Answer:amplitude
Explanation:
When a single charge q is placed on one corner of a square, the electric field at the center of the square is F/q. If three other equal charges are placed on the other corners, the electric field at the center of the square due to these four equal charges is
a) F/(2q)
b) F/(4q)
c) 4F/q
d) F/q
e) zero
Please explain how you came to the correct answer.
The answer to the question is option (c) 4F/q.
The electric field due to a single charge q at the center of a square is given by
E = F/qWhere F = 1/4πε₀ * q / r²,
where r is the distance between the charge and the center of the square. The electric field of the single charge q has a magnitude of F/q when it is at the center of the square.
If three other charges, each of the same magnitude, are placed on the other corners of the square, the resultant electric field at the center of the square is the vector sum of the electric fields due to the four charges.Let the distance between the charges and the center of the square be a.
The force due to each charge is given by
F' = 1/4πε₀ * q / a²
The electric field due to each charge at the center of the square is given by
E' = F'/q = 1/4πε₀ * q / a²q.
The electric field at the center of the square due to these four charges is the vector sum of the electric fields due to the four charges. Since the charges are placed at the corners of a square and are equidistant from the center, the angle between any two fields is 90°.
Hence, the resultant electric field is given by
E = 2E' sin 45° + 2E' sin 135°= 2E' /√2 + 2E' /-√2= 4E' /√2= 4 (1/4πε₀ * q / a²q) /√2= 4F /q.
Hence, the option (c) is the correct answer.
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What are advantages of charge-coupled devices (CCDs) over photographic film? Choose all that apply. a. The large CCD cameras that astronomers use can collect almost all the photons of incoming light that strike the chip, which greatly reduces exposure times. b. CCD chips can be designed to detect wavelengths of infrared and ultraviolet light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light. c. CCD chips can be designed to block out infrared and ultraviolet light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light. d.The large CCD cameras that astronomers use can collect almost all the photons of incoming light that strike the chip, which greatly increases exposure times. e. CCD chips can be designed to detect wavelengths of X-ray and microwave light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light.
Charge-coupled devices (CCDs) are electronic devices used for detecting electromagnetic radiation (light) and converting it into an electrical signal. The advantages of CCDs over photographic film include the following:
(a) The large CCD cameras that astronomers use can collect almost all the photons of incoming light that strike the chip, which greatly reduces exposure times. This option a is correct.
(b) CCD chips can be designed to detect wavelengths of infrared and ultraviolet light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light is also correct. Thus option b is correct. Additionally, option d is incorrect because large CCD cameras can't increase exposure times but instead, reduce them. Option e is incorrect because CCD chips are not designed to detect wavelengths of X-ray and microwave light but instead for infrared and ultraviolet light as stated in option b. Thus, the options that are correct include options a and b.
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1. A boy lifts a 5.1-kg block vertically 6.0 m at constant speed. The work done (in Joules) by the boy is Round of your answer to 1 decimal place. Do not include the unit.
2. A 54.2 kg diver jumps from a height of 2.2 m with an initial speed of 2.5 m/s. What is his speed (in m/s) entering the water?
(1)A boy lifts a 5.1-kg block vertically 6.0 m at constant speed the work done by the boy is approximately 299.9 J.(2) the speed of the diver entering the water is approximately 2.5 m/s.
1): The work done by the boy can be calculated using the formula:
Work = Force × Distance
Since the block is lifted vertically at a constant speed, the force applied by the boy must be equal to the weight of the block.
Weight = mass × acceleration due to gravity
Weight = 5.1 kg × 9.8 m/s^2 (acceleration due to gravity)
Weight ≈ 49.98 N
Therefore, the work done by the boy is:
Work = Force × Distance
Work = 49.98 N × 6.0 m
Work ≈ 299.9 Joules
Rounded to 1 decimal place, the work done by the boy is approximately 299.9 J.
(2) To find the speed of the diver entering the water, we can use the principle of conservation of energy. The initial potential energy of the diver at the top of the dive can be converted into kinetic energy just before entering the water.
The potential energy at the top of the dive is given by:
Potential energy = mass × gravity × height
Potential energy = 54.2 kg × 9.8 m/s^2 × 2.2 m
Potential energy ≈ 1198.36 J
The initial kinetic energy just before entering the water can be calculated as:
Kinetic energy = 0.5 × mass × velocity^2
Kinetic energy = 0.5 × 54.2 kg × (2.5 m/s)^2
Kinetic energy ≈ 169.75 J
According to the conservation of energy, the potential energy at the top should be equal to the kinetic energy just before entering the water. Therefore:
Potential energy = Kinetic energy
1198.36 J = 169.75 J
To find the speed (velocity) of the diver, we can rearrange the equation:
Velocity^2 = (2 × Kinetic energy) / mass
Velocity^2 = (2 × 169.75 J) / 54.2 kg
Velocity^2 ≈ 6.254
Taking the square root of both sides, we find:
Velocity ≈ 2.5 m/s
Therefore, the speed of the diver entering the water is approximately 2.5 m/s.
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A girl on a swing may increase the amplitude of the swing's oscillations if she moves her legs at the natural frequency of the swing. This is an example of: ________
A girl on a swing may increase the amplitude of the swing's oscillations if she moves her legs at the natural frequency of the swing. This is an example of: This phenomenon is an example of resonance.
Resonance occurs when an external force is applied to a system at its natural frequency, resulting in a significant increase in the system's amplitude. In the case of the girl on a swing, the natural frequency of the swing is determined by its length and the force of gravity. When the girl moves her legs at the natural frequency of the swing, she applies periodic impulses to the swing, synchronizing her motion with the natural oscillations of the swing. As a result, the amplitude of the swing's oscillations increases. This happens because the energy transferred to the swing with each leg movement is added constructively to the existing oscillations, leading to a cumulative effect. Resonance can be observed in various systems, from musical instruments to bridges, and it is often desirable in specific applications. Understanding the concept of resonance allows us to manipulate and control systems by applying forces at their natural frequencies to achieve desired outcomes.
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Open and Closed circuit similarities
Answer:
An open circuit is one where the continuity has been broken by an interruption in the path for current to flow. A closed circuit is one that is complete, with good continuity throughout
a sim,ple elctrical ciurucuiot contains a battery a light bulb and a properly copnnected ammeter the ammeter has a very low internal resistance because it is connected in
In a simple electrical circuit, if it contains a battery, a light bulb, and a properly connected ammeter, the ammeter has a very low internal resistance because it is connected in series with the circuit.
An electrical circuit is made up of a combination of resistors, voltage sources, and current sources that are interconnected in a closed loop. It is used to generate an electric current in a complete circuit and can be as straightforward as a battery connected to a bulb or as complicated as a full-scale electronic circuit.
Ammeters are measuring devices that are used to measure current in a circuit. The ammeter should be connected in series with the circuit to allow current to flow through it. An ammeter with a very low internal resistance should be used since any extra resistance in the ammeter can change the current being measured.
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Using Newton's revision of Kepler's third law, calculate the mass (in solar masses) of a star where an Earth-like planet orbits it with a semi-major axis of 2 AU and a period of 1.73 Earth-years. Recall that for an Earth-like planet, its mass is negligible compared to that of the star.
Kepler's third law states that the square of the period of a planet's orbit around the Sun is proportional to the cube of the semi-major axis of its orbit. Newton's revision of Kepler's third law states that the sum of the masses of two objects in orbit around each other is proportional to the cube of the semi-major axis of their orbit and inversely proportional to the square of their orbital period.
Using Newton's revision of Kepler's third law, the mass of a star can be calculated as follows:
1. Convert the semi-major axis of the Earth-like planet's orbit from AU to meters.1 AU = 149,597,870,700 meters2 AU = 2 × 149,597,870,700 meters = 299,195,741,400 meters.
2. Convert the period of the Earth-like planet's orbit from Earth years to seconds.1 year = 31,557,600 seconds1.73 Earth-years = 1.73 × 31,557,600 seconds = 54,592,128 seconds
3. Substitute the values into Newton's revision of Kepler's third law and solve for the mass of the star.(m1 + m2) = (4π²a³)/(G T²), where m1 is the mass of the star, m2 is the mass of the Earth-like planet (which is negligible), a is the semi-major axis of the planet's orbit, T is the period of the planet's orbit, and G is the gravitational constant. G = 6.674 × 10⁻¹¹ N m²/kg²(m1 + 0) = (4π² × 299,195,741,400³)/(6.674 × 10⁻¹¹ × 54,592,128²)(m1 + 0) = 2.476 × 10³⁰ kg.
The mass of the star is 2.476 × 10³⁰ kg, which is approximately 1.24 solar masses since the mass of the Sun is 1.99 × 10³⁰ kg.
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Question 4 (1 point) Diffraction gratings provide much brighter interference patterns since more light passes through them compared with double slits. O True O False Question 5 (1 point) A When the th
False statement regarding Diffraction and True statement regarding reflected light
4) False. **Diffraction gratings** do not provide much brighter interference patterns compared to double slits. In fact, more light passes through double slits than through diffraction gratings. Diffraction gratings consist of multiple closely spaced slits that diffract and spread out the light, resulting in individual interference maxima and minima that are less intense. On the other hand, double slits allow more light to pass through, resulting in brighter interference patterns.
5) True. When the thickness of a film in air is such that reflected light undergoes destructive interference, the statement is true. Destructive interference occurs when the path length difference between the two reflected rays is an odd multiple of half the wavelength of light. This leads to the cancellation of certain wavelengths of light, resulting in reduced or no reflected light. Therefore, if the film thickness satisfies the condition for destructive interference, the reflected light will be significantly diminished.
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rubbing two sticks together can make a fire. in the right circumstances, radioactive atoms can create an explosion. what action causes both of these two different events?
Both the process of rubbing two sticks together to cause a fire and the explosion caused by radioactive atoms involve the release of energy.
The friction between the two matchsticks generates heat, which can ignite the surfaces of the matchsticks and cause combustion and a fire. In this example the energy released appears as heat and light. Nuclear decay, which occurs in the case of radioactive atoms, is a process in which unstable atomic nuclei undergo spontaneous changes and release energy in the form of radiation. A chain reaction that results in a sudden release of energy and an explosion may be triggered under specific circumstances, as in a nuclear reactor or nuclear weapon.
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2. (6 points] The energy levels for the hydrogen atom are given by the formula E-136V, At what temperature would a gas of hydrogen have, on average, one atom in the first excited state for every milli
The energy levels for the hydrogen atom are given by the formula E-136V. The temperature that a gas of hydrogen have, on average, one atom in the first excited state for every million atoms in the ground state is 121,600 K.
The formula for energy levels for the hydrogen atom is given by: E=-13.6/n²
where,
n is the principal quantum number of the energy level
Here, the first excited state is n = 2 and the ground state is n = 1.
Therefore, the energy difference between the first excited state and the ground state can be calculated by:
E(2) - E(1) = (-13.6 / 2²) - (-13.6 / 1²) = -3.4 eV
The probability that an atom will be in the first excited state is proportional to the Boltzmann factor:
exp(-ΔE/kT) = 10⁶ / 1
We know, ΔE = 3.4 eV.
k is the Boltzmann constant which is equal to 8.617×10⁻⁵ eV/K.
Let's solve for the temperature, T:
T = ΔE / (k ln(10⁶))= (3.4 eV) / (8.617×10⁻⁵ eV/K ln(10⁶))= 121,600 K
Therefore, at a temperature of 121,600 K, a gas of hydrogen would have, on average, one atom in the first excited state for every million atoms in the ground state.
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what is the angular distance (in degrees) from the north celestial pole to the point on the sky called the summer solstice?
The angular distance from the North Celestial Pole to the point on the sky known as the summer solstice is approximately 23.5 degrees.
The summer solstice marks the point in the Earth's orbit around the Sun when the Northern Hemisphere experiences its longest day and shortest night. During this time, the North Pole is tilted towards the Sun at its maximum angle of 23.5 degrees. As a result, the Sun appears to reach its highest point in the sky for observers in the Northern Hemisphere. The North Celestial Pole, also known as the North Star or Polaris, is the point in the sky directly above the Earth's North Pole. It serves as a fixed reference point for celestial navigation. The summer solstice occurs when the Sun's declination reaches its maximum positive value of +23.5 degrees. This angle represents the tilt of the Earth's axis in relation to its orbit around the Sun. Therefore, the angular distance from the North Celestial Pole to the summer solstice point is approximately 23.5 degrees.
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A vertical straight wire 35.0 cm in length carries a current. You do not know either the magnitude of the current or whether the current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0350 T that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 N. Find the magnitude of the current. Express your answer with the appropriate units. HÅ B ? I= Value Units Submit Request Answer Part D Find the direction of the current. The current is traveling horizontally leftward. O The current is traveling vertically upward. The current is traveling horizontally rightward. The current is traveling vertically downward. Submit Request Answer
A vertical straight wire 35.0 cm in length carries a current. You do not know either the magnitude of the current or whether the current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0350 T that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 N. The magnitude of the current in the wire is approximately 0.073 A.The direction of the current is vertically downward.
To find the magnitude of the current in the wire, we can use the formula for the magnetic force on a current-carrying wire in a magnetic field:
F = B * I * L * sin(theta)
where F is the magnetic force, B is the magnetic field strength, I is the current, L is the length of the wire, and theta is the angle between the wire and the magnetic field.
Given:
B = 0.0350 T (magnetic field strength)
L = 35.0 cm = 0.35 m (length of the wire)
F = 0.0180 N (magnetic force)
Rearranging the formula, we can solve for the current I:
I = F / (B * L * sin(theta))
Since the wire experiences a horizontal magnetic force to the west, the angle theta between the wire and the magnetic field is 90 degrees (perpendicular).
I = 0.0180 N / (0.0350 T * 0.35 m * sin(90°))
Using the given values and evaluating the expression:
I ≈ 0.073 A (amperes)
Therefore, the magnitude of the current in the wire is approximately 0.073 A.
To determine the direction of the current, we can use the right-hand rule. When the magnetic force is directed to the west and the magnetic field is pointing due north, the current in the wire must be traveling vertically downward.
Thus, the direction of the current is vertically downward.
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A toy boat made of solid wood has a mass of 4.78 kgkg and density of 0.5 g/cm3g/cm3 . It is placed in water. Some steel weights are placed on top of it so it floats with 10 %% of its volume above the surface. Find the mass of steel placed on top of the boat
Therefore, the mass of steel placed on top of the boat is 0.000956 kg or 0.956g
To solve this problem, we need to determine the volume of the toy boat and the volume of water it displaces when floating.
Since the boat floats with 10% of its volume above the water surface, we can calculate the total volume of the boat using the given density.
Density is defined as mass divided by volume.
Rearranging the equation, we can find the volume of the boat:
Volume of the boat = Mass of the boat / Density of the boat
Volume of the boat = 4.78 kg / (0.5 g/cm³ * 1000 cm³/g) [Converting the density from g/cm³ to kg/cm³]
Volume of the boat = 4.78 kg / 0.5 kg/cm³
Volume of the boat = 9.56 cm³
Now, since 10% of the boat's volume is above the water surface, we can calculate the volume of water it displaces:
Volume of water displaced = 10% * Volume of the boat
Volume of water displaced = 10% * 9.56 cm³
Volume of water displaced = 0.1 * 9.56 cm³
Volume of water displaced = 0.956 cm³
To find the mass of the steel placed on top of the boat, we need to consider that the density of water is 1 g/cm³.
Mass of steel = Density of water * Volume of water displaced
Mass of steel = 1 g/cm³ * 0.956 cm³
Mass of steel = 0.956 g
Since we want the mass in kilograms, we convert grams to kilograms:
Mass of steel = 0.956 g / 1000 g/kg
Mass of steel = 0.000956 kg
Therefore, the mass of steel placed on top of the boat is 0.000956 kg or .0956g (rounded to three decimal places).
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DRAW
the figure and show complete solution on each problem. please
answer all
IV. Find the momentum of a 60-g bullet whose kinetic energy is 270 J. V. A 60-ton car moving at 1.2 mile per hour is instantaneously coupled to a stationary 40-ton car. What is the speed of the couple
The momentum of the car is 5.7 Kgm/s
The final velocity is 0.3 m/s
What is the momentum?
Momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object's mass and its velocity.
p = m * v
Where;
KE = 1/2mv^2
v = √2KE/M
v = √2 * 270/0.06
v = 95 m/s
Then;
p = mv
p = 0.06 * 95 m/s
= 5.7 Kgm/s
V = 1.2 m/hr or 0.5 m/s
mass = 60 ton and 40 ton or 54431.1 Kg and 36287.4 Kg
Momentum before collision = Momentum after collision
(54431.1 * 0.5 m/s) + 0 = (54431.1 + 36287.4 )v
v = 27215.55/90718.5
v = 0.3 m/s
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a rock hits the ground at a speed of 15 m/s and leaves a hold 50 cm deep. after it hits the ground, what is the magnitude of the rock's (assumed) uniform acceleration?
The magnitude of the rock's (assumed) uniform acceleration is v² - 225.
Initial speed, u = 15 m/s
Displacement, s = 50 cm = 0.5 m
Magnitude of acceleration, a = ?
We know, v² - u² = 2as
Let's substitute the given values into the above formula. v² - u² = 2as (v is the final velocity)
Final velocity, v = ?u = 15 m/s (Initial velocity)
s = 0.5 m (Displacement)
a = ?
v² - u² = 2as (v² - u²)/2s = a(v+u)/2(a = (v² - u²)/2s)
(a = (v² - u²)/2s)(a = (v² - (15 m/s)²)/2(0.5 m))(a = (v² - 225)/1)(a = v² - 225)
Therefore, the magnitude of the rock's uniform acceleration is v² - 225, given that a rock hits the ground at a speed of 15 m/s and leaves a hold 50 cm deep after it hits the ground.
The magnitude of the rock's uniform acceleration is v² - 225.
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Find the center of mass of the following particles (drawn large so they can be seen)
It is possible to suspend an object at rest from its center of gravity, and gravity won't induce it to begin rotating no matter how it is positioned.
Thus, The center of gravity of an object will be located somewhere along a vertical line that passes through the point of suspension if you suspend it from any point and allow it to come to rest.
The gravitational acceleration is (almost) constant near the surface of the earth, where the center of mass also lies.
A place that represents the average or typical location of an object's mass, as if all of the mass were contained there, is known as the center of mass (CM). The geometric center of a uniform sphere serves as its center of mass. The barycenter is another name for the CM.
Thus, It is possible to suspend an object at rest from its center of gravity, and gravity won't induce it to begin rotating no matter how it is positioned.
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you are driving a vehicle at 55 mph on dry pavement, about how much total stopping distance do you need to bring it to a stop?
When driving at 55 mph on dry pavement, you'll need approximately 420 feet of total stopping distance to bring your vehicle to a stop.
The total stopping distance that you need to bring a vehicle to a stop when driving at 55 mph on dry pavement depends on many variables, including the following:
Reaction time, vehicle weight, braking system efficiency, tires, road conditions, and so on.
According to a generally accepted formula, the total stopping distance can be calculated as follows:
TSD = (DR + BD) + (RD × DD)Where:TSD = Total Stopping DistanceDR = Distance Traveled during Driver Reaction TimeBD = Braking Distance (Distance required to stop a vehicle, from the point where brakes are applied)RD = Rolling Distance (Distance traveled by the vehicle while its brakes are in operation)
DD = Distance Lost Due to Perception Time or Delay Time
For a vehicle traveling at 55 mph on dry pavement, the driver's reaction time is around 1.5 seconds (according to some sources).
DR is estimated to be approximately 203 feet (60 m), which is equivalent to 1.5 seconds of time at 55 mph. If the brakes on the vehicle are well-maintained, the BD for a typical car is around 216 feet (66 m).
Rolling distance is determined by the friction between the tires and the road surface and is determined by the tire tread and type, as well as the road surface texture.
It ranges from 155 to 175 feet (47 to 53 m) on dry pavement in most cases. If we use the upper limit of 175 feet for rolling distance, the total stopping distance can be calculated as follows:TSD = (DR + BD) + (RD × DD)TSD = (203 + 216) + (175 × 1.5)TSD = 419.5 feet
Therefore, when driving at 55 mph on dry pavement, you'll need approximately 420 feet of total stopping distance to bring your vehicle to a stop.
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you go outside on a hot, sunny summer day and you feel the warm sun on your skin. what causes this warmth?
The warmth that you feel on your skin when you go outside on a hot and sunny summer day is caused by the heat of the sun.
The sun is the ultimate source of energy and it emits heat in the form of electromagnetic radiation. The sun's heat is transferred to the earth through radiation, conduction, and convection. Radiation is the process by which the sun emits heat through space and it travels to the earth in the form of electromagnetic waves.
Conduction is the process by which heat is transferred from one object to another through direct contact, such as the ground warming up when the sun's rays hit it. Convection is the process by which heat is transferred through the movement of fluids, such as the heating of air that rises and causes cooler air to flow in and replace it.
In conclusion, the warmth that you feel on your skin on a hot and sunny summer day is caused by the heat of the sun, which is transferred to the earth through radiation, conduction, and convection.
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car mass 1000kg is travelling along a straight horizontal road at a speed of 20m/s when it brakes sharply then skids. Friction brings the car to
,rest. If the friction force between the tires and road is 9000N. Calculate the distance travelled by car befor it comes to rest
A car mass 1000kg is traveling along a straight horizontal road at a speed of 20m/s when it brakes sharply then skids. Friction brings the car to,rest. If the friction force between the tires and road is 9000N. The distance traveled by the car before it comes to rest (while skidding due to braking) is approximately 22.22 meters.
To calculate the distance traveled by the car before it comes to rest, we can use the equations of motion.
First, we need to find the acceleration of the car when it brakes sharply. The friction force acting on the car is equal to the product of the mass of the car and its acceleration:
Friction force = mass × acceleration
9000 N = 1000 kg × acceleration
acceleration = 9000 N / 1000 kg
acceleration = 9 m/s^2
Next, we can use the equation of motion that relates initial velocity, final velocity, acceleration, and distance:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car comes to rest)u = initial velocity (20 m/s)a = acceleration (-9 m/s^2, negative because it acts in the opposite direction to the car's motion)s = distance traveledPlugging in the values, we get:
0^2 = 20^2 + 2(-9)s
0 = 400 - 18s
18s = 400
s = 400 / 18
s ≈ 22.22 m
Therefore, the distance traveled by the car before it comes to rest (while skidding due to braking) is approximately 22.22 meters.
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Consider a car speeding up as it drives along a level road. what is an action-reaction pair (from newton’s third law)?
Newton's third law of motion states that for every action, there is an equal and opposite reaction. The action-reaction pair is used to describe the interaction between two objects. Therefore, when a car is speeding up as it drives along a level road, an action-reaction pair occurs.
An action-reaction pair is a pair of forces that are equal in strength and opposite in direction. When an object exerts a force on another object, the second object exerts an equal and opposite force on the first object. The action-reaction pair when a car speeds up as it drives along a level road can be explained as follows: Action force: The car exerts a force on the road in the forward direction. This is the action force. Reaction force: The road exerts a force on the car in the backward direction. This is the reaction force.
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A tank contains 100 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 13 L/min.
Let y be the number of kg of salt in the tank after t minutes.
The differential equation for this situation would be:
dy
y(0) =
A tank contains 60 kg of salt and 1000 L of water. A solution of a concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min. The solution is mixed and drains from the tank at the same rate.
Let y be the number of kg of salt in the tank after t minutes.
Write the differential equation for this situation
dy =
y(0) = 60
y' + ty^1/3 = tan(t), y(3) = – 5
a) Rewrite the differential equation, if necessary, to obtain the form y' = f(t, y)
F(t, x) = _______
b) Compute the partial derivative of f with respect to y. Determine where in the ty-plane both f(t, y) and its derivative are continuous.
c) Find the largest open rectangle in the ty-plane on which the solution of the initial value problem above is certain to exist for the initial condition. (Enter oo for infinity)
t interval is
y interval is
The interval of y is (-5,∞), as the solution exists until the concentration of salt reaches zero. The solution to the differential equation is given by,dy/dt = rate in - rate out. There is initially 100 kg of salt and 2000 L of water in the tank.
This means that the initial concentration of salt in the tank is:
100 kg / (1000 L + 2000 L)
= 0.03333 kg/L
As pure water is added to the tank at the rate of 12 L/min, and the mixed solution is drained out of the tank at the rate of 13 L/min. The volume of the solution in the tank after t minutes is given by,
(2000 + 12t) - 13t = 2000 - t.
The concentration of salt in the tank after t minutes is given by,
y = (100 - t)(0.03333)
The differential equation for this situation would be,
dy/dt = (0.03)(9) - (y/2000)(13)dy/dt + (13/2000)y
= 0.27
Rewrite the differential equation, if necessary, to obtain the form
y' = f(t, y):
dy/dt + (13/2000)y
= 0.27 - y(0)
= 60
The given differential equation is,
y' + ty^(1/3) = tan(t),
y(3) = -5a)
Rewriting the differential equation, we gety' = -ty^(1/3) + tan(t) - y...[1]b) The partial derivative of f with respect to y is,
df/dy = (1/3)ty^(-2/3) - 1
f and its derivative are continuous everywhere except for y = 0.
c) The largest open rectangle in the ty-plane on which the solution of the initial value problem above is certain to exist for the initial condition is obtained as follows:
The interval of t is (3,∞), as the initial condition is given at t = 3.
The interval of y is (-5,∞), as the solution exists until the concentration of salt reaches zero.
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VP 28.2.2 ▼ Part E A proton (charge +1.60 × 10-¹⁹ C) and an electron (charge -1.60 × 10-¹⁹ C) are both moving in the xy-plane with the same speed, 4.20 × 105 m/s. The proton is moving in the +y-direction along the line x = 0, and the electron is moving in the -y-direction along the line x = +4.00 mm. At the instant when the proton and electron are at their closest approach, what is the magnitude of the magnetic force that the proton exerts on the electron? Express your answer with the appropriate units. μA ? Fon e- = Value Units
a) The distance of closest approach is2.00 mm
b) The magnitude of the magnetic force that the proton exerts on the electron: 1.60 × 10⁻¹⁹ N
a) To find the distance of closest approach between the proton and electron, we need to determine the y-coordinate of the electron when it is at x = 0.
Given that the electron is moving along the line x = +4.00 mm, at the instant of closest approach, the y-coordinate of the electron will be equal to the negative of its x-coordinate.
Therefore, the distance of closest approach is 4.00 mm or 2.00 mm (taking the absolute value).
b) The magnetic force between two moving charges can be calculated using the formula
F = (|q₁| * |q₂| * v * B) / r,
where F is the force,
|q₁| and |q₂| are the magnitudes of the charges,
v is the velocity of the charge,
B is the magnetic field, and
r is the distance between the charges.
In this case, the proton and electron have equal magnitudes of charge |q₁| = |q₂| = 1.60 × 10⁻¹⁹ C, and they are moving with the same speed v = 4.20 × 10⁵ m/s. The magnetic force depends on the magnetic field B, which is not given in the question.
Therefore, we cannot calculate the exact magnitude of the magnetic force without knowing the value of the magnetic field.
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in an electrical charge, some charges would repel one another. which is an example of electrical charges that would repel one another?
In an electrical charge, like charges would repel one another.
An example of electrical charges that would repel one another is two positive charges or two negative charges. Like charges have the same polarity, that is, they have the same charge, and they would repel one another. It's essential to understand the concept of electrical charges in order to understand this concept.
Electric charge is a fundamental property of matter, and it can exist in two types: positive and negative. Like charges repel one another, whereas opposite charges attract one another. The reason why charges attract or repel is that they produce electric fields that interact with other charges.
It is important to note that the magnitude of the electric field is inversely proportional to the square of the distance between two charges.
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A heat pump defrost control must do three things: turn on auxiliary heat, turn off the condenser fan motor, and:
Answer:
Shift the reversing valve from heat to cool.
A jumbo jet of mass 4 x 10² kg travelling at a speed 5000 m/s lands on the airport. It takes 2 minutes t come to rest. Calculate the average force applied by th ground on the aeroplane. 12 Ans: -1.67x107
Answer:
I got 16680N
Explanation:
We can use this formula; F = ma
F = ma
F = (4 × 10²) × ((0-5000)/120)
F = (4 × 10²) × (-41.7)
F = -16680N
Therefore, Fₐᵥ = 16680N.
I n it in 20 minutes
JL.61 A producer of refrigerator compressors wants to implement
a just-in-time production line to support demand from a neighboring
appliance manufacturer. Demand from the applian
The producer of refrigerator compressors needs to determine the number of kanbans required for implementing a just-in-time production line to meet the demand for 150 compressors per day from a neighboring appliance manufacturer.
To calculate the number of kanbans required, we need to consider the production lead time, safety stock factor, and optimal production quantity. The production lead time is 5 days, which means that it takes 5 days to produce a batch of compressors once the production process starts.
The safety stock factor is 17%, indicating that the producer wants to maintain an additional 17% of the daily demand as safety stock to mitigate any unforeseen fluctuations. The optimal production quantity is 95 units, which is the batch size that minimizes setup costs.
To determine the number of kanbans, we first need to calculate the total demand during the production lead time. Since the demand is 150 compressors per day and the lead time is 5 days, the total demand during the lead time is 150 compressors/day * 5 days = 750 compressors. Adding the safety stock, the total demand becomes 750 compressors + (17% * 150 compressors) = 750 compressors + 25.5 compressors = 775.5 compressors.
Next, we divide the total demand by the optimal production quantity to get the number of kanbans required. The number of kanbans is calculated as 775.5 compressors / 95 compressors per kanban = 8.16 kanbans. Since we cannot have a fraction of a kanban, we round up to the nearest whole number. Therefore, the producer of compressors requires 9 kanbans to meet the demand of 150 compressors per day from the neighboring appliance manufacturer.
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The complete question is:
A producer of refrigerator compressors wants to implement a just-in-time production line to support demand from a neighboring appliance manufacturer. Demand from the appliance manufacturer is for 150 compressors a day. The production lead time is 5 days and the producer wants to have a 17% safety stock factor. This producer has also cut setup costs such that the optimal production quantity is 95 units. How many kanbans does this producer of compressors require?
Two vectors are given by a = 6.7î + 5.3ĵ and b = 2.6î + 3.9ĵ. Find (a)|à × b¦.(b) à · b‚(c)(a + b). b, and (d) the component of a along the direction of b? (a) Number i Units (b) Number i U
Two vectors are given by a = 6.7î + 5.3ĵ and b = 2.6î + 3.9ĵ(a) |a × b| ≈ 18.243(b)a · b = 38.09(c) (a + b) × b = 12.35(d)the component of a along the direction of b is approximately 8.319.
Let's solve each part of the question step by step:
(a) To find the magnitude of the cross product between vectors a and b, we can use the formula:
|a × b| = |a| |b| sin(θ)
where |a| represents the magnitude of vector a, |b| represents the magnitude of vector b, and θ represents the angle between the two vectors.
In this case, |a| = √(6.7^2 + 5.3^2) ≈ 8.502, and |b| = √(2.6^2 + 3.9^2) ≈ 4.826.
To find the angle θ, we can use the dot product:
a · b = |a| |b| cos(θ)
Plugging in the values, we have:
6.7 * 2.6 + 5.3 * 3.9 = 8.502 * 4.826 * cos(θ)
Simplifying the equation gives us:
θ ≈ arc cos((6.7 * 2.6 + 5.3 * 3.9) / (8.502 * 4.826)) ≈ 0.304 radians.
Now, we can find the magnitude of the cross product:
|a × b| = |a| |b| sin(θ) = 8.502 * 4.826 * sin(0.304) ≈ 18.243.
Therefore, |a × b| ≈ 18.243.
(b) To find the dot product of vectors a and b, we use the formula:
a · b = (6.7 * 2.6) + (5.3 * 3.9) = 17.42 + 20.67 = 38.09.
Therefore, a · b = 38.09.
(c) To find the cross product (a + b) × b, we need to first find the sum of vectors a and b:
(a + b) = (6.7 + 2.6)î + (5.3 + 3.9)ĵ
= 9.3î + 9.2ĵ.
Then we can calculate the cross product:
(a + b) × b = (9.3 * 3.9) - (9.2 * 2.6)
= 36.27 - 23.92
= 12.35.
Therefore, (a + b) × b = 12.35.
(d) To find the component of vector a along the direction of vector b, we can use the formula:
Component of a along b = |a| cos(θ)
where θ is the angle between vectors a and b.
We already calculated θ in part (a) to be approximately 0.304 radians. Therefore:
Component of a along b = |a| cos(θ) = 8.502 * cos(0.304) ≈ 8.319.
Therefore, the component of a along the direction of b is approximately 8.319.
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2. a car traveling at 27 m/s runs out of gas while traveling up a slope. if the car coasts 85 m up the slope before starting to roll back down, what is the angle of incline?
The angle of inclination is 25.92°. The gravitational force is equal and opposite to the force component parallel to the inclined surface on a block placed on an inclined surface.
We can find the angle of inclination by using this information. This can be illustrated with the following formula;
mg sin θ = f
Here, m = Mass of car = 1,000 kg
g = acceleration due to gravity = 9.8 ms⁻²
sin θ = Opposite / Hypotenuse
= Height / Length
f = Force component parallel to slope
= Weight × sin θ
= mg sin θ
We'll need to utilize the following data:
Initial velocity, u = 27 m/s
Displacement, S = 85 m
Acceleration, a = -9.8 m/s²
By using the kinematic equation of motion, the time it takes to travel up the slope can be calculated;
v² = u² + 2as,
Where, v = Final velocity = 0 m/s
u = Initial velocity = 27 m/s
a = Acceleration = -9.8 m/s²
s = Displacement = 85 m.
After calculating for t, we can use this time value to calculate the angle of inclination by utilizing the aforementioned formula.
mg sin θ = f
Here, m = 1,000 kg,
g = 9.8 ms⁻²,
f = mgsinθsinθ = f / mg.
Now, solve for f, f = ma
Therefore, f = 1,000 kg × 9.8 m/s² × sin θ
The angle of inclination can now be calculated:
sinθ = f / mg
= 1,000 kg × 9.8 m/s² × sin θ / 1,000 kg × 9.8 m/s²
= sin θ= (u² - v²) / 2as
= (27 m/s)² / 2(-9.8 m/s²)(85 m)
= 25.92°
Therefore, the angle of inclination is 25.92°.
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Which of these is NOT a form of physical weathering
a.
Thermal stress
b.
Abrasion
c.
Ice wedging
d.
Hydrolysis
Hydrolysis is NOT a form of physical weathering,
The form of physical weathering that is NOT included in the given options is hydrolysis. Physical weathering refers to the breakdown of rocks and minerals into smaller pieces without any change in their chemical composition.
It is caused by various physical processes that act on the rocks. The options provided in the question all represent different forms of physical weathering, except for hydrolysis. Thermal stress is a form of physical weathering that occurs when rocks expand and contract due to changes in temperature, causing them to crack and break apart. Abrasion refers to the process of rocks being worn down and broken into smaller fragments by the action of external forces like wind, water, or ice. Ice wedging is a type of physical weathering that occurs when water seeps into cracks in rocks, freezes, and expands, causing the rocks to break apart.
Hydrolysis, on the other hand, is a form of chemical weathering rather than physical weathering. It involves the reaction of minerals in rocks with water, leading to the breakdown and alteration of the rock's composition. This process usually occurs over a longer period and involves the dissolution and transformation of minerals through chemical reactions. In summary, the option d) hydrolysis is not a form of physical weathering but rather a type of chemical weathering that involves the reaction of minerals in rocks with water. The other options a) thermal stress, b) abrasion, and c) ice wedging are all examples of physical weathering processes.
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Problem 1: - = 0.029 kg of ice at -10°C are mixed with 0.051 kg of water at 20°C. The water and ice are mixed in a calorimeter so that no heat escapes the system. The specific heat of water is Cw 4186 J/(kg° C), the latent heat of fusion is Lƒ = 3.33 × 105 J/kg, and the specific heat of ice is c¿ 2090 J/(kg.° C). (a) What is the final temperature of the system, when thermal equilibrium is reached? (b) How many kg of ice remain when thermal equilibrium is reached? (c) How many kg of water remain when thermal equilibrium is reached? (d) What is the change in entropy of the system?
0.029 kg of ice at -10°C are mixed with 0.051 kg of water at 20°C. The water and ice are mixed in a calorimeter so that no heat escapes the system. The specific heat of water is [tex]C_w[/tex] 4186 J/(kg ° C), the latent heat of fusion is [tex]L_f[/tex] = 3.33 × 10⁵ J/kg, and the specific heat of ice is [tex]C_i[/tex] 2090 J/(kg ° C).
(a) The final temperature of the system, when thermal equilibrium is reached is 49.9°C.
(b) 0.000607 kg of ice remain when thermal equilibrium is reached.
(c) 0.0504 kg of water remain when thermal equilibrium is reached.
(d) The change in entropy of the system is 0.
To solve this problem, we can apply the principle of conservation of energy and consider the heat gained or lost by each substance.
(a) To find the final temperature of the system, we need to calculate the heat gained by the ice and the water. The heat gained by the ice is used to raise its temperature from -10°C to the final temperature, and the heat gained by the water is used to lower its temperature from 20°C to the final temperature. At thermal equilibrium, the heat gained by the ice is equal to the heat lost by the water.
Heat gained by the ice: [tex]Q_i_c_e=m_i_c_e*c_i_c_e*(T_f_i_n_a_l-T_i_c_e)[/tex]
Heat lost by the water: [tex]Q_w_a_t_e_r=m_w_a_t_e_r*c_w_a_t_e_r*(T_w_a_t_e_r -T_f_i_n_a_l)[/tex]
Since [tex]Q_i_c_e=Q_w_a_t_e_r[/tex],
(0.029 kg) * (2090 J/(kg °C)) * ([tex]T_f_i_n_a_l[/tex] - (-10°C)) = (0.051 kg) * (4186 J/(kg °C)) * (20°C - [tex]T_f_i_n_a_l[/tex])
[tex]T_f_i_n_a_l[/tex] ≈ 4268.508 / (0.06061 + 86.715)
[tex]T_f_i_n_a_l[/tex] ≈ 49.9°C
Therefore, the final temperature of the system, when thermal equilibrium is reached, is approximately 49.9°C.
(b) To determine how many kilograms of ice remain when thermal equilibrium is reached, we need to calculate the heat gained by the ice, which is equal to the heat lost by the water. We can use the equation:
[tex]Q_i_c_e=m_i_c_e*L_f_u_s_i_o_n[/tex]
[tex]m_i_c_e*L_f_u_s_i_o_n=m_w_a_t_e_r*C_w_a_t_e_r*(T_w_a_t_e_r-T_f_i_n_a_l)[/tex]
[tex]m_i_c_e[/tex] = (0.051 kg * 4186 J/(kg °C) * (20°C - 49.9°C)) / (3.33 x 10⁵ J/kg)
[tex]m_i_c_e[/tex] ≈ 0.000607 kg
Therefore, approximately 0.000607 kg of ice remain when thermal equilibrium is reached.
(c) To determine how many kilograms of water remain when thermal equilibrium is reached, we can subtract the mass of the remaining ice from the initial mass of water:
[tex]m_w_a_t_e_r _r_e_m_a_i_n=m_w_a_t_e_r-m_i_c_e[/tex]
[tex]m_w_a_t_e_r_r_e_m_a_i_n[/tex] = 0.051 kg - 0.000607 kg
[tex]m_w_a_t_e_r_r_e_m_a_i_n[/tex] ≈ 0.0504 kg
Therefore, approximately 0.0504 kg of water remain when thermal equilibrium is reached.
(d) The change in entropy of the system can be determined using the formula:
ΔS = Q/T
where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature. Since there is no heat transfer in the system (no heat escapes), the change in entropy is zero:
ΔS = 0
Therefore, the change in entropy of the system is zero.
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