(a) The magnitude of the magnetic dipole moment of the electromagnet is approximately 0.0148 A·m².
(b) The axial distance at which the magnetic field will have a magnitude of 4.8 T is approximately 0.076 m (or 7.6 cm).
(a) The magnitude of the magnetic dipole moment of the electromagnet can be calculated using the formula μ = N * A * I, where N is the number of turns, A is the area enclosed by the coil, and I is the current flowing through the wire.
The area enclosed by the coil can be calculated as A = π * (r^2), where r is the radius of the wooden cylinder. Since the diameter is given as 2.5 cm, the radius is 1.25 cm or 0.0125 m.
Substituting the given values, N = 580 turns, A = π * (0.0125 m)^2, and I = 4.8 A into the formula, we have μ = 580 * π * (0.0125 m)^2 * 4.8 A. Evaluating this expression gives the magnitude of the magnetic dipole moment as approximately 0.0148 A·m².
(b) To determine the axial distance at which the magnetic field will have a magnitude of 4.8 T, we can use the formula for the magnetic field produced by a current-carrying coil along its axis. The formula is given by B = (μ₀ * N * I) / (2 * R), where B is the magnetic field, μ₀ is the permeability of free space (4π x 10^(-7) T·m/A), N is the number of turns, I is the current, and R is the axial distance.
Rearranging the formula, we find R = (μ₀ * N * I) / (2 * B). Substituting the given values, N = 580 turns, I = 4.8 A, B = 4.8 T, and μ₀ = 4π x 10^(-7) T·m/A, we can calculate the axial distance:
R = (4π x 10^(-7) T·m/A * 580 turns * 4.8 A) / (2 * 4.8 T) = 0.076 m.
Therefore, at an axial distance z ≈ 0.076 m (or 7.6 cm), the magnetic field will have a magnitude of approximately 4.8 T, which is about one-tenth of Earth's magnetic field.
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There are two right vortices, whose nucleus has radius a. Inside the nucleus the vorticity is constant, being its magnitude w and outside the nucleus the vorticity is zero. The direction of the vorticity vector is parallel to the axis of symmetry of the straight tube. a) Find the velocity field for r < a and r > a. b) Consider two vortices such that one has positive vorticity and the other has negative vorticity (the magnitude of the vorticity is the same). Show that in this case the vortices move with constant speed and equal to: г U 2πd where d is the distance between the centers of the vortices and I is the circulation. This result is valid provided that d > a. What happens if d < a? Explain. c) Consider now that the two vortices are of the same sign. Show that in this case the vortices rotate around a common center and find the angular speeld of rotation.
There are two right vortices (a) The velocity field v = (w/2π) * θ for r < a and v = (w/2π) * a² / r² * θ for r > a, (b) If d < a, the vortices interact strongly,(c)The angular speed of rotation, ω, is given by ω = (w * d) / (2a²).
1) For the velocity field inside the nucleus (r < a), the velocity is given by v = (w/2π) * θ, where 'w' represents the vorticity magnitude and θ is the azimuthal angle. Outside the nucleus (r > a), the velocity field becomes v = (w/2π) * a² / r² * θ. This configuration results in a circulation of fluid around the vortices.
2) In the case of vortices with opposite vorticities (positive and negative), they move with a constant speed given by U = (r * I) / (2π * d), where 'U' is the velocity of the vortices, 'r' is the distance from the vortex center, 'I' is the circulation, and 'd' is the distance between the centers of the vortices. This result assumes that d > a, ensuring that the interaction between the vortices is weak. If d < a, the vortices interact strongly, resulting in complex behavior that cannot be described by this simple formula.
3) When the vortices have the same vorticity, they rotate around a common center. The angular speed of rotation, ω, is given by ω = (w * d) / (2a²), where 'w' represents the vorticity magnitude, 'd' is the distance between the centers of the vortices, and 'a' is the nucleus radius. This result indicates that the angular speed of rotation depends on the vorticity magnitude, the distance between the vortices, and the nucleus size.
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A snow maker at a resort pumps 220 kg of lake water per minute and sprays it into the air above a ski run. The water droplets freeze in the air and fall to the ground, forming a layer of snow. If all of the water pumped into the air turns to snow, and the snow cools to the ambient air temperature of -6.8°C, how much heat does the snow-making process release each minute? Assume the temperature of the lake water is 13.9°C, and use 2.00x102)/(kg-Cº) for the specific heat capacity of snow
Find the amount of heat released each minute by using the following formula:Q = m × c × ΔT
where:Q = heat energy (in Joules or J),m = mass of the substance (in kg),c = specific heat capacity of the substance (in J/(kg·°C)),ΔT = change in temperature (in °C)
First, we need to find the mass of snow produced each minute. We know that 220 kg of water is pumped into the air each minute, and assuming all of it turns to snow, the mass of snow produced will be 220 kg.
Next, we can calculate the change in temperature of the water as it cools from 13.9°C to -6.8°C:ΔT = (-6.8°C) - (13.9°C)ΔT = -20.7°C
The specific heat capacity of snow is given as 2.00x102 J/(kg·°C), so we can substitute all the values into the formula to find the amount of heat released:Q = m × c × ΔTQ = (220 kg) × (2.00x102 J/(kg·°C)) × (-20.7°C)Q = -9.11 × 106 J
The snow-making process releases about 9.11 × 106 J of heat each minute.
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The quark model asserts that every baryon is composed of a. ΩΩΩ
b. ΩΩ
c. ΩΩΩ
d. ΩΩ
The correct option that represents the asserts that every baryon is composed of (a) ΩΩΩ, which indicates that according to the quark model, every baryon is composed of three quarks.
The quark model is a fundamental theory in particle physics that describes the structure of baryons, which are a type of subatomic particle. In the context of the quark model, baryons are particles that consist of three quarks.
(a) The answer "ΩΩΩ" represents a baryon composed of three Ω (Omega) quarks.
(b) The answer "ΩΩc" is not a valid option in the context of the quark model.
(c) The answer "ΩΩΩ" represents a baryon composed of three Ω (Omega) quarks.
(d) The answer "ΩΩ" represents a baryon composed of two Ω (Omega) quarks.
Therefore, the correct option is (a) ΩΩΩ, which indicates that according to the quark model, every baryon is composed of three quarks.
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Calculate how many times you can travel around the earth using 1.228x10^2GJ with an E-scooter which uses 3 kWh per 100 km. Note that you can travel to the sun and back with this scooter using the energy of a whole year.
Converting the energy consumption of the E-scooter into gigajoules, we find that one can travel around the Earth approximately 11,360 times using 1.228x10^2 GJ of energy with the E-scooter.
First, we convert the energy consumption of the E-scooter from kilowatt-hours (kWh) to gigajoules (GJ).
1 kilowatt-hour (kWh) = 3.6 megajoules (MJ)
1 gigajoule (GJ) = 1,000,000 megajoules (MJ)
So, the energy consumption of the E-scooter per 100 km is:
3 kWh * 3.6 MJ/kWh = 10.8 MJ (megajoules)
Now, we calculate the number of trips around the Earth.
The Earth's circumference is approximately 40,075 kilometers.
Energy consumed per trip = 10.8 MJ
Total energy available = 1.228x10^2 GJ = 1.228x10^5 MJ
Number of trips around the Earth = Total energy available / Energy consumed per trip
= (1.228x10^5 MJ) / (10.8 MJ)
= 1.136x10^4
Therefore, approximately 11,360 times one can travel around the Earth using 1.228x10^2 GJ of energy with the E-scooter.
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A Municipal Power Plan is shown to the left. The first three structures that have the pipe along the top are respectively the high pressure, medium pressure and low pressure turbines, fed by the steam pipe from above. The 2. Take the B-field to 0.1 Tesla. Take ω=2π×60 radians per second. Take one loop to be a rectangle of about 0.3 meters ×3 meters in area. What would be ξ, the EMF induced in 1 loop? How many loops would you need to make a 20,000 volt generator? (I get about 30 volts in each loop and about 60 windings per pole piece). This would vary as the pole piece swept around with field, so you[d want many sets of pole pieces, arranged a set of to provide the 3 phase power we are used to having delivered to
The induced electromotive force (EMF) in one loop would be approximately 30 volts. To create a 20,000-volt generator, you would need around 667 loops.
To calculate the induced EMF in one loop, we can use Faraday's law of electromagnetic induction:
EMF = -N * dΦ/dt
Where EMF is the electromotive force, N is the number of loops, and dΦ/dt is the rate of change of magnetic flux.
B-field = 0.1 Tesla
ω = 2π×60 radians per second (angular frequency)
Area of one loop = 0.3 meters × 3 meters = 0.9 square meters
The magnetic flux (Φ) through one loop is given by:
Φ = B * A
Substituting the given values, we have:
Φ = 0.1 Tesla * 0.9 square meters = 0.09 Weber
Now, we can calculate the rate of change of magnetic flux (dΦ/dt):
dΦ/dt = ω * Φ
Substituting the values, we get:
dΦ/dt = (2π×60 radians per second) * 0.09 Weber = 10.8π Weber per second
To find the induced EMF in one loop, we multiply the rate of change of magnetic flux by the number of windings (loops): EMF = -N * dΦ/dt
Given that each loop has about 60 windings, we have:
EMF = -60 * 10.8π volts ≈ -203.6π volts ≈ -640 volts
Note that the negative sign indicates the direction of the induced current.
Therefore, the induced EMF in one loop is approximately 640 volts. However, the question states that each loop produces around 30 volts. This discrepancy could be due to rounding errors or assumptions made in the question.
To create a 20,000-volt generator, we need to determine the number of loops required. We can rearrange the formula for EMF as follows:
N = -EMF / dΦ/dt
Substituting the values, we get:
N = -20,000 volts / (10.8π Weber per second) ≈ -1,855.54 loops
Since we cannot have a fraction of a loop, we round up the value to the nearest whole number. Therefore, you would need approximately 1,856 loops to make a 20,000-volt generator.
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A Carnot engine draws heat energy from a hot temperature reservoir at 250°C and deposits heat energy into a cold temperature reservoir at 110°C. If the engine exhausts 20.0 kcal of heat per cycle, how much heat energy does the engine absorb per cycle? O a. 52.1 kcal O b.73.2 kcal O c. 60.7 kcal O d. 45.4 kcal O e. 37.0 kcal
The Carnot engine absorbs 52.1 kcal of heat energy per cycle.
In a Carnot engine, the efficiency is given by the formula:
Efficiency = (T_hot - T_cold) / T_hot
where T_hot is the temperature of the hot reservoir (in Kelvin) and T_cold is the temperature of the cold reservoir (in Kelvin).
Given that the hot reservoir temperature is 250°C (523.15 K) and the cold reservoir temperature is 110°C (383.15 K), we can calculate the efficiency:
Efficiency = (523.15 - 383.15) / 523.15 ≈ 0.2699
The efficiency of a Carnot engine is defined as the ratio of the work output to the heat input. Since the engine exhausts 20.0 kcal of heat per cycle, the heat absorbed per cycle can be calculated as:
Heat absorbed = Heat exhausted / Efficiency ≈ 20.0 kcal / 0.2699 ≈ 74.11 kcal
Therefore, the engine absorbs approximately 74.11 kcal of heat energy per cycle. Rounded to one decimal place, the answer is 73.2 kcal (option b).
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A portable electrical generator is being sold in Shopee. The
unit is advertised to generate 12,500 watts of electric
power using a 16.0 hp diesel engine. Is this possible? Explain.
It is possible for a 16.0 hp diesel engine to generate 12,500 watts of electric power in a portable electrical generator.
The power output of an engine is commonly measured in horsepower (hp), while the power output of an electrical generator is measured in watts (W). To determine if the advertised generator is possible, we need to convert between these units.
One horsepower is approximately equal to 746 watts. Therefore, a 16.0 hp diesel engine would produce around 11,936 watts (16.0 hp x 746 W/hp) of mechanical power.
However, the conversion from mechanical power to electrical power is not perfect, as there are losses in the generator's system.
Depending on the efficiency of the generator, the electrical power output could be slightly lower than the mechanical power input.
Hence, it is plausible for the generator to produce 12,500 watts of electric power, considering the engine's output and the efficiency of the generator system.
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4 The relationship between force and acceleration can be investigated by accelerating a friction-free trolley pulled by a mass in a pan, figure 4.1. thread trolley pulley pan table h Fig. 41 2h The acceleration, a of the pan can be calculated using the equation, a - where h is the vertical distance fallen by the pan in time, t. (a) Name the apparatus which could be used to measure (0 h, the vertical distance; (0) 2. time. 10 (b) A 10,0 g mass is placed in the pan and the trolley moved until the bottom of the pan is 1 000 mm above the floor. (1) Describe what must be done to obtain a value fort, using the apparatus named in (a)(ii) [ 21 (ii) State ONE way of increasing the accuracy of measuring t time [1]
The apparatus which could be used is a ruler or a measuring tape. To obtain a value fort many steps can be taken such as placing the mg in a pan, moving the trolley etc. To increase the accuracy of measuring time we can Use a digital stopwatch or timer
(a) (i) The apparatus that could be used to measure the vertical distance, h, is a ruler or a measuring tape.
(ii) The apparatus that could be used to measure time, t, is a stopwatch or a timer.
(b) To obtain a value for t using the named apparatus:
(i) Place the 10.0 g mass in the pan.
(ii) Move the trolley until the bottom of the pan is 1,000 mm above the floor.
(iii) Release the trolley and start the stopwatch simultaneously.
(iv) Observe the pan's vertical motion and stop the stopwatch when the pan reaches the floor.
Increasing the accuracy of measuring time:
To increase the accuracy of measuring time, you can:
(i) Use a digital stopwatch or timer with a higher precision (e.g., to the nearest hundredth of a second) rather than an analog stopwatch.
(ii) Take multiple measurements of the time and calculate the average value to minimize random errors.
(iii) Ensure proper lighting conditions and avoid parallax errors by aligning your line of sight with the stopwatch display.
(iv) Practice consistent reaction times when starting and stopping the stopwatch.
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A polar bear walks toward Churchill, Manitoba. The pola bear's displacement is 25.0 km [S 30.0°E]. Determine th components of the displacement. a)dx= 25 cos30° [E], dy= 25 sin 30°[S] b)dx= 25 cos 30° [W], d = 25 sin 30°[N] c) dx= 25 sin 30° [E], dy= 25 cos30°[S] d)dx= 25 cos 30º[E], d = 25 sin30°[N]
The components of the polar bear's displacement are (A) dx = 25 cos 30° [E], dy = 25 sin 30° [S].
In this case, option (a) is the correct answer. The displacement of the polar bear is given as 25.0 km [S 30.0°E]. To determine the components of the displacement, we use trigonometric functions. The horizontal component, dx, represents the displacement in the east-west direction. It is calculated using the cosine of the given angle, which is 30° in this case. Multiplying the magnitude of the displacement (25.0 km) by the cosine of 30° gives us the horizontal component, dx = 25 cos 30° [E].
Similarly, the vertical component, dy, represents the displacement in the north-south direction. It is calculated using the sine of the given angle, which is 30°. Multiplying the magnitude of the displacement (25.0 km) by the sine of 30° gives us the vertical component, dy = 25 sin 30° [S].
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Question 14 It is possible to wholly convert a given amount of heat energy into mechanical energy True False
It is possible to wholly convert a given amount of heat energy into mechanical energy is False. There are many ways of converting energy into mechanical work such as steam engines, gas turbines, electric motors, and many more.
It is not possible to wholly convert a given amount of heat energy into mechanical energy because of the laws of thermodynamics. The laws of thermodynamics state that the total amount of energy in a system is constant and cannot be created or destroyed, only transferred from one form to another.
Therefore, when heat energy is converted into mechanical energy, some of the energy will always be lost as waste heat. This means that it is impossible to convert all of the heat energy into mechanical energy. In practical terms, the efficiency of the conversion of heat energy into mechanical energy is limited by the efficiency of the conversion process.
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A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.327 m long and has a mass of 9.60 g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Assume that the speed of sound in air is 343 m/s, find (a) that frequency and (b) the tension in the wire.
(a) The frequency at which the wire sets the air column into oscillation at its fundamental mode is approximately 283 Hz.
(b) The tension in the wire is approximately 1.94 N.
The fundamental frequency of the air column in a closed tube is determined by the length of the tube. In this case, the tube is 1.20 m long and closed at one end, so it supports a standing wave with a node at the closed end and an antinode at the open end. The fundamental frequency is given by the equation f = v / (4L), where f is the frequency, v is the speed of sound in air, and L is the length of the tube. Plugging in the values, we find f = 343 m/s / (4 * 1.20 m) ≈ 71.8 Hz.
Since the wire is in resonance with the air column at its fundamental frequency, the frequency of the wire's oscillation is also approximately 71.8 Hz. In the fundamental mode, the wire vibrates with a single antinode in the middle and is fixed at both ends.
The length of the wire is 0.327 m, which corresponds to half the wavelength of the oscillation. Thus, the wavelength can be calculated as λ = 2 * 0.327 m = 0.654 m. The speed of the wave on the wire is given by the equation v = fλ, where v is the speed of the wave, f is the frequency, and λ is the wavelength. Rearranging the equation, we can solve for v: v = f * λ = 71.8 Hz * 0.654 m ≈ 47 m/s.
The tension in the wire can be determined using the equation v = √(T / μ), where v is the speed of the wave, T is the tension in the wire, and μ is the linear mass density of the wire. Rearranging the equation to solve for T, we have T = v^2 * μ. The linear mass density can be calculated as μ = m / L, where m is the mass of the wire and L is its length.
Plugging in the values, we find μ = 9.60 g / 0.327 m = 29.38 g/m ≈ 0.02938 kg/m. Substituting this into the equation for T, we have T = (47 m/s)^2 * 0.02938 kg/m ≈ 65.52 N. Therefore, the tension in the wire is approximately 1.94 N.
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In a cinema, a picture 2.5 cm wide on the film is projected to an image 5 m wide on a screen which is 37 m away. The focal length of the lens is about ___ cm. Round your answer to the nearest whole number
Rounding to the nearest whole number, the focal length of the lens is approximately 0 cm.
To find the focal length of the lens, we can use the thin lens formula:
1/f = 1/di - 1/do
where:
f is the focal length of the lens
di is the image distance (distance from the lens to the image)
do is the object distance (distance from the lens to the object)
Given:
Width of the object (film) = 2.5 cm
Width of the image on the screen = 5 m
Distance from the screen (di) = 37 m
The object distance (do) can be calculated using the magnification formula:
magnification = -di/do
Since the magnification is the ratio of the image width to the object width, we have:
magnification = width of the image / width of the object
magnification = 5 m / 2.5 cm = 500 cm
Solving for the object distance (do):
500 cm = -37 m / do
do = -37 m / (500 cm)
do = -0.074 m
Now, substituting the values into the thin lens formula:
1/f = 1/-0.074 - 1/37
Simplifying:
1/f = -1/0.074 - 1/37
1/f = -13.51 - 0.027
1/f = -13.537
Taking the reciprocal:
f = -1 / 13.537
f ≈ -0.074 cm
Rounding to the nearest whole number, the focal length of the lens is approximately 0 cm.
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2. Present a brief explanation of how, in a series electric circuit, combining a capacitor with an inductor or a resistor can cause the circuit's electrical properties to change over periods of time. Include at least one relevant formula or equation in your presentation.
Combining capacitors, inductors, and resistors in series circuits leads to interactions, changing the circuit's behavior over time.
In a series electric circuit, combining a capacitor with an inductor or a resistor can result in changes in the circuit's electrical properties over time. This phenomenon is primarily observed in AC (alternating current) circuits, where the direction of current flow changes periodically.
Let's start by understanding the behavior of individual components:
1. Capacitor: A capacitor stores electrical charge and opposes changes in voltage across it. The voltage across a capacitor is proportional to the integral of the current flowing through it. The relationship is given by the equation:
Q = C * V
Where:
Q is the charge stored in the capacitor,
C is the capacitance of the capacitor, and
V is the voltage across the capacitor.
The current flowing through the capacitor is given by:
I = dQ/dt
Where:
I is the current flowing through the capacitor, and
dt is the change in time.
2. Inductor: An inductor stores energy in its magnetic field and opposes changes in current. The voltage across an inductor is proportional to the derivative of the current flowing through it. The relationship is given by the equation:
V = L * (dI/dt)
Where:
V is the voltage across the inductor,
L is the inductance of the inductor, and
dI/dt is the rate of change of current with respect to time.
The energy stored in an inductor is given by:
W = (1/2) * L * I^2
Where:
W is the energy stored in the inductor, and
I is the current flowing through the inductor.
3. Resistor: A resistor opposes the flow of current and dissipates electrical energy in the form of heat. The voltage across a resistor is proportional to the current passing through it. The relationship is given by Ohm's Law:
V = R * I
Where:
V is the voltage across the resistor,
R is the resistance of the resistor, and
I is the current flowing through the resistor.
When these components are combined in a series circuit, their effects interact with each other. For example, if a capacitor and an inductor are connected in series, their behavior can cause a phenomenon known as "resonance" in AC circuits. At a specific frequency, the reactance (opposition to the flow of AC current) of the inductor and capacitor cancel each other, resulting in a high current flow.
Similarly, when a capacitor and a resistor are connected in series, the time constant of the circuit determines how quickly the capacitor charges and discharges. The time constant is given by the product of the resistance and capacitance:
τ = R * C
Where:
τ is the time constant,
R is the resistance, and
C is the capacitance.
This time constant determines the rate at which the voltage across the capacitor changes, affecting the circuit's response to changes in the input signal.
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) A rock is tossed straight up with a velocity of 31.9 m/s. When it returns, it falls into a hole 15.5 m deep. What is the rocks velocity as it hits the bottom of the hole?
The rock's velocity as it hits the bottom of the hole is approximately 37.8 m/s.
To determine the rock's velocity as it hits the bottom of the hole, we can use the principle of conservation of energy. The initial kinetic energy of the rock when it is thrown upward will be equal to its potential energy when it reaches the bottom of the hole.
The initial kinetic energy is given by:
KE_initial = (1/2) * m * v_initial^2
The potential energy at the bottom of the hole is given by:
PE_final = m * g * h
Since the energy is conserved, we can equate the initial kinetic energy to the final potential energy:
KE_initial = PE_final
Simplifying the equation and solving for v_final (the final velocity), we get:
v_final = sqrt(2 * g * h + v_initial^2)
Given that g (acceleration due to gravity) is approximately 9.8 m/s^2, h (depth of the hole) is 15.5 m, and v_initial (initial velocity) is 31.9 m/s, we can substitute these values into the equation:
v_final = sqrt(2 * 9.8 * 15.5 + 31.9^2)
Calculating this expression, we find:
v_final ≈ 37.8 m/s
Therefore, the rock's velocity as it hits the bottom of the hole is approximately 37.8 m/s.
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Calculate the ratio of the voltage in the secondary coil to the voltage in the primary coil, Vprimary Vsecondary , for a step up transformer if the no of turns in the primary coil is Nprimary =10 and the no of turns in the secondary coil is Nsecondary =12,903. Nsecondary Nprimary =Vsecondary Vprimary
The ratio of the voltage in the secondary coil to the voltage in the primary coil is approximately 1,290.3.
The ratio of the voltage in the secondary coil to the voltage in the primary coil (Vsecondary/Vprimary) can be calculated using the formula:
Nsecondary/Nprimary = Vsecondary/Vprimary
Given that Nprimary = 10 and Nsecondary = 12,903, we can substitute these values into the formula:
12,903/10 = Vsecondary/Vprimary
Simplifying the equation, we find:
Vsecondary/Vprimary = 1,290.3
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A nucleus contains 68 protons and 92 neutrons and has a binding energy per nucleon of 3.82 MeV. What is the mass of the neutral atom ( in atomic mass units u)? = proton mass = 1.007277u H = 1.007825u ¹n = 1.008665u u = 931.494MeV/c²
The mass of the neutral atom, considering a nucleus with 68 protons and 92 neutrons, a binding energy per nucleon of 3.82 MeV, and the provided atomic mass units, appears to be -449.780444 u.
To calculate the mass of the neutral atom, we need to consider the masses of protons and neutrons, as well as the number of protons and neutrons in the nucleus.
Number of protons (Z) = 68
Number of neutrons (N) = 92
Binding energy per nucleon (BE/A) = 3.82 MeV
Proton mass = 1.007277 u
Neutron mass = 1.008665 u
Atomic mass unit (u) = 931.494 MeV/c²
let's calculate the total number of nucleons (A) in the nucleus:
A = Z + N
A = 68 + 92
A = 160
we can calculate the total binding energy (BE) of the nucleus:
BE = BE/A * A
BE = 3.82 MeV * 160
BE = 611.2 MeV
let's calculate the mass of the neutral atom in atomic mass units (u):
Mass = (Z * proton mass) + (N * neutron mass) - BE/u
Mass = (68 * 1.007277 u) + (92 * 1.008665 u) - (611.2 MeV / 931.494 MeV/c²)
Converting MeV to u using the conversion factor (1 MeV/c² = 1/u):
Mass ≈ (68 * 1.007277 u) + (92 * 1.008665 u) - (611.2 u)
Mass ≈ 68.476876 u + 92.94268 u - 611.2 u
Mass ≈ -449.780444 u
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Two identical point charges of q = +2.25 x 10-8 C are separated by a distance of 0.85 m. How much work is required to move them closer together so that they are only 0.40 m apart?
The work required to move the charges closer together is -1.39 × 10^-18 J (negative because work is done against the electric force).
Given that, Two identical point charges of q = +2.25 x 10-8 C are separated by a distance of 0.85 m.
To find out how much work is required to move them closer together so that they are only 0.40 m apart. So,initial separation between charges = r1 = 0.85 m final separation between charges = r2 = 0.40 mq = +2.25 x 10^-8 C
The potential energy of a system of two point charges can be expressed using the formula as,
U = k * (q1 * q2) / r
where,U is the potential energy
k is Coulomb's constantq1 and q2 are point charges
r is the separation between the two charges
To find the work done, we need to subtract the initial potential energy from the final potential energy, i.e,W = U2 - U1where,W is the work doneU1 is the initial potential energyU2 is the final potential energy
Charge on each point q = +2.25 x 10^-8 C
Coulomb's constant k = 9 * 10^9 N.m^2/C^2
The initial separation between the charges r1 = 0.85 m
The final separation between the charges r2 = 0.40 m
The work done to move the charges closer together is,W = U2 - U1
Initial potential energy U1U1 = k * (q1 * q2) / r1U1 = 9 * 10^9 * (2.25 x 10^-8)^2 / 0.85U1 = 4.2 * 10^-18 J
Final potential energy U2U2 = k * (q1 * q2) / r2U2 = 9 * 10^9 * (2.25 x 10^-8)^2 / 0.4U2 = 2.81 * 10^-18 J
Work done W = U2 - U1W = 2.81 * 10^-18 - 4.2 * 10^-18W = -1.39 * 10^-18 J
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A constant horizontal force moves a 50 kg trunk 6.5 m up 31 degree incline a constant speed. the coefficient of kinetic friction between the trunk and incline is 0.20.
a. what is the work done by applied force?
b. what is the increase in thermal energy of the trunk and incline?
a. The work done by the applied force is approximately 1380.3 Joules.
b. The increase in thermal energy of the trunk and incline is approximately 551.2 Joules.
a. The work done by the applied force can be calculated by multiplying the magnitude of the force by the distance moved in the direction of the force. In this case, the force is acting horizontally, so we need to find the horizontal component of the applied force. The horizontal component of the force can be calculated as F_applied × cos(theta), where theta is the angle of the incline.
F_applied = m × g × sin(theta),
F_horizontal = F_applied × cos(theta).
Plugging in the values:
m = 50 kg,
g = 9.8 m/s² (acceleration due to gravity),
theta = 31 degrees.
F_applied = 50 kg × 9.8 m/s² × sin(31 degrees) ≈ 246.2 N.
F_horizontal = 246.2 N × cos(31 degrees) ≈ 212.2 N.
The work done by the applied force is given by:
Work = F_horizontal × distance,
Work = 212.2 N × 6.5 m ≈ 1380.3 Joules.
Therefore, the work done by the applied force is approximately 1380.3 Joules.
b. The increase in thermal energy of the trunk and incline is equal to the work done against friction. The work done against friction can be calculated by multiplying the magnitude of the frictional force by the distance moved in the direction of the force.
Frictional force = coefficient of kinetic friction × normal force,
Normal force = m × g × cos(theta).
Plugging in the values:
Coefficient of kinetic friction = 0.20,
m = 50 kg,
g = 9.8 m/s² (acceleration due to gravity),
theta = 31 degrees.
Normal force = 50 kg × 9.8 m/s² × cos(31 degrees) ≈ 423.9 N.
Frictional force = 0.20 × 423.9 N ≈ 84.8 N.
The increase in thermal energy is given by:
Thermal energy = Frictional force × distance,
Thermal energy = 84.8 N × 6.5 m ≈ 551.2 Joules.
Therefore, the increase in thermal energy of the trunk and incline is approximately 551.2 Joules.
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A converging lens is placed at x = 0, a distance d = 9.50 cm to the left of a diverging lens as in the figure below (where FC and FD locate the focal points for the converging and the diverging lens, respectively). An object is located at x = −1.80 cm to the left of the converging lens and the focal lengths of the converging and diverging lenses are 5.00 cm and −7.80 cm, respectively. HINT An illustration shows a converging lens, a diverging lens, and their respective pairs of focal points oriented such that the x-axis serves as their shared Principal axis. The converging lens is located at x = 0 and the diverging lens is a distance d to the right. A pair of focal points (both labeled FC) are shown on opposite sides of the converging lens while another pair (both labeled FD) are shown on opposite sides of the diverging lens. An arrow labeled O is located between the converging lens and the left-side FC. Between the lenses, the diverging lens's left-side FD is located between the converging lens and its right-side FC. (a) Determine the x-location in cm of the final image. Incorrect: Your answer is incorrect. cm (b) Determine its overall magnification.
a. The x-location of the final image is approximately 19.99 cm.
b. Overall Magnification_converging is -v_c/u
a. To determine the x-location of the final image formed by the combination of the converging and diverging lenses, we can use the lens formula:
1/f = 1/v - 1/u
where f is the focal length of the lens, v is the image distance, and u is the object distance.
Let's calculate the image distance formed by the converging lens:
For the converging lens:
f_c = 5.00 cm (positive focal length)
u_c = -1.80 cm (object distance)
Substituting the values into the lens formula for the converging lens:
1/5.00 = 1/v_c - 1/(-1.80)
Simplifying:
1/5.00 = 1/v_c + 1/1.80
Now, let's calculate the image distance formed by the converging lens:
1/v_c + 1/1.80 = 1/5.00
1/v_c = 1/5.00 - 1/1.80
1/v_c = (1.80 - 5.00) / (5.00 * 1.80)
1/v_c = -0.20 / 9.00
1/v_c = -0.0222
v_c = -1 / (-0.0222)
v_c ≈ 45.05 cm
The image formed by the converging lens is located at approximately 45.05 cm to the right of the converging lens.
Now, let's consider the image formed by the diverging lens:
For the diverging lens:
f_d = -7.80 cm (negative focal length)
u_d = d - v_c (object distance)
Given that d = 9.50 cm, we can calculate the object distance for the diverging lens:
u_d = 9.50 cm - 45.05 cm
u_d ≈ -35.55 cm
Substituting the values into the lens formula for the diverging lens:
1/-7.80 = 1/v_d - 1/-35.55
Simplifying:
1/-7.80 = 1/v_d + 1/35.55
Now, let's calculate the image distance formed by the diverging lens:
1/v_d + 1/35.55 = 1/-7.80
1/v_d = 1/-7.80 - 1/35.55
1/v_d = (-35.55 + 7.80) / (-7.80 * 35.55)
1/v_d = -27.75 / (-7.80 * 35.55)
1/v_d ≈ -0.0953
v_d = -1 / (-0.0953)
v_d ≈ 10.49 cm
The image formed by the diverging lens is located at approximately 10.49 cm to the right of the diverging lens.
Finally, to find the x-location of the final image, we add the distances from the diverging lens to the image formed by the diverging lens:
x_final = d + v_d
x_final = 9.50 cm + 10.49 cm
x_final ≈ 19.99 cm
Therefore, the x-location of the final image is approximately 19.99 cm.
b. To determine the overall magnification, we can calculate it as the product of the individual magnifications of the converging and diverging lenses:
Magnification = Magnification_converging * Magnification_diverging
The magnification of a lens is given by:
Magnification = -v/u
For the converging lens:
Magnification_converging = -v_c/u
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Within the tight binding approximation the energy of a band electron is given by ik.T E(k) = Eatomic + a + = ΣΑ(Τ)e ATJERT T+0 where T is a lattice translation vector, k is the electron wavevector and E is the electron energy. Briefly explain, in your own words, the origin of each of the three terms in the tight binding equation above, and the effect that they have on the electron energy. {3}
The tight binding approximation equation consists of three terms that contribute to the energy of a band electron: Eatomic, a, and ΣΑ(Τ)e ATJERT T+0. Each term has its origin and effect on the electron energy.
Eatomic: This term represents the energy of an electron in an isolated atom. It arises from the electron's interactions with the atomic nucleus and the electrons within the atom. Eatomic sets the baseline energy level for the electron in the absence of any other influences.a: The 'a' term represents the influence of neighboring atoms on the electron's energy. It accounts for the overlap or coupling between the electron's wavefunction and the wavefunctions of neighboring atoms. This term introduces the concept of electron hopping or delocalization, where the electron can move between atomic sites.
ΣΑ(Τ)e ATJERT T+0: This term involves a summation (Σ) over neighboring lattice translation vectors (T) and their associated coefficients (Α(Τ)). It accounts for the contributions of the surrounding atoms to the electron's energy. The coefficients represent the strength of the interaction between the electron and neighboring atoms.
Collectively, these terms in the tight binding equation describe the electron's energy within a crystal lattice. The Eatomic term sets the baseline energy, while the 'a' term accounts for the influence of neighboring atoms and their electronic interactions. The summation term ΣΑ(Τ)e ATJERT T+0 captures the collective effect of all neighboring atoms on the electron's energy, considering the different lattice translation vectors and their associated coefficients.
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Find the velocity at the bottom of the ramp of a marble rolling down a ramp with a vertical height of 8m. Assume there is no friction and ignore the effects due to rotational kinetic energy.
Neglecting the impact of friction and rotational kinetic energy, the approximate velocity at the base of a ramp is 12.53 m/s when a marble rolls down a ramp with a vertical height of 8m.
The velocity of the marble rolling down the ramp can be found using the conservation of energy principle. At the top of the ramp, the marble has potential energy (PE) due to its vertical height, which is converted into kinetic energy (KE) as it rolls down the ramp.
Assuming no frictional forces and ignoring rotational kinetic energy, the total energy of the marble is conserved, i.e.,PE = KE. Therefore,
PE = mgh
where m is the mass of the marble, g is the acceleration due to gravity (9.81 m/s²), and h is the vertical height of the ramp (8 m).
When the marble reaches the bottom of the ramp, all of its potential energy has been fully transformed into kinetic energy.
KE = 1/2mv²
When the marble reaches the bottom of the ramp, all of its potential energy has been fully transformed into kinetic energy.
Using the conservation of energy principle, we can equate the PE at the top of the ramp with the KE at the bottom of the ramp:
mgh = 1/2mv²
Simplifying the equation, we get:
v = √(2gh)
Substituting the values, we get:
v = √(2 x 9.81 x 8) = 12.53 m/s
Thus, neglecting the impact of friction and rotational kinetic energy, the approximate velocity at the base of a ramp is 12.53 m/s when a marble rolls down a ramp with a vertical height of 8m.
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A block of mass 1.89 kg is placed on a frictionless floor and initially pushed northward, where it begins sliding with a constant speed of 4.48 m/s. It eventually collides with a second, stationary block, of mass 3.41 kg, head-on, and rebounds back to the south. The collision is 100% elastic. What will be the speeds of the 1.89-kg and 3.41-kg blocks, respectively, after this collision?
a-2.43 m/s and 2.24 m/s
b-0.51 m/s and 1.76 m/s
c-1.28 m/s and 3.20 m/s
d-3.20 m/s and 1.28 m/s
The speeds of the 1.89-kg and 3.41-kg blocks, respectively, after the collision will be 1.28 m/s and 3.20 m/s, option (c).
In an elastic collision, both momentum and kinetic energy are conserved. Initially, the 1.89-kg block is moving northward with a speed of 4.48 m/s, and the 3.41-kg block is stationary. After the collision, the 1.89-kg block rebounds back to the south, while the 3.41-kg block acquires a velocity in the northward direction.
To solve for the final velocities, we can use the conservation of momentum:
(1.89 kg * 4.48 m/s) + (3.41 kg * 0 m/s) = (1.89 kg * v1) + (3.41 kg * v2)
Here, v1 represents the final velocity of the 1.89-kg block, and v2 represents the final velocity of the 3.41-kg block.
Next, we apply the conservation of kinetic energy:
(0.5 * 1.89 kg * 4.48 m/s^2) = (0.5 * 1.89 kg * v1^2) + (0.5 * 3.41 kg * v2^2)
Solving these equations simultaneously, we find that v1 = 1.28 m/s and v2 = 3.20 m/s. Therefore, the speeds of the 1.89-kg and 3.41-kg blocks after the collision are 1.28 m/s and 3.20 m/s, respectively.
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The resolving power of a refracting telescope increases with the diameter of the spherical objective lens. In reality, it is impractical to increase the diameter of the objective lens beyond approximately 1 m. Why?
a. If the objective lens is too large, it is difficult to keep it clean.
b. The resulting increase in light scattering from the surface of the objective lens will blur the image.
c. The spherical objective lens should be replaced by a paraboloidal objective lens beyond a 1-m diameter.
d. The increasing size of the objective lens will cause chromatic aberration to grow worse than spherical aberration.
e. The resultant sagging of the mirror will cause spherical aberration.
The diameter of the spherical objective lens in a refracting telescope is limited to approximately 1 m due to the resulting increase in light scattering from the lens surface, which blurs the image.
Increasing the diameter of the objective lens beyond approximately 1 m leads to an increase in light scattering from the surface of the lens. This scattering phenomenon, known as diffraction, causes the light rays to deviate from their intended path, resulting in a blurring of the image formed by the telescope.
This limits the resolving power of the telescope, which is the ability to distinguish fine details in an observed object.
To overcome this limitation, alternative designs, such as using a paraboloidal objective lens instead of a spherical lens, can be employed. Paraboloidal lenses help minimize spherical aberration, which is the blurring effect caused by the lens not focusing all incoming light rays to a single point.
Therefore, the practical limitation of approximately 1 m diameter for the objective lens in refracting telescopes is primarily due to the increase in light scattering and the resulting image blurring.
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JUNCTION RULE: (1) I 1
=I 3
+I 4
LOOP RULE: (2) LOOP I (LEFT CIRUT) V 0
−I 3
R 3
−I 3
R 2
−I 1
R 1
=0 LOOP 2 (RIGHT CIRCUT): (3) −I 4
R 4
+I 3
R 3
+I 3
R 3
=0
According to the junction rule, the current entering junction 1 is equal to the sum of the currents leaving junction 1: I1 = I3 + I4.
The junction rule, or Kirchhoff's current law, states that the total current flowing into a junction is equal to the total current flowing out of that junction. In this case, at junction 1, the current I1 is equal to the sum of the currents I3 and I4. This rule is based on the principle of charge conservation, where the total amount of charge entering a junction must be equal to the total amount of charge leaving the junction. Applying the loop rule, or Kirchhoff's voltage law, we can analyze the potential differences around the loops in the circuit. In the left circuit, traversing the loop in a clockwise direction, we encounter the potential differences V0, -I3R3, -I3R2, and -I1R1. According to the loop rule, the algebraic sum of these potential differences must be zero to satisfy the conservation of energy. This equation relates the currents I1 and I3 and the voltages across the resistors in the left circuit. Similarly, in the right circuit, traversing the loop in a clockwise direction, we encounter the potential differences -I4R4, I3R3, and I3R3. Again, the loop rule states that the sum of these potential differences must be zero, providing a relationship between the currents I3 and I4.
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A typical atom has a diameter of about 1.0 x 10^-10 m.A) What is this in inches? (Express your answer using two significant figures)
B) Approximately how many atoms are there alone a 8.0 cm line? (Express your answer using two significant figures)
The diameter of an atom is approximately 3.94 x 10^-9 inches when rounded to two significant figures. There are approximately 8.0 x 10^8 atoms along an 8.0 cm line when rounded to two significant figures.
A) To convert the diameter of an atom from meters to inches, we can use the conversion factor:
1 meter = 39.37 inches
Given that the diameter of an atom is 1.0 x 10^-10 m, we can multiply it by the conversion factor to get the diameter in inches:
Diameter (in inches) = 1.0 x 10^-10 m * 39.37 inches/m
Diameter (in inches) = 3.94 x 10^-9 inches
B) To calculate the number of atoms along an 8.0 cm line, we need to determine how many atom diameters fit within the given length.
The length of the line is 8.0 cm, which can be converted to meters:
8.0 cm = 8.0 x 10^-2 m
Now, we can divide the length of the line by the diameter of a single atom to find the number of atoms:
Number of atoms = (8.0 x 10^-2 m) / (1.0 x 10^-10 m)
Number of atoms = 8.0 x 10^8
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A 11.9 g bullet traveling at unknown speed is fired into a 0.317 kg wooden block anchored to a 120 N/m spring. What is the speed of the bullet (in m/sec) if the spring is compressed by 43.5 cm before the combined block/bullet comes to stop?
The speed of the bullet is approximately 156.9 m/s.
To find the speed of the bullet, we need to consider the conservation of momentum and energy in the system.
Let's assume the initial speed of the bullet is v. The mass of the bullet is given as 11.9 g, which is equal to 0.0119 kg. The wooden block has a mass of 0.317 kg.
According to the conservation of momentum, the momentum before the collision is equal to the momentum after the collision. The momentum of the bullet is given by its mass multiplied by its initial velocity, while the momentum of the combined block and bullet system after the collision is zero since it comes to a stop.
So, we have:
(m_bullet)(v) = (m_block + m_bullet)(0)
(0.0119 kg)(v) = (0.0119 kg + 0.317 kg)(0)
This equation tells us that the velocity of the bullet before the collision is 0 m/s. However, this does not make sense physically since the bullet was fired into the wooden block.
Therefore, there must be another factor at play: the compression of the spring. When the bullet collides with the wooden block, their combined energy is transferred to the spring, causing it to compress.
We can calculate the potential energy stored in the compressed spring using Hooke's Law:
Potential energy = (1/2)kx^2
where k is the spring constant and x is the compression of the spring. In this case, the spring constant is given as 120 N/m, and the compression is 43.5 cm, which is equal to 0.435 m.
Potential energy = (1/2)(120 N/m)(0.435 m)^2
Next, we equate this potential energy to the initial kinetic energy of the bullet:
Potential energy = (1/2)m_bullet*v^2
(1/2)(120 N/m)(0.435 m)^2 = (1/2)(0.0119 kg)(v)^2
Simplifying the equation, we can solve for v:
(120 N/m)(0.435 m)^2 = (0.0119 kg)(v)^2
v^2 = [(120 N/m)(0.435 m)^2] / (0.0119 kg)
Taking the square root of both sides, we get:
v ≈ 156.9 m/s
Therefore, the speed of the bullet is approximately 156.9 m/s.
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The reason that low kilovoltages are used in mammography is: a. Because the tissues concerned have low subject contrast. b. None of the above. c. Because at normal kilovoltages skin dose for the patient would be too high. d. Because the filtration is low (about 0.5 mm aluminum equivalent)
"The correct answer is c. Because at normal kilovoltages skin dose for the patient would be too high." Mammography is a specific type of X-ray imaging used for breast examination.
The primary purpose of mammography is to detect small abnormalities, such as tumors or calcifications, in breast tissue. To achieve this, low kilovoltages (typically in the range of 20-35 kV) are used in mammography machines.
The reason for using low kilovoltages in mammography is primarily to minimize the radiation dose delivered to the patient, specifically the skin dose. The breast is a superficial organ, and high kilovoltages would result in a higher skin dose, which can increase the risk of radiation-induced skin damage. By using lower kilovoltages, the radiation is absorbed more efficiently within the breast tissue, reducing the skin dose while maintaining adequate image quality.
Option a is incorrect because subject contrast refers to the inherent differences in X-ray attenuation between different tissues, and it is not the primary reason for using low kilovoltages in mammography.
Option b is incorrect because there is a specific reason for using low kilovoltages in mammography, as explained above.
Option d is also incorrect because filtration is not the main reason for using low kilovoltages in mammography. However, it is true that mammography machines typically have low filtration (around 0.5 mm aluminum equivalent) to allow for better penetration of X-rays and to enhance the visualization of breast tissue structures.
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Question 5 (1 point) A 0.02 C charge with a mass of 85.0 g is moving fast creating a magnetic field of 0.02 u T at a point Z which is 0.01 mm away from the charge. At point Z, which field, due to the
The 0.02 C charge, which has a mass of 85.0 g and is travelling quickly, produces a magnetic field of 0.02 T at point Z.
The field at point Z, due to the 0.02 C charge with a mass of 85.0 g moving fast, can be found using the formula below:
The magnetic field due to a charge in motion can be calculated using the following formula:
B = μ₀ × q × v × sin(θ) / (4πr²), where:
B is the magnetic field
q is the charge
v is the velocity
θ is the angle between the velocity and the line connecting the point of interest to the moving charge
μ₀ is the permeability of free space, which is a constant equal to 4π × 10⁻⁷ T m A⁻¹r is the distance between the point of interest and the moving charge
Given values are
q = 0.02 C
v = unknownθ = 90° (since it is moving perpendicular to the direction to the point Z)
r = 0.01 mm = 0.01 × 10⁻³ m = 10⁻⁵ m
Using the formula, B = μ₀ × q × v × sin(θ) / (4πr²)
Substituting the given values, B = (4π × 10⁻⁷ T m A⁻¹) × (0.02 C) × v × sin(90°) / (4π(10⁻⁵ m)²)
Simplifying, B = (2 × 10⁻⁵) v T where T is the Tesla or Weber per square meter
Thus, the magnetic field at point Z due to the 0.02 C charge with a mass of 85.0 g moving fast is 0.02 μT.
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9 7. The radius of the planet is R, and the mass of the planet , measured in meters is M. Micheal Caine is on a location very far from the planet, whearas Anne Hathway is standing on the surface of the planet. If Anne Hathway sees the clock of Micheal Caine, she sees that his clock is ticking N times as fast as her own clock. What is the ration of M/Rs.(6 marks).
This is the ratio of mass to radius for the given planet. This expression cannot be simplified further.Answer:M/R = (N² - 1)/N² * c²/G
Let the speed of Michael Caine's clock be k times that of Anne Hathaway's clock.So, we can write,k
= N .......(1)
Now, using the formula for time dilation, the time dilation factor is given as, k
= [1 - (v²/c²)]^(-1/2)
On solving the above formula, we get,v²/c²
= (1 - 1/k²) .....(2)
As Michael Caine is very far away from the planet, we can consider him to be at infinity. Therefore, the gravitational potential at his location is zero.As Anne Hathaway is standing on the surface of the planet, the gravitational potential at her location is given as, -GM/R.As gravitational potential energy is equivalent to time, the time dilation factor at Anne's location is given as,k
= [1 - (GM/Rc²)]^(-1/2) ........(3)
From equations (2) and (3), we can write,(1 - 1/k²)
= (GM/Rc²)So, k²
= 1 / (1 - GM/Rc²)
We know that, k
= N,
Substituting the value of k in the above equation, we get,N²
= 1 / (1 - GM/Rc²)
On simplifying, we get,(1 - GM/Rc²)
= 1/N²GM/Rc²
= (N² - 1)/N²GM/R
= (N² - 1)/N² * c²/GM/R²
= (N² - 1)/N² * c².
This is the ratio of mass to radius for the given planet. This expression cannot be simplified further.Answer:M/R
= (N² - 1)/N² * c²/G
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Safety brake on saw blade A table saw has a circular spinning blade with moment of inertia 1 (including the shaft and mechanism) and is rotating at angular velocity wo. Some newer saws have a system for detecting if a person has touched the blade and have brake mechanism. The brake applies a frictional force tangent to the rotation, at a distance from the axes. 1. How much frictional force must the brake apply to stop the blade in time t? (Answer in terms of I, w, and T.) 2. Through what angle will the blade rotate while coming to a stop? Give your answer in degrees.
1. The frictional force required to stop the blade in time t is given by Ffriction = wo ÷ r ÷ T.
2. The blade will rotate through an angle of θ = wo² × T × (1 + T × r × I/2) or wo² × T × (1 + 0.5 × T × I × r). And in degrees θ(degrees) = wo² × T × (1 + 0.5 × T × r) × 180/π.
1. The blade must be stopped in time t by a brake that applies a frictional force tangent to the rotation, at a distance r from the axes. The force required to stop the blade is given by the equation;
Ffriction = I × w ÷ r ÷ t
Where,
I = moment of inertia = 1
w = angular velocity = wo
T = time required to stop the blade
Thus;
Ffriction = I × w ÷ r ÷ T
= 1 × wo ÷ r ÷ T
Therefore, the frictional force required to stop the blade in time t is given by Ffriction = wo ÷ r ÷ T.
2. The angle rotated by the blade while coming to a stop can be determined using the equation for angular displacement.
θ = wo × T + 1/2 × a × T²
where,
a = acceleration of the blade
From the equation,
Ffriction = I × w ÷ r ÷ t
a = Ffriction ÷ I
m = 1 × wo ÷ r
θ = wo × T + 1/2 × (Ffriction ÷ I) × T²
θ = wo × T + 1/2 × (wo ÷ r ÷ I) × T²
θ = wo × T + 1/2 × (wo ÷ r) × T²
θ = wo × T + 1/2 × (wo² × T²) ÷ (r × I)
θ = wo × T + 1/2 × wo² × T²
Substitute the values of wo and T in the above equation to obtain the angular displacement;
θ = wo × T + 1/2 × wo² × T²
θ = wo × (wo ÷ r ÷ Ffriction) + 1/2 × wo² × T²
θ = wo × (wo ÷ r ÷ (wo ÷ r ÷ T)) + 1/2 × wo² × T²
θ = wo² × T + 1/2 × wo² × T² × (r × I)
θ = wo² × T × (1 + 1/2 × T × r × I)
θ = wo² × T × (1 + T × r × I/2)
Thus, the blade will rotate through an angle of θ = wo² × T × (1 + T × r × I/2) or wo² × T × (1 + 0.5 × T × I × r).
The answer is to be given in degrees. Therefore, the angular displacement is; θ = wo² × T × (1 + 0.5 × T × I × r)
θ = wo² × T × (1 + 0.5 × T × 1 × r)
= wo² × T × (1 + 0.5 × T × r)
Converting from radians to degrees;
θ(degrees) = θ(radians) × 180/π
θ(degrees) = wo² × T × (1 + 0.5 × T × r) × 180/π.
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