The orbital period of an exoplanet using a light curve is calculated using the length of time between each dip in the light curve, represented by a line that drops below the normal light intensity.
How to solvePlanet | Mass of parent star (relative to sun) | Orbital Period (days) | Distance from parent star (AU) | Distance from parent star (km)
Kepler-5b | 1.37 Ms | 3.55 | 0.05064 | 7,580,000
Kepler-6b | 1.21 Ms | 3.23 | 0.04559 | 6,820,000
Kepler-7b | 1.36 Ms | 4.89 | 0.06250 | 9,350,000
Kepler-8b | 1.21 Ms | 3.52 | 0.04828 | 7,220,000
Kepler-9b | 1.04 Ms | 384.84 | 1.046 | 156,500,000
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displays a 12.0 V battery and 3 uncharged capacitors of capacitances C1 = 4 mu F, C2 = 6 mu F, and C3 = 3 mu F. The switch is thrown to the left side until capacitor 1 is fully charged. Then the switch is thrown to the right. What is the final charge on (a) capacitor 1, (b) capacitor 2, and (c) capacitor 3?
(a) The final charge on capacitor 1 is 48 microcoulombs.
(b) The final charge on capacitor 2 is 36 microcoulombs.
(c) The final charge on capacitor 3 is 24 microcoulombs.
When the switch is thrown to the left side, capacitor 1 charges to 12 volts. Then, when the switch is thrown to the right, capacitors 1 and 2 are in parallel with each other and in series with capacitor 3. The total capacitance in the circuit is 2.4 microfarads. Using the equation [tex]Q = CV,[/tex]where Q is the charge, C is the capacitance, and V is the voltage, the final charge on capacitor 1 is [tex](4/2.4) x 12 = 48[/tex] microcoulombs, on capacitor 2 is [tex](6/2.4) x 12 = 36[/tex] microcoulombs, and on capacitor 3 is[tex](3/2.4) x 12 = 24[/tex] microcoulombs.
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Resistors of 5 ohms and 10 ohms are connected in series with a battery supplying 3 volts. What is the total resistance?
The total resistance of the 5 ohm and 10 ohm resistors connected in series is 15 ohms.
Answer:
Therefore, the total resistance of the circuit is 15 ohms and the current flowing through the circuit is 0.2 amperes.
Explanation:
When resistors are connected in series, their resistances add up to give the total resistance of the circuit. So, in this case, the total resistance of the circuit is:
Total resistance = 5 ohms + 10 ohms = 15 ohms
Now that we know the total resistance of the circuit, we can use Ohm's law to calculate the current flowing through the circuit:
Current = Voltage / Resistance
where Voltage is the voltage of the battery and Resistance is the total resistance of the circuit.
Plugging in the values, we get:
Current = 3 volts / 15 ohms = 0.2 amperes
Therefore, the total resistance of the circuit is 15 ohms and the current flowing through the circuit is 0.2 amperes.
although many blind individuals have difficulty maintaining circadian rhythms, some are able to do so through ___
Although many blind individuals have difficulty maintaining circadian rhythms, some are able to do so through non-visual photoreception. This involves specialized cells in the retina called intrinsically photosensitive retinal ganglion cells (ipRGCs), which can detect light and help regulate the body's internal clock.
Although many blind individuals have difficulty maintaining circadian rhythms, some are able to do so through the use of light therapy. This involves using bright lights in the morning and evening to help regulate sleep-wake cycles and improve overall sleep quality.
Additionally, some blind individuals may also use social cues and routine to help maintain a consistent sleep schedule.
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What is the magnitude and direction of the force exerted on a 3.50 µC charge by a 250 N/C electric field that points due east?
The magnitude of the force exerted on a 3.50 µC charge by a 250 N/C electric field that points due east can be found using the equation F = q E, where F is the force, q is the charge, and E is the electric field.
Plugging in the values, we get:
F = (3.50 × 10^-6 C) × (250 N/C)
F = 0.875 N
Therefore, the magnitude of the force is 0.875 N.
The direction of the force can be found using the right-hand rule. If you point your fingers in the direction of the electric field (due east in this case) and then curl your fingers towards the direction of the charge (which we'll assume is positive), then your thumb will point in the direction of the force. In this case, the force will be pointing upwards (perpendicular to the electric field and the charge's motion). So, the direction of the force is upwards.
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A 14.0-foot-long, nearsighted python is stretched out perpendicular to a plane mirror, admiring its reflected image. If the greatest distance to which the snake can see clearly is 22.0 ft, how close must its head be to the mirror for it to see a clear image of its tail?
______ ft.
The python's head must be [tex]4.0 feet[/tex] away from the mirror to see a clear image of its tail. The answer is [tex]4.0 feet[/tex].
The python's head must be to the mirror to see a clear image of its tail, The concept of the virtual image formed by the mirror.
Given:
Length of the python [tex](L)= 14.0 feet[/tex]
Greatest distance to which the snake can see clearly (distance of clear vision) = [tex]22.0 feet[/tex]
Let's assume that the python's head is at a distance x feet from the mirror.
According to the concept of the virtual image formed by the mirror, the image distance is equal to the object distance:
[tex]d_{(image)} = d_{(object)}[/tex]
The calculation is as follows:
[tex]x = (22.0- 14.0) - x\\2x = 22.0 - 14.0\\x = 8.0 / 2\\x = 4.0\ feet[/tex]
So, the python's head must be [tex]4.0 feet[/tex] away from the mirror to see a clear image of its tail.
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How much work is done by the electric force during the motion of the proton J?
DeltaU =121
V1-35
V2-156
The work done by the electric force during the motion of the proton J is calculated using the formula W = qΔV, where W is the work done, q is the charge of the proton, and ΔV is the potential difference.
ΔV = ΔU = 121V1 - 35V2 - 156
To calculate the work done, first find the potential difference:
ΔV = 121V1 - 35V2 - 156
Then, multiply the potential difference by the charge of a proton (q = 1.602 × 10⁻¹⁹ C):
W = (1.602 × 10⁻¹⁹ C) × (121V1 - 35V2 - 156)
The work done by the electric force during the motion of proton J depends on the values of V1 and V2. Plug in the respective values for V1 and V2 to get the final answer for W.
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what is the water pressure as it exits into the air? express your answer with the appropriate units.
Water pressure varies, but is typically measured in psi or kPa and ranges from a few to several hundred.
The water tension as it exits out of sight relies upon a few variables, including the level of the water source and the size of the opening. The strain can be determined utilizing the Bernoulli condition, which expresses that the tension of a liquid declines as its speed increments.
Expecting a water source at ground level and dismissing any frictional misfortunes, the tension can be approximated as
[tex]P = 0.5rhov^2,[/tex]
where P is the strain in Dad, rho is the thickness of water (1000 [tex]kg/m^3[/tex]), and v is the speed in m/s. For instance, on the off chance that the water is leaving at a speed of 10 m/s, the tension can be determined as P = 0.51000([tex]10^2[/tex]) = 50,000 Dad, or roughly 7.25 psi. Notwithstanding, the genuine strain can fluctuate generally contingent upon the particular conditions.
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The complete question is:
Part A Water flows from the pipe shown in the figure with a speed of 8.0 m/s . (Fiaure 1) What is the water pressure as it exits into the air? Express your answer to two significant figures and include the appropriate units. HA Value Units p= 1 Value Units.
for research purposes a sonic buoy is tethered to the ocean floor and emits an infrasonic pulse of sound (speed = 1522 m/s). the period of this sound is 58 ms. determine the wavelength of the sound.
The wavelength of the sound emitted by the sonic buoy is 88.3 meters
wavelength = speed / frequency
In this case, the speed of sound in water is given as 1522 m/s, and the period (T) of the sound is 58 ms. The period is the time it takes for one complete cycle of the sound wave, and is related to the frequency (f) by the formula:
T = 1/f
Therefore, we can solve for the frequency:
f = 1/T = 1/0.058 s = 17.24 Hz
Now we can use the formula for wavelength:
wavelength = speed / frequency = 1522 m/s / 17.24 Hz = 88.3 m
So the wavelength of the sound emitted by the sonic buoy is 88.3 meters. This sound is considered infrasonic, which means it has a frequency below the range of human hearing (20 Hz).
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The wavelength of the sound emitted by the sonic buoy is 88.3 meters
wavelength = speed / frequency
In this case, the speed of sound in water is given as 1522 m/s, and the period (T) of the sound is 58 ms. The period is the time it takes for one complete cycle of the sound wave, and is related to the frequency (f) by the formula:
T = 1/f
Therefore, we can solve for the frequency:
f = 1/T = 1/0.058 s = 17.24 Hz
Now we can use the formula for wavelength:
wavelength = speed / frequency = 1522 m/s / 17.24 Hz = 88.3 m
So the wavelength of the sound emitted by the sonic buoy is 88.3 meters. This sound is considered infrasonic, which means it has a frequency below the range of human hearing (20 Hz).
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The preexponential and activation energy for the diffusion of chromium in nickel are 1.1x10
−
4
m
2
/s and 272000 J/mol, respectively. At what temperature will the diffusion coeffcient have a value of 1.2x10
−
14
m
2
/s?
The temperature at which the diffusion coefficient has a value of 1.2x10⁻¹⁴ m²/s is 803 K.
What is temperature?Temperature is a physical property of matter that can be measured with thermometers. It is a measure of the average kinetic energy of the particles that make up a substance. Temperature is expressed in degrees Celsius, Fahrenheit, or Kelvin.
The diffusion coefficient can be calculated using the Arrhenius equation:
D = [tex]A\times e ^ {(-Ea/RT)[/tex]
where D is the diffusion coefficient, A is the preexponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature.
In this case, A = 1.1x10⁻⁴ m²/s and Ea = 272000 J/mol. We can rearrange the equation to solve for T:
T = (−Ea/R)⋅ln(D/A)
Plugging in the given values, we get:
T = (−272000 J/mol/8.314 J/mol/K)⋅ln(1.2x10⁻¹⁴ m²/s/1.1x10⁻⁴ m²/s)
T ≈ 803 K
Therefore, the temperature at which the diffusion coefficient has a value of 1.2x10⁻¹⁴ m²/s is 803 K.
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The temperature at which the diffusion coefficient has a value of 1.2x10⁻¹⁴ m²/s is 803 K.
What is temperature?Temperature is a physical property of matter that can be measured with thermometers. It is a measure of the average kinetic energy of the particles that make up a substance. Temperature is expressed in degrees Celsius, Fahrenheit, or Kelvin.
The diffusion coefficient can be calculated using the Arrhenius equation:
D = [tex]A\times e ^ {(-Ea/RT)[/tex]
where D is the diffusion coefficient, A is the preexponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature.
In this case, A = 1.1x10⁻⁴ m²/s and Ea = 272000 J/mol. We can rearrange the equation to solve for T:
T = (−Ea/R)⋅ln(D/A)
Plugging in the given values, we get:
T = (−272000 J/mol/8.314 J/mol/K)⋅ln(1.2x10⁻¹⁴ m²/s/1.1x10⁻⁴ m²/s)
T ≈ 803 K
Therefore, the temperature at which the diffusion coefficient has a value of 1.2x10⁻¹⁴ m²/s is 803 K.
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11. let ∼ be an equivalence relation on a. prove that if a ∼b, then [a] = [b].
If a ∼ b, then [a] = [b] by the definition of equivalence relation.
An equivalence relation is a relation that satisfies three properties: reflexivity, symmetry, and transitivity. If a ∼ b, then by the reflexivity property, a is related to itself, which implies that a is in the equivalence class [a].
Similarly, by the symmetry property, if a is related to b, then b is related to a, which implies that b is also in the equivalence class [a]. Therefore, [a] contains both a and b.
Now, using the transitivity property, since a is related to b and b is related to a, it follows that a is related to a, which implies that a is in the equivalence class [b]. Therefore, [a] = [b].
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consider an infinite sheet of parallel wires. the sheet lies in the xy plane. a current i runs in the -y direction through each wire. there are n/a wires per unit length in the x direction.
The magnetic field is proportional to the current and inversely proportional to the number of wires per unit length in the x direction.
The magnetic field produced by an infinite sheet of parallel wires can be determined using Ampere's Law. Since the current is running in the -y direction through each wire, the magnetic field lines will circulate around each wire in the clockwise direction when viewed from above. The magnitude of the magnetic field at a point above the sheet will depend on the distance from the sheet, as well as the number of wires per unit length in the x direction.
Using Ampere's Law, the integral of the magnetic field around a closed loop will be equal to μ₀ times the current enclosed by the loop. For a rectangular loop with sides of length L and H, the magnetic field along the sides parallel to the wires will be constant and equal to μ₀ times the current per unit length (i/n) times the width of the loop (L), while the field along the sides perpendicular to the wires will be zero. Thus, the integral of the magnetic field around the loop will be 2 times the magnetic field along one of the parallel sides, or 2μ₀(i/n)L.
Setting this equal to μ₀ times the current enclosed by the loop (iLH), we can solve for the magnetic field at a point above the sheet:
B = μ₀i/2n
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Solenoids and Toroids If a current is 2.0 A, how many turns per centimeter must be wound on a solenoid in order to produce a magnetic field of within it?
318 turns per centimeter must be wound on a solenoid in order to produce a magnetic field of within it If a current is 2.0 A.
To determine the number of turns per centimeter needed to produce a magnetic field within a solenoid with a current of 2.0 A, we need to know the desired strength of the magnetic field. Additionally, it's important to note that solenoids are cylindrical coils of wire that produce a magnetic field when a current passes through them. Toroids, on the other hand, are donut-shaped coils of wire that also produce a magnetic field.
Assuming we want a magnetic field strength of 1 tesla within the solenoid, we can use the equation
B = μ[tex]_0[/tex] × n × I
where B is the magnetic field strength, μ[tex]_0[/tex] is the permeability of free space (4π x [tex]10^-^7[/tex]Tm/A), n is the number of turns per unit length, and I is the current.
Rearranging this equation to solve for n, we get n = B / (μ[tex]_0[/tex] × I).
Plugging in the values given, we get
n = (1 T) / (4π x [tex]10^-^7[/tex] Tm/A × 2.0 A) = 3.18 x [tex]10^6[/tex] turns/meter.
To convert this to turns per centimeter, we divide by 100, which gives us 3.18 x [tex]10^4[/tex] turns/cm.
Therefore, to produce a magnetic field of 1 tesla within a solenoid with a current of 2.0 A, we need to wind approximately 31,800 turns per meter, or 318 turns per centimeter.
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a 2.0 kg-ball moving at 3.0 m/s perpendicular to a wall rebounds from the wall at 2.5 m/s. the change in the momentum of the ball is
The change in momentum of the ball after it rebounds from the wall at 2.5 m/s is -1.0 kg.m/s.
The change in the momentum of the ball can be calculated using the formula:
change in momentum = final momentum - initial momentum
To find the initial momentum, we multiply the mass of the ball (2.0 kg) by its initial velocity (3.0 m/s):
initial momentum = 2.0 kg × 3.0 m/s = 6.0 kg*m/s
To find the final momentum, we multiply the mass of the ball (2.0 kg) by its final velocity (2.5 m/s):
final momentum = 2.0 kg × 2.5 m/s = 5.0 kg*m/s
Therefore, the change in momentum of the ball is:
change in momentum = final momentum - initial momentum
change in momentum = 5.0 kg*m/s - 6.0 kg*m/s
change in momentum = -1.0 kg*m/s
The negative sign indicates that the momentum of the ball decreased during the collision with the wall.
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according to faraday’s law and lenz’s law, what should happen to the current in a coil of wire when the north pole of a bar magnet is moved toward it?
The current in the coil of wire should be induced, flowing in a direction that opposes the motion of the magnet.
According to Faraday's Law, when a magnetic field is changed, an electromotive force (EMF) is induced in a nearby conductor. Lenz's Law states that the direction of the induced current creates a magnetic field that opposes the change in magnetic flux that induced it. Therefore, as the north pole of the magnet approaches the coil, a magnetic flux is created and the induced current flows in a direction that produces a magnetic field in the opposite direction. This opposes the motion of the magnet, slowing it down. When the magnet is moved away, the opposite happens and the induced current flows in the opposite direction.
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What charge is stored in a 180 µF capacitor when 120 V is applied to it?
The charge stored in the 180 µF capacitor when 120 V is applied to it is 0.0216 coulombs
The charge stored in a 180 µF capacitor when 120 V is applied to it can be calculated using the formula Q = CV, where Q is the charge stored, C is the capacitance in farads, and V is the voltage applied. Plugging in the given values, we get [tex]Q = (180 * 10^(-6) F)[/tex] x (120 V) = 21.6 µC (microcoulombs). Therefore, 21.6 µC of charge is stored in the capacitor.
To find the charge stored in a 180 µF capacitor when 120 V is applied to it, we can use the formula:
Q = C × V
Where:
Q = charge stored (in coulombs),
C = capacitance (in farads),
V = voltage applied (in volts).
Step 1: Convert the capacitance to farads:
[tex]180 µF = 180 * 10^(-6) F = 0.00018 F[/tex]
Step 2: Plug the capacitance and voltage values into the formula:
Q = 0.00018 F * 120 V
Step 3: Calculate the charge stored:
Q = 0.0216 C
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A vertical column load, P = 600 kN, is applied to a rigid concrete foundation with dimensions B = 1 m and L = 2 m. The foundation rests at a depth Df = 0.75 m on a uniform dense sand with the following properties: average modulus of elasticity, Es = 20,600 kN/m2, and Poisson’s ratio, μs = 0.3. Estimate the elastic settlement due to the net applied pressure, Δσ, on the foundation. Given: H = 5 m.
The estimated elastic settlement due to the net applied pressure of 600 kN on the foundation is 1.86 mm.
To estimate the elastic settlement due to the net applied pressure, Δσ, on the foundation, we can use the following equation:
Δs = (Δσ / Es) * ((1 - μs) / (1 + μs)) * (B / (2 * (Df + H)))
Where Δs is the elastic settlement, Δσ is the net applied pressure, Es is the average modulus of elasticity of the sand, μs is the Poisson's ratio of the sand, B is the width of the foundation, Df is the depth of the foundation, and H is the height of the sand layer.
Substituting the given values, we get:
Δs = (600 / 20600) * ((1 - 0.3) / (1 + 0.3)) * (1 / (2 * (0.75 + 5))) = 0.00186 m or 1.86 mm
Therefore, the estimated elastic settlement due to the net applied pressure of 600 kN on the foundation is 1.86 mm.
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the depth of water behind the hoover dam in nevada is 145 m. what is the water pressure at a depth of 145 m? the weight density of water is 9800 n/m3 . answer in units of n/m2 .
The water pressure at a depth of 145 m behind the Hoover Dam in Nevada is 13,940,010 N/m²
To find the water pressure at a depth of 145 m behind the Hoover Dam in Nevada, we will use the water pressure formula, which includes the terms "water pressure" and "density".
The water pressure formula is: P = ρgh
Where:
P = water pressure (in N/m²)
ρ = density of water (in N/m³)
g = acceleration due to gravity (9.81 m/s²)
h = depth of water (in meters)
Given the weight density of water is 9800 N/m³, and the depth (h) is 145 m:
Step 1: Plug in the given values into the formula:
P = (9800 N/m³)(9.81 m/s²)(145 m)
Step 2: Multiply the values:
P = 9800 x 9.81 x 145
Step 3: Calculate the final result:
P = 13,940,010 N/m²
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how many grams of lithium (atomic mass of 6.91 g/mol) are in a lithium-ion battery that produces 4.00 a·h of electricity?
There are approximately 0.0000413 grams of lithium in a lithium-ion battery that produces 4.00 a·h of electricity.
grams of lithium = (4.00 a·h) x (1 mole of electrons / 96,485 C) x (1 mole of lithium / 1 mole of electrons) x (6.91 g / 1 mole of lithium)
where:
- 4.00 a·h is the amount of electricity produced by the battery
- 96,485 C is the Faraday constant, which relates the amount of electricity to the number of electrons involved in the reaction
- 1 mole of electrons is the number of electrons that flow through the battery during the reaction
- 1 mole of lithium is the amount of lithium involved in the reaction
- 6.91 g is the atomic mass of lithium, which tells us how many grams are in one mole of the element
Plugging in the numbers, we get:
grams of lithium = (4.00 a·h) x (1 mole of electrons / 96,485 C) x (1 mole of lithium / 1 mole of electrons) x (6.91 g / 1 mole of lithium)
grams of lithium = (4.00 a·h / 96,485 C) x 6.91 g
grams of lithium = 0.0000413 g
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Inside a nuclear power plant, energy is liberated as nuclear reactions proceed inside the core. As this happens, the mass of the nuclei
Decreases.
Stays the same.
Increases.
Inside a nuclear power plant, energy is liberated as nuclear reactions proceed inside the core. As this happens, the mass of the nuclei decreases.
This decrease in mass occurs because some of the mass is converted into energy according to Einstein's famous equation, E=mc², where E is the energy, m is the mass, and c is the speed of light. Nuclear reactions are processes in which one or more nuclides are produced from the collisions between two atomic nuclei or one atomic nucleus and a subatomic particle. The nuclides produced from nuclear reactions are different from the reacting nuclei (commonly referred to as the parent nuclei). E = mc2 is the key to understanding why and how energy is released in nuclear reactions. Two concepts are central to both nuclear fission and fusion: First, the mass of a nucleus is less than the sum of the masses the nucleons would have if they were free. This is called the mass defect.
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a power pack charging cell phone battery has an output of 0.40A at 5.2 V (both are rms). how do I find the rms current at the 120 V/60Hz wall outlet where the power pack is plugged in?
The RMS current at the 120 V/60Hz wall outlet where the power pack is plugged in is 17.3 mA.
To find the RMS current at the 120 V/60Hz wall outlet where the power pack is plugged in, first, determine the power consumed by the charging cell phone battery. Power (P) is calculated using the formula P = VI, where V is the voltage and I is the current.
In this case, the power pack output is 0.40A (RMS current) and 5.2V (RMS voltage). Therefore, the power consumed by the charging cell phone battery is:
P = (0.40A) × (5.2V)
= 2.08 watts
Now, assume the power pack is 100% efficient (which is not true in reality, but it simplifies the calculation), the same amount of power will be drawn from the 120V/60Hz wall outlet. Using the power formula again, rearrange it to find the RMS current at the wall outlet:
I = P / V
Where V is the voltage at the wall outlet (120V) and P is the power (2.08 watts). The RMS current at the wall outlet is:
I = 2.08 watts / 120V
≈ 0.0173A or 17.3 mA
So, the RMS current at the 120V/60Hz wall outlet where the power pack is plugged in is approximately 17.3 mA.
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at a distance of 5 km from a radio transmitter the amplitude of electric field strength is measured to be 0.35 v/m. what is the total power emitted by the transmitter?
The total power emitted by the transmitter is approximately 3.802 kW.
How much the total power emitted by the transmitter?The total power emitted by a radio transmitter can be calculated using the formula:
P = 4πr²⁰
where P is the total power emitted, r is the distance from the transmitter, and σ is the power density (in W/m²) at that distance.
First, we need to calculate the power density at a distance of 5 km from the transmitter. The electric field strength (E) is related to the power density (σ) by the formula:
E = sqrt(2σ/μ0)
where μ0 is the permeability of free space (4π × 10⁻⁷ H/m).
Solving for σ, we get:
σ = (E²μ⁰)/2
σ = (0.35 V/m)² × 4π × 10⁻⁷ H/m
σ = 1.221 × 10⁻⁹ W/m²
Now that we have the power density at 5 km, we can calculate the total power emitted by the transmitter:
P = 4πr²⁰
P = 4π(5 km)² × 1.221 × 10⁻⁹ W/m²
P = 3.802 kW (kilowatts)
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a soccer ball with mass 0.440 kg is initially moving with speed 2.30 m/s. a soccer player kicks the ball, exerting a constant force of magnitude 43.0 n in the same direction as the ball's motion.Over what distance must her foot be in contact with the ball to increase the ball's speed to 6.00m/s ?
The distance required for the foot to be in contact with the ball in order to increase its speed from 2.30 m/s to 6.00 m/s is 2.42 meters.
How to find distance required for the foot to be in contact with the ball?We can use the equation for work done by a constant force, which is given by:
W = Fd cos(θ )
where W is the work done, F is the force applied, d is the distance over which the force is applied, and θ is the angle between the force and the displacement.
In this case, the force is applied in the same direction as the displacement, so θ = 0. Therefore, we can simplify the equation to:
W = Fd
We can also use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy:
W = (1/2)mv2 - (1/2)mv1
where m is the mass of the object, v1 is its initial velocity, and v2 is its final velocity.
Combining these equations, we have:
Fd = (1/2)mv2 - (1/2)mv1
Solving for d, we get:
d = (1/2F)mv2 - (1/2F)mv1
Plugging in the given values, we get:
d = (1/2 x 43.0 N) x 0.440 kg x (6.00 m/s - 2.30 m/s)
d = 2.42 m
Therefore, the foot must be in contact with the ball over a distance of 2.42 meters to increase the ball's speed to 6.00 m/s.
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a 3.77 uf capacitor is connected to a 240v ac source with a frequency of 447 hz. What is the rms current in the capacitor?
A 3.77 [tex]\mu f[/tex] capacitor is connected to a 240v ac source with a frequency of 447 hz. The rms current in the capacitor is 0.224 A.
Plugging in the given values, we have:
Xc = 1 / (2π * 447 Hz * 3.77 μF) = 758.1 Ω
Now, we need to calculate the root-mean-square voltage of the AC source, which is given by:
Vrms = Vpeak / √2
where Vpeak is the peak voltage of the AC source, which is given by:
Vpeak = 240 V
So, we have:
Vrms = 240 V / √2 = 169.7 V
Finally, we can calculate the rms current in the capacitor using the formula:
Irms = Vrms / Xc = 169.7 V / 758.1 Ω = 0.224 A (rounded to three significant figures)
Frequency refers to the number of times that a particular event occurs within a given time frame. It can be used to describe a wide range of phenomena, from the number of times a particular word appears in a text to the number of waves that pass a particular point in a second. In physics, frequency is often used to describe the rate at which a wave oscillates, which is measured in Hertz (Hz). This is important in fields such as acoustics and optics, where the frequency of a wave determines its pitch or color, respectively.
Frequency is also an important concept in statistics, where it is used to describe the distribution of values within a dataset. The frequency of a particular value refers to the number of times that value occurs within the dataset. This information can be used to create frequency distributions, which provide a visual representation of how often different values appear within a dataset.
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An experimentalist claims to have raised the temperature of a small amount of water to 150C by transferring heat from high-pressure steam at 120C. Is this a reasonable claim? Why? Assume no refrigerator or heat pump is used in the process
When compared to its source, which is at 120 degrees Celsius, the water's temperature rises to 150 degrees Celsius. 120 ∘ C . This is an infraction of the second law of thermodynamics, which states that heat cannot be moved from a low to a high temperature without producing an outside impact, such as a heat pump.
What exactly are reservoirs of thermal energy?When a significant amount of heat is added to or removed from a thermal reservoir, also known as a thermal energy reservoir or thermal bath, the temperature of the reservoir varies only slightly.
What is thermal energy, and how can it be used?Molecules moving within an object or substance are said to be moving with thermal energy. The thermal energy of any material or item.
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Laser light of wavelength 632.8 nmnm falls normally on a slit that is 0.0220 mmmm wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.70 W/m2W/m2Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all?.
There are approximately 34 dark fringes on either side of the central bright fringe, for a total of 68 dark fringes.
To find the maximum number of totally dark fringes on the screen, we can use the formula:
n = (w/d) * (D/λ)
Where n is the number of fringes, w is the width of the slit, d is the distance from the slit to the screen, D is the distance from the slit to the light source, and λ is the wavelength of the laser light.
Plugging in the given values, we get:
n = (0.0220 mm / 1) * (1 / 632.8 nm)
n = 34.37
This means there are approximately 34 dark fringes on either side of the central bright fringe, for a total of 68 dark fringes. However, this assumes that the screen is large enough to show all the fringes. If the screen is too small, some of the fringes may not be visible.
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a force of 600n compress a spring by 0.5m. what is the spring constant for this spring
The spring constant can be calculated by using the formula: Spring Constant = Force / Compression. Plugging in the given values we get the spring constant for this spring is 1200 N/m.
Hooke's Law states that the force exerted by a spring is proportional to its displacement from its equilibrium position, which is given by the equation: F = -k * x Where: - F is the force applied to the spring (in Newtons) - k is the spring constant (in N/m) - x is the displacement of the spring from its equilibrium position (in meters).
In this problem, you are given a force (F) of 600 N and a displacement (x) of 0.5 m.
Then find the spring constant (k).
To do this, we'll rearrange the formula to solve for k: k = F / x
Now, we can plug in the values:
k = 600 N / 0.5 m
k = 1200 N/m
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Calculate the rotational inertia of a meter stick, with mass 0.78 kg, about an axis perpendicular to the stick and located at the 29 cm mark. (Treat the stick as a thin rod.) ____ kg. m^2 From the table of some rotational inertias, determine the rotational inertia for a thin rod about the center. Then use the parallel-axis theorem. How far is the rotation axis shifted from the center of the rod?
The rotating axis is 0.5 metres away from the rod's centre.
To calculate the rotational inertia of the meter stick about an axis perpendicular to the stick and located at the 29 cm mark, we can use the formula for the rotational inertia of a thin rod:
I = (1/3) * M * [tex]L^2[/tex],
where I is the rotational inertia, M is the mass of the rod, and L is the length of the rod.
Given that the mass of the meter stick is 0.78 kg and its length is 1 meter (100 cm), we can substitute these values into the formula:
I = (1/3) * 0.78 kg * [tex](100 cm)^2.[/tex]
Simplifying the equation gives:
I = 2600kg.[tex]cm^2[/tex].
Converting the units to kg·[tex]m^2[/tex], we divide by 10,000:
I = 0.26 kg·[tex]m^2[/tex].
So, the rotational inertia of the meter stick about the axis located at the 29 cm mark is 0.26 kg·[tex]m^2[/tex].
To determine the rotational inertia for a thin rod about its center, we can use the formula:
I_center = (1/12) * M *[tex]L^2,[/tex]
where I_center is the rotational inertia about the center. Using the same mass (0.78 kg) and length (1 meter), we substitute these values into the formula:
I_center = (1/12) * 0.78 kg * [tex](100 cm)^2.[/tex]
Simplifying the equation gives:
I_center = 650 kg·[tex]cm^2.[/tex]
Converting the units to kg·[tex]m^2,[/tex]we divide by 10,000:
I_center = 0.065 kg·[tex]m^2.[/tex]
According to the parallel-axis theorem, the rotational inertia about an axis parallel to and displaced a distance 'd' from the center is given by:
I_displaced = I_center + M *[tex]d^2.[/tex]
We know that I_displaced is equal to 0.26 kg·[tex]m^2[/tex](from the previous calculation). Substituting the values into the equation:
0.26 kg·[tex]m^2[/tex]= 0.065 kg·[tex]m^2[/tex]+ 0.78 kg * [tex]d^2.[/tex]
Rearranging the equation, we get:
0.195 kg·[tex]m^2[/tex] = 0.78 kg * [tex]d^2.[/tex]
Solving for 'd', we have:
d^2 = 0.195 kg·[tex]m^2[/tex] / 0.78 kg.
d^2 = 0.25[tex]m^2.[/tex]
Taking the square root of both sides:
d = 0.5 m.
Therefore, the rotation axis is shifted 0.5 meters from the center of the rod.
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A hollow copper wire with an inner diameter of 1.1 mm and an outer diameter of 1.8 mm carries a current of 15 A. What is the current density in the wire? Please show work. I got 3.9 *10^7 A/m^2, but I was wrong. Thanks in advance.
The current density is:9.41*10⁶ A/m²
Current density in wire is :
J= I/A
J: current density
I: Current
A: cross sectional area of wire
cross sectional area of wire= Π [{(1.8*10⁻³)/2}² - {(1.1*10⁻³)/2}]²
= 1.59*10⁻⁶ m²
hence, J= 15/1.59*10⁻⁶ = 9.41*10⁶ A/m²\
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A 2 GHz radar antenna of effective area 6.0 m2 transmits 100 kW. If a target with a 12 m2 radar cross section is 100 km away, (a) what is the round-trip travel time for the return radar pulse (b) what is the received power (c) what is the maximum detectable range if the radar system has a minimum detectable power of 2.0 pW. (1 pw = 10-12 W)
A. The round-trip travel time of the radar pulse is 6.7 x 10⁻⁴ s.
B. The received power is 1.6 x 10⁻⁵ W.
C. The maximum detectable range of the radar system is 16.2 km.
The round-trip travel time for the return radar pulse can be calculated using the formula for the speed of light, c = 3 x 10⁸ m/s, and the distance of 100 km, as t = 2 x 100 km/3 x 10⁸ m/s = 6.7 x 10⁻⁴ s.
The received power can be calculated by using the radar equation, Pr = PtxGtxAe/4πr², where Pr is the received power, Ptx is the transmitted power of 100 kW, Gtx is the antenna gain, Ae is the effective area of 6.0 m2, r is the distance of 100 km, and 4π is a constant. Substituting these values gives Pr = 1.6 x 10⁻⁵ W.
The maximum detectable range can be calculated using the minimum detectable power, Pmin, of 2.0 pW and the radar equation, as rmax = sqrt(PtxGtxAe/4πPmin). Substituting these values in the equation gives rmax = 16.2 km.
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a force f = (-30, 50) n acts on a mass of 10 kg. at time t = 0 s, the mass has a velocity v0 = (-2, -5) m/s. what are the (x,y) components of the velocity vector (in m/s) after 4 seconds?
We'll add the initial velocity (v0 = (-2, -5) m/s) to the result: vf = v0 + a * t = (-2, -5) m/s + (-12, 20) m/s = (-14, 15) m/s So, the (x, y) components of the velocity vector after 4 seconds are (-14, 15) m/s.
To find the (x,y) components of the velocity vector after 4 seconds, we need to use the formula:
v = v0 + (f/m)t
where v is the final velocity, v0 is the initial velocity, f is the force acting on the mass, m is the mass, and t is the time elapsed.
Plugging in the given values, we have:
f = (-30, 50) N
m = 10 kg
v0 = (-2, -5) m/s
t = 4 s
To find the x-component of the velocity vector, we can use:
vx = v0x + (fx/m)t
where vx is the x-component of the velocity vector, v0x is the initial x-component of the velocity, fx is the x-component of the force, and t is the time elapsed.
Plugging in the values, we have:
vx = -2 + (-30/10) x 4
vx = -2 - 12
vx = -14 m/s
To find the y-component of the velocity vector, we can use:
vy = v0y + (fy/m)t
where vy is the y-component of the velocity vector, v0y is the initial y-component of the velocity, fy is the y-component of the force, and t is the time elapsed.
Plugging in the values, we have:
vy = -5 + (50/10) x 4
vy = -5 + 20
vy = 15 m/s
Therefore, the (x,y) components of the velocity vector (in m/s) after 4 seconds are (-14, 15).
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