's Second Law, Fn = mv2/r, where m is the mass of the car, v is the speed, and r is the radius of the curve.
Therefore, the maximum speed with which the car can make the turn without sliding can be calculated as follows:
v2 = (μsm2)/(rFn) = (μs15002)/(20 * 1500 * 9.8) = 0.68 m/s2
v = √(0.68 m/s2) = 0.823 m/s
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a portable cd player uses a current of 7.5 ma at a potential diference of 3.5 v. how much energy does the player use in 35 s?
A portable CD player uses 7.5mA of current at a potential difference of 3.5V. Since it is running for 35 seconds, the total energy consumed in that time is calculated by the product of potential difference, current and time consumed and it is solved as 918.75mJ.
The amount of energy used by the portable CD player can be calculated using the formula:
E = VIt
where E is the energy, V is the potential difference, I is the current and t is the time.
The portable CD player uses a current of 7.5 mA at a potential difference of 3.5 V.
Thus, the energy used by the player in 35 seconds can be calculated as follows:
[tex]E = VIt\\ = 3.5 V \times 7.5 mA \times35 s \\= 918.75 mJ[/tex]
Therefore, the portable CD player uses 918.75 mJ of energy in 35 seconds.
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2. how would an atmosphere of helium effect the pitch of an organ pipe? how would it effect the pitch of a turning fork? explain. g
An atmosphere of helium would have a large effect on the pitch of instruments. For an organ pipe, helium would cause the pitch of the instrument's sound to be higher than normal.
This is because helium is a lighter gas than air, and therefore has a higher speed of sound through it. This increases the frequency of the sound waves, resulting in a higher pitch. For a tuning fork, helium would also cause the pitch of the sound to be higher than normal, however, the effect would be less significant.
This is because the frequency of the sound waves produced by a tuning fork is determined more by the size and shape of the fork, rather than the gas that it is in. Therefore, while the pitch of a tuning fork would be higher in a helium atmosphere, the effect would not be as great as with an organ pipe.
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suppose you have an atwood machine with two different masses m and m. what are the external forces acting on this system?
The external forces acting on this system are: gravity and the tension in the string.
An Atwood machine is a system consisting of two masses, m, and m, connected by a string that passes over a pulley. In this system, the external forces are gravity and the tension in the string. Gravity pulls both masses downward, while the tension in the string acts in opposite directions on the two masses, pulling the heavier one down and the lighter one up.
The tension in the string is determined by the masses m and m and the acceleration of the system. If m is the heavier mass and m is the lighter mass, the tension in the string will be greater than if both masses had the same weight. This is because the tension must balance the gravitational forces on the two masses. The greater the mass, the greater the gravitational force, and the greater the tension in the string must be to balance it.
The acceleration of the system is determined by the masses, the tension in the string, and the amount of friction in the system. The greater the tension, the greater the acceleration, and the smaller the mass, the greater the acceleration. Friction acts against the acceleration, reducing the net acceleration of the system.
In summary, the external forces acting on an Atwood machine with two different masses m and m are gravity and the tension in the string. The tension in the string is determined by the masses and the acceleration of the system, while the acceleration is determined by the masses, the tension in the string, and the amount of friction in the system.
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explain why adding charge to a bee causes hairs on the bee's body to bend more in response to a field.
Insects use their hair-like mechanoreceptors to detect electrical fields around them. This electrical charge causes the hairs to move and bend towards the source of the field.
Explanation:
Bee hair is also sensitive to static electricity. As bees collect pollen, they become negatively charged, causing the pollen to become positively charged. This charge difference makes it easier for pollen to stick to the bees' hairy bodies and legs.
Therefore, when a bee is charged with static electricity, it becomes easier for its hairs to detect the electrical fields around it. As a result, the hairs on the bee's body will bend more in response to the electrical field.
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Which are true about Halogens? (please choose all that apply)
a) Reactive
b) Group 7A
c) Non-metals
d) Metals
e) Non-reactive
Answer:
the answer is a number
Explanation:
support me
ay telescope mirrors are arranged in a way very similar to optical telescope mirrors. group of answer choices true false
Telescope mirrors are arranged in a way very similar to optical telescope mirrors is a True statement.
The Telescopes and optical telescope can also be used as
Telescopes: Telescopes, including optical telescopes, use mirrors or lenses to collect and focus light. Optical telescope: Optical telescope mirrors are usually curved in shape to reflect light from distant objects and form an image at the focal point.Similarly, other types of telescopes such as radio telescopes, infrared telescopes, and ultraviolet telescopes also use mirrors to collect and focus the corresponding wavelengths of light they are designed to detect.
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which of the following is not an example of mechanical action? group of answer choices electrically charged beaters of the bells on the clavecin electrique digital sampling machines levers and hammers inside an acoustic piano rotors (tonewheels) of a telharmonium
Digital sampling machines are not an example of mechanical action.
What is mechanical action?
Mechanical action refers to the physical movement of mechanical components to produce a sound or perform a specific function. This can include a wide range of actions, such as the striking of hammers on strings in a piano, the rotation of tonewheels in a Hammond organ, or the movement of valves in a trumpet.
Mechanical action can also be found in other types of machinery and equipment, and tools such as engines, gears, and levers, where physical movement is used to perform a specific task or function.
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while a car travels around a circular track at a constant speed, its a) acceleration is non-zero and along the path b) acceleration is non zero and inward toward the center c) acceleration is zero d) acceleration is non-zero and outward from the center
While a car travels around a circular track at a constant speed, its (d) acceleration is non-zero and outward from the center.
When a car travels around a circular track at a constant speed, it is constantly changing direction, and therefore, constantly accelerating. This acceleration is known as centripetal acceleration and is directed towards the center of the circle.
However, according to Newton's third law, every action has an equal and opposite reaction. In this case, the car also experiences an equal and opposite acceleration, known as the centrifugal acceleration, which is directed outward from the center of the circle.
This is the non-zero acceleration experienced by the car, and it acts to counterbalance the centripetal acceleration, keeping the car moving in a circular path.
Therefore, the correct answer is (d) acceleration is non-zero and outward from the center.
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true or false: electrically neutral objects have no positive or negative charges, that is why they are neutral
True, electrically neutral objects have no positive or negative charges, that is why they are neutral.
Electrically neutral objects have an equal number of protons and electrons, so the total charge on the object is zero. This is why we refer to the object as being ‘neutral’. It has no charge.
In a chemical reaction, protons are transferred between particles, such as atoms or ions. When an atom loses or gains protons, its charge changes, either becoming more negative or more positive.
An atom which has the same number of protons and electrons has a net charge of zero and is neutral.
In an electric field, a neutral object experiences no force due to the absence of electric charge on its surface. When placed in a magnetic field, however, a neutral object still experiences a force due to its electrons.
Electrons have a magnetic moment, so they are affected by the magnetic field, causing the neutral object to move.
Electrically neutral objects have no positive or negative charges, that is why they are neutral.
Their charge is zero, so they are not affected by electric fields, but still experience a force in a magnetic field due to the magnetic moment of their electrons.
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Patients with anterograde amnesia were taught to solve the Tower of Hanoi problem. It was later found that they _________.a. remembered solving the problem and could do so againb. didn't remember the problem and couldn't solve itc. remembered solving the problem but couldn't do it againd. didn't remember solving the problem but could do it again
Patients with anterograde amnesia were taught to solve the Tower of Hanoi problem. It was later found that they "didn't remember solving the problem but could do it again" which is option d.
Anterograde amnesia is a type of amnesia that affects the ability to generate new memories following the onset of the condition. This condition is generally caused by damage to the hippocampus or adjacent structures, and it typically affects a person's ability to learn and remember new information.
Therefore, patients with anterograde amnesia are unable to form new memories and often rely on implicit memory systems to perform specific tasks over time.
Thus the correct answer is option d.
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