When the switch in the circuit is closed after being open for a long time, the circuit becomes steady, and a current of
i = ϵ / (R1 + R2) flows through the circuit. the charge stored on the capacitor a long time after the switch is closed is 85.75 μC. Answer: 85.75 μC.
The charge stored on the capacitor is given by the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.
Let's first calculate the voltage across the capacitor. Since the switch has been open for a long time, the capacitor would have been discharged and would act as a short circuit. Therefore, the voltage across the capacitor after the switch is closed is given by the following equation:
Vc = (R2 / (R1 + R2)) * ϵ
= (6 / 22) * 9
= 2.45V
Now, using the formula Q = CV, we can calculate the charge stored on the capacitor.
Q = C * Vc
= 35 * 10^-6 * 2.45
= 85.75 μC
Therefore, the charge stored on the capacitor a long time after the switch is closed is 85.75 μC. Answer: 85.75 μC.
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For a certain diatomic species, the first two lines of the R
branch appear at 8.7129 x 1013 Hz and 8.7715 x 1013 Hz. Determine
the position of the band gap.
The position of the band gap for the diatomic species is approximately 5.875 x [tex]10^{11[/tex]Hz. To determine the position of the band gap, we need to calculate the frequency difference between the two lines of the R branch. The band gap corresponds to the energy difference between two electronic states in the diatomic species.
The frequency difference can be calculated using the formula:
Δν = ν₂ - ν₁
where Δν is the frequency difference, ν₁ is the frequency of the lower-energy line, and ν₂ is the frequency of the higher-energy line.
Given the frequencies:
ν₁ = 8.7129 x [tex]10^{13[/tex] Hz
ν₂ = 8.7715 x [tex]10^{13[/tex] Hz
Let's calculate the frequency difference:
Δν = 8.7715 x [tex]10^{13[/tex] Hz - 8.7129 x [tex]10^{13[/tex] Hz
Δν ≈ 5.875 x[tex]10^{11[/tex] Hz
Therefore, the position of the band gap for the diatomic species is approximately 5.875 x [tex]10^{11[/tex]Hz.
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quick answer please
QUESTION 7 4 points Sove a A conducting wire loop of radius 12 cm, that contains a 4.0-0 resistor, is in the presence of a uniform magnetic field of strength 3.0 T that is perpendicular to the plane o
The magnitude of the current induced in the conducting wire loop is 0.003375 A.
The magnitude of the current induced in the conducting wire loop can be determined using Faraday's law of electromagnetic induction. According to Faraday's law, the magnitude of the induced emf in a closed conducting loop is equal to the rate of change of magnetic flux passing through the loop. In this case, the magnetic field is uniform and perpendicular to the plane of the loop.
Therefore, the magnetic flux is given by:
φ = BA
where B is the magnetic field strength and A is the area of the loop.
Since the loop is circular, its area is given by:
A = πr²
where r is the radius of the loop. Thus,
φ = Bπr²
Using the given values,
φ = (3.0 T)(π)(0.12 m)² = 0.0135 Wb
The induced emf is then given by:
ε = -dφ/dt
Since the magnetic field is constant, the rate of change of flux is zero. Therefore, the induced emf is zero as well. However, when there is a resistor in the loop, the induced emf causes a current to flow through the resistor.
Using Ohm's law, the magnitude of the current is given by:
I = ε/R
where R is the resistance of the resistor. Thus,
I = (0.0135 Wb)/4.0 Ω
I = 0.003375 A
This is the current induced in the loop.
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Which of the following statemtents is inconsistent with the second law of thermodynamics? a. The entropy of the universe tends to increase. b. Perpetual motion machines are impossible. c. The arrow of time moves in the forward direction. d. A refrigerator cycle is a spontaneous process. e. Heat spontaneously flows from high temperature to low temperature regions.
The statement that is inconsistent with the second law of thermodynamics is “A refrigerator cycle is a spontaneous process.”Why is it inconsistent with the second law of thermodynamics?The second law of thermodynamics states that heat naturally flows from hotter objects to colder objects.
The other statements listed are consistent with the second law of thermodynamics. For example, the entropy of the universe always tends to increase. Entropy is a measure of disorder or randomness. The universe’s entropy is constantly increasing because it is moving from a state of order to a state of disorder, in which everything becomes evenly distributed. Perpetual motion machines, which produce more energy than they consume, are impossible because they violate the second law of thermodynamics.
The arrow of time moves in the forward direction because the universe is always moving towards disorder, not order. Heat naturally flows from high temperature to low temperature regions due to the second law of thermodynamics.
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wo narrow, parallel slits separated by 0.850 mm are illuminated by 570−nm light, and the viewing screen is 2.60 m away from the slits. (a) What is the phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe? rad (b) What is the ratio of the intensity at this point to the intensity at the center of a bright fringe? I max I =
Answer:The phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 1.31 radians.The ratio of the intensity at this point to the intensity at the center of a bright fringe is I_max/I = 1.90 or I = 1.90 I_max.
(a) The phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 1.31 radian.
We can use the formula:δ = (2π/λ)dsinθFor a bright fringe, the angle θ is very small, so we can use the approximation sinθ = θ, where θ is in radians.
δ = (2π/λ)dsinθ
= (2π/570 x 10⁻⁹ m) x 0.850 x 10⁻³ m x (2.50 x 10⁻³ m/2.60 m)
= 1.31 radian
(b) The ratio of the intensity at this point to the intensity at the center of a bright fringe is
Imax/I = cos²(δ/2)
= cos²(0.655)
= 0.526.
Therefore, I/Imax = 1.90 or
I = 1.90 I max.
More explanation:Two narrow parallel slits separated by 0.850 mm are illuminated by 570−nm light and the screen is 2.60 m away from the slits.
Let the angle between the central bright fringe and the point be θ.The phase difference between the two waves at the point on the screen is given by
δ = (2π/λ)dsinθ
We can assume that sinθ is approximately equal to θ in radians because the angle is very small.From the equation given above, we know that
δ = (2π/λ)dsinθ
We have the values as
λ = 570−nm
= 570 x 10⁻⁹ m.
θ = (2.50 mm/2.60 m)
= 2.50 x 10⁻³ m.
From the above equation, we can get the value ofδ = 1.31 radians.The intensity at a distance x from the center of the central bright fringe is given by:
I = I_max cos²πd sinθ/λ
Where d is the separation of the slits and I_max is the intensity of the bright fringe at the center.
From the equation given above, we know thatI = I_max cos²πd sinθ/λ We have the values as
d = 0.850 mm
= 0.850 x 10⁻³ m,
λ = 570−nm
= 570 x 10⁻⁹ m and
θ = (2.50 mm/2.60 m)
= 2.50 x 10⁻³ m.
On substituting the values in the equation, we get,I/I_max = 0.526.
Therefore, I_max/I = 1.90 or
I = 1.90 I_max.
Therefore,The phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 1.31 radians.The ratio of the intensity at this point to the intensity at the center of a bright fringe is I_max/I = 1.90 or
I = 1.90 I_max.
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Instruction: Indicate in the space provided whether the statement is true or false. If the statement is
false, change the underlined words) to make the statement true. 1. The direction of the current is the same as the flow of the negative charges.
2. The electric field inside a conductor is zero if the charges are already in motion.
3. It is possible to allow current to flow from lower potential to higher potential through the
influence of an electromotive force.
4. The amount of current flowing per unit area increases when the electric field on that area
increases.
1. False. The direction of the current is the opposite of the flow of the negative charges.
2. True. The electric field inside a conductor is zero if the charges are already in motion.
3. False. It is impossible to allow current to flow from lower potential to higher potential through the influence of an electromotive force.
4. True. The amount of current flowing per unit area increases when the electric field on that area increases.
An electric current is a flow of electric charge. It is measured in amperes (A). Electric current flows in conductors, which are materials that allow charges to move freely. The movement of electrons in a conductor causes an electric current to flow.
1. False. The direction of the current is the opposite of the flow of the negative charges.
2. True. The electric field inside a conductor is zero if the charges are already in motion.
3. False. It is impossible to allow current to flow from lower potential to higher potential through the influence of an electromotive force.
4. True. The amount of current flowing per unit area increases when the electric field on that area increases.
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Consider a ray of light passing between two mediums, as shown in the figure. The distance h between points A and B is 2.00 cm. Assume the index of refraction ni in medium 1 is 1.00. Medium 1 n = 1.00 45 Medium 2 А n, = ? h B C Determine the index of refraction nz for medium 2 if the distance d between points B and C in the figure is 0.950 cm. n2 = If instead n2 = 1.54, calculate the distance d between points B and C. d = cm
1. The index of refraction, n₂ for medium 2 is 1.65
2. The distance, d between points B and C is 0.984 cm
1. How do i determine the index of refraction, n₂ for medium 2?First, we shall obtain the angle in medium 2. Details below:
Opposite (d) = 0.950 cmAdjacent (h) = 2 cmAngle θ = ?Tan θ = Opposite / Adjacent
Tan θ = 0.95 / 2
Take the inverse of Tan
θ = Tan⁻¹ (0.95 / 2)
= 25.4°
Finally, we shall obtain the index of refraction, n₂ for medium 2. Details below:
Index of refraction for medium 1 (n₁) = 1Angle of medium 1 (θ₁) = 45°Angle of refraction (θ₂) = 25.4°Index of refraction for medium 2 (n₂) =?n₁ × Sine θ₁ = n₂ × Sine θ₂
1 × Sine 45 = n₂ × Sine 25.4
Divide both sides by Sine 25.4
n₂ = (1 × Sine 45) / Sine 25.4
= 1.65
Thus, the index of refraction, n₂ for medium 2 is 1.65
2. How do i determine the distance, d between points B and C?First, we shall obtain the angle in medium 2. Details below:
Index of refraction for medium 1 (n₁) = 1Angle of medium 1 (θ₁) = 45°Index of refraction for medium 2 (n₂) = 1.6Angle of medium 2 (θ₂) =?n₁ × Sine θ₁ = n₂ × Sine θ₂
1 × Sine 45 = 1.6 × Sine θ₂
Divide both sides by 1.6
Sine θ₂ = (1 × Sine 45) / 1.6
Sine θ₂ = 0.4419
Take the inverse of Sine
θ₂ = Sine⁻¹ 0.4419
= 26.2°
Finally, we shall obtain the distance, d. Details below:
Angle θ = 26.2°Adjacent (h) = 2 cmOpposite = Distance (d) =?Tan θ = Opposite / Adjacent
Tan 26.2 = d / 2
Cross multiply
d = 2 × Tan 26.2
= 0.984 cm
Thus, the distance, d is 0.984 cm
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Complete question:
See attached photo
Given a 32.0 V battery and 20.00 and 72.00 resistors, find the current (in A) and power (in W) for each when connected in series.
The answer is power dissipated across the resistor with resistance R1 is 2.42 W, and the power dissipated across the resistor with resistance R2 is 8.62 W.
Potential difference V = 32V Resistance R1 = 20.00Ω Resistance R2 = 72.00Ω. The two resistors are connected in series. Total resistance in the circuit is given by R = R1 + R2 = 20.00 Ω + 72.00 Ω = 92.00 Ω
Current I in the circuit can be calculated as, I = V/R= 32V/92.00 Ω= 0.348A
Power P dissipated across the resistor can be calculated as P = I²R= 0.348² × 20.00 Ω = 2.42 W
The power dissipated across the resistor with resistance R2 is, P2 = I²R2= 0.348² × 72.00 Ω = 8.62 W
Therefore, the current through the circuit is 0.348 A.
The power dissipated across the resistor with resistance R1 is 2.42 W, and the power dissipated across the resistor with resistance R2 is 8.62 W.
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A solid sphere (I = 2MR2/5) rolls without slipping down a plane inclined at 29◦ relative to horizontal. What type of friction acts and what is the coefficient of friction? The answers are rounded to two significant digits.
The answers are rounded to two significant digits:* Type of friction: rolling friction* Coefficient of friction: 0.02
The type of friction that acts on a solid sphere rolling without slipping down a plane inclined at 29° relative to horizontal is rolling friction. Rolling friction is a type of friction that occurs when two surfaces are in contact and one is rolling over the other.
It is much less than static friction, which is the friction that occurs when two surfaces are in contact and not moving relative to each other.
The coefficient of rolling friction is denoted by the Greek letter mu (μ). The coefficient of rolling friction is always less than the coefficient of static friction.
The exact value of the coefficient of rolling friction depends on the materials of the two surfaces in contact.
For a solid sphere rolling without slipping down a plane inclined at 29° relative to horizontal, the coefficient of rolling friction is approximately 0.02. This means that the force of rolling friction is approximately 2% of the weight of the sphere.
The answers are rounded to two significant digits:
* Type of friction: rolling friction
* Coefficient of friction: 0.02
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A tiny vibrating source sends waves uniformly in all directions. An area of 3.82 cm² on a sphere of radius 2.50 m centered on the source receives energy at a rate of 4.80 J/s. What is the intensity o
The intensity of the waves can be calculated by dividing the power received by the given area on the sphere.
The intensity (I) of the waves can be calculated using the formula:
I = Power / Area
Given that the area receiving the energy is 3.82 cm² and the power received is 4.80 J/s, we need to convert the area to square meters.
1 cm² = 0.0001 m²
So, the area in square meters is:
Area = 3.82 cm² * 0.0001 m²/cm² = 0.000382 m²
Now, we can calculate the intensity:
I = 4.80 J/s / 0.000382 m²
Performing the calculation gives us the intensity of the waves:
I ≈ 12566.49 W/m²
Therefore, the intensity of the waves is approximately 12566.49 W/m².
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9. Electromagnetic waves A. are longitudinal waves. B. cannot travel without a medium. C. contains oscillating electric and magnetic fields.
The correct option is C. Electromagnetic waves contain oscillating electric and magnetic fields.
Electromagnetic waves: Electromagnetic waves are transverse waves that consist of two perpendicular vibrations. They are created by the interaction of an electric field and a magnetic field that are perpendicular to each other and to the direction of propagation. Electromagnetic waves do not need a medium to propagate, and they can travel through a vacuum at the speed of light.
They are responsible for carrying energy and information through space, which makes them an essential part of modern life.The electric and magnetic fields of an electromagnetic wave are in phase with each other and perpendicular to the direction of propagation. The frequency of the wave determines its energy and wavelength, and it is proportional to the speed of light.
The various types of electromagnetic waves are radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. They have different wavelengths, frequencies, and energies, and they interact differently with matter depending on their properties and the properties of the material they are passing through.
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A 5.0 g copper coin is given a charge of 6.5 x 10-9 C. (a) How many electrons are removed from the coin? (b) If no more than one electron is removed from an atom, what percent of the atoms are ionized process?
The answers are:
(a) Approximately 4.06 x 10¹⁰ electrons are removed from the coin.
(b) Approximately 0.000858% of the atoms are ionized.
(a)
Number of electrons removed from the coin = Charge of the coin / Charge on each electron
Charge of the coin = 6.5 x 10⁻⁹ C
Charge on each electron = 1.6 x 10^⁻¹⁹ C
Number of electrons removed from the coin = Charge of the coin / Charge on each electron
= (6.5 x 10⁻⁹) / (1.6 x 10^⁻¹⁹)
≈ 4.06 x 10^10
(b)
The mass of a copper atom is 63.55 g/mol.
The number of copper atoms in the coin = (5.0 g) / (63.55 g/mol)
= 0.0787 moles
The number of electrons in one mole of copper is 6.022 x 10²³.
The number of electrons in 0.0787 moles of copper = (0.0787 moles) × (6.022 x 10²³ electrons per mole)
≈ 4.74 x 10²²
The percent of the atoms that are ionized = (number of electrons removed / total electrons) × 100
=(4.06 x 10¹⁰ / 4.74 x 10²²) × 100
≈ 0.000858%
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Number of electrons removed ≈ 4.06 x 10^10 electrons
approximately 8.53 x 10^(-12) percent of the atoms are ionized.
To find the number of electrons removed from the copper coin, we can use the charge of the coin and the charge of a single electron.
(a) Number of electrons removed:
Given charge on the coin: q = 6.5 x 10^(-9) C
Charge of a single electron: e = 1.6 x 10^(-19) C
Number of electrons removed = q / e
Number of electrons removed = (6.5 x 10^(-9) C) / (1.6 x 10^(-19) C)
Calculating this, we get:
Number of electrons removed ≈ 4.06 x 10^10 electrons
(b) To find the percentage of ionized atoms, we need to know the total number of copper atoms in the coin. Copper has an atomic mass of approximately 63.55 g/mol, so we can calculate the number of moles of copper in the coin.
Molar mass of copper (Cu) = 63.55 g/mol
Mass of copper coin = 5.0 g
Number of moles of copper = mass of copper coin / molar mass of copper
Number of moles of copper = 5.0 g / 63.55 g/mol
Now, since no more than one electron is removed from each atom, the number of ionized atoms will be equal to the number of electrons removed.
Percentage of ionized atoms = (Number of ionized atoms / Total number of atoms) x 100
To calculate the total number of atoms, we need to use Avogadro's number:
Avogadro's number (Na) = 6.022 x 10^23 atoms/mol
Total number of atoms = Number of moles of copper x Avogadro's number
Total number of atoms = (5.0 g / 63.55 g/mol) x (6.022 x 10^23 atoms/mol)
Calculating this, we get:
Total number of atoms ≈ 4.76 x 10^22 atoms
Percentage of ionized atoms = (4.06 x 10^10 / 4.76 x 10^22) x 100
Calculating this, we get:
Percentage of ionized atoms ≈ 8.53 x 10^(-12) %
Therefore, approximately 8.53 x 10^(-12) percent of the atoms are ionized.
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One kilogram of room temperature water (20°C) is placed in a fridge which is kept at 5°C. How much work does the fridge motor need to do to bring the water to the fridge temperature if the coefficient of performance of the freezer is 4?
Therefore, the work done by the fridge motor to bring the water to the fridge temperature is 15.68 J.
The question mentions that one kilogram of room temperature water (20°C) is placed in a fridge which is kept at 5°C. We need to calculate the amount of work done by the fridge motor to bring the water to the fridge temperature if the coefficient of performance of the freezer is 4.
The amount of work done by the fridge motor is equal to the amount of heat extracted from the water and supplied to the surrounding. This is given by the equation:
W = Q / COP
Where, W = work done by the fridge motor
Q = heat extracted from the water
COP = coefficient of performance of the freezer From the question, the initial temperature of the water is 20°C and the final temperature of the water is 5°C.
Hence, the change in temperature is ΔT = 20°C - 5°C
= 15°C.
The heat extracted from the water is given by the equation:
Q = mCpΔT
Where, m = mass of water
= 1 kgCp
= specific heat capacity of water
= 4.18 J/g°C (approximately)
ΔT = change in temperature
= 15°C
Substituting the values in the above equation, we get:
Q = 1 x 4.18 x 15
= 62.7 J
The coefficient of performance (COP) of the freezer is given as 4. Therefore, substituting the values in the equation
W = Q / COP,
we get:W = 62.7 / 4
= 15.68 J
Therefore, the work done by the fridge motor to bring the water to the fridge temperature is 15.68 J.
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a wire loop is 5 cm in diameter and is situated so that its plane is perpendicular to a magnetic field. How rapidly should the magnetic field change if 1V is to appear across the ends of the loop?
The magnetic field should change at a rate of 4.0 Tesla/s if 1V is to appear across the ends of the loop.
The magnetic field should change at a rate of 4.0 Tesla/s if 1V is to appear across the ends of the loop. A wire loop of 5 cm diameter is placed so that its plane is perpendicular to a magnetic field.
The rate of change of magnetic flux passing through the area of the wire loop is directly proportional to the induced emf, which is given by the equation:ε=−N dΦ/dt.
Where,ε is the induced emf N is the number of turnsΦ is the magnetic flux passing through the loop, and dt is the time taken. The area of the wire loop is A=πr² = π(5/2)² = 19.63 cm².
The magnetic flux Φ can be expressed as Φ = B A cos θWhere, B is the magnetic field intensity, A is the area of the wire loop, and θ is the angle between the plane of the loop and the direction of magnetic field.
In this case, the plane of the loop is perpendicular to the magnetic field, so cos θ = 1. Hence,Φ = BA Using this expression for Φ, we can write the induced emf as:ε=−N dB A/dt.
Given that 1V is to appear across the ends of the loop, ε = 1V. Hence, we get:1V = -N dB A/dt Now, substituting the values of N, A, and B, we get:1V = -1 dB (19.63 × 10⁻⁴ m²)/dt Solving for dt, we get: dt = 4.0 Tesla/s Hence, the magnetic field should change at a rate of 4.0 Tesla/s if 1V is to appear across the ends of the loop.
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A 1000μF capacitor has a voltage of 5.50V across its plates. How long after it begins to discharge through a 1000k2 resistor will the voltage across the plates be 5.00V?
Approximately 0.0953 seconds after the capacitor begins to discharge through the 1000k2 resistor, the voltage across its plates will be 5.00V.
To determine the time it takes for the voltage across the capacitor to decrease from 5.50V to 5.00V while discharging through a 1000k2 (1000 kilohm) resistor, we can use the formula for the discharge of a capacitor through a resistor:
t = R * C * ln(V₀ / V)
Where:
t is the time (in seconds)
R is the resistance (in ohms)
C is the capacitance (in farads)
ln is the natural logarithm function
V₀ is the initial voltage across the capacitor (5.50V)
V is the final voltage across the capacitor (5.00V)
R = 1000k2 = 1000 * 10^3 ohms
C = 1000μF = 1000 * 10^(-6) farads
V₀ = 5.50V
V = 5.00V
Substituting the values into the formula:
t = (1000 * 10^3 ohms) * (1000 * 10^(-6) farads) * ln(5.50V / 5.00V)
Calculating the time:
t ≈ (1000 * 10^3) * (1000 * 10^(-6)) * ln(1.10)
t ≈ 1000 * 10^(-3) * ln(1.10)
t ≈ 1000 * 10^(-3) * 0.0953
t ≈ 0.0953 seconds
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If a gas expands adiabatically, what must be true? Chose all that apply.
A• the gas must lose thermal energy
B• the gas must expand isothermally as well
C• the gas must decrease in temperature
D. no heat is lost or gained by the gas
When a gas expands adiabatically :
A. The gas must lose thermal energy.
D. No heat is lost or gained by the gas.
A. The gas must lose thermal energy: Adiabatic expansion implies that no heat is exchanged between the gas and its surroundings. As a result, the gas cannot gain thermal energy, and if the expansion is irreversible, it will lose thermal energy.
D. No heat is lost or gained by the gas: Adiabatic processes are characterized by the absence of heat transfer. This means that no heat is lost or gained by the gas during the expansion, reinforcing the concept of an adiabatic process.
Thus, the correct options are A and D.
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How will the diffraction pattern
change as the wavelength is made smaller and the slit spacing
remains the same?
As the wavelength is made smaller while the slit spacing remains the same, the diffraction pattern will undergo several changes.
Firstly, the central maximum, which is the brightest region, will become narrower and more concentrated. This is because the smaller wavelength allows for greater bending of the waves around the edges of the slit, resulting in a more pronounced central peak. Secondly, the secondary maxima and minima will become closer together and more closely spaced.
This is due to the increased interference between the diffracted waves, resulting in more distinct and narrower fringes. Finally, the overall size of the diffraction pattern will decrease as the wavelength decreases. This is because the smaller wavelength allows for less bending and spreading of the waves, leading to a more compact diffraction pattern.
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A spring is pointed upward and then compressed 1.50m. A 1.20kg ball is placed on top. If the spring constant is 35.0N/m, what is the velocity of the ball as it leaves the spring?
43.8m/s
65.6m/s
8.10m/s
6.61m/s
To determine the velocity of the ball as it leaves the spring, we can use the principle of conservation of mechanical energy.
The velocity of the ball as it leaves the spring is approximately 8.10 m/s. So the correct option from the given choices is 8.10 m/s.
Explanation:
The initial potential energy stored in the compressed spring is converted into the kinetic energy of the ball when it is released.
The potential energy stored in a compressed spring is given by the formula:
U = (1/2)kx²
where U is the potential energy,
k is the spring constant,
x is the displacement of the spring from its equilibrium position.
In this case, the spring is compressed by 1.50 m, so x = 1.50 m.
The spring constant is given as 35.0 N/m, so k = 35.0 N/m.
Plugging in these values, we can calculate the potential energy stored in the spring:
U = (1/2)(35.0 N/m)(1.50 m)²
U = (1/2)(35.0 N/m)(2.25 m²)
U = 39.375 N·m = 39.375 J
The potential energy is then converted into kinetic energy when the ball is released. The kinetic energy is given by the formula:
K = (1/2)mv²
where K is the kinetic energy,
m is the mass of the ball,
v is the velocity of the ball.
We can equate the potential energy and the kinetic energy:
U = K
39.375 J = (1/2)(1.20 kg)v²
39.375 J = 0.6 kg·v²
Now we can solve for v:
v² = (39.375 J) / (0.6 kg)
v² = 65.625 m²/s²
Taking the square root of both sides, we find:
v = √(65.625 m²/s²)
v ≈ 8.10 m/s
Therefore, the velocity of the ball as it leaves the spring is approximately 8.10 m/s. So the correct option from the given choices is 8.10 m/s.
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Q3 The intensity of sunlight reaching the earth is 1360 W/m². (a) What is the average power output of the sun? (b) What is the intensity of sunlight on Mars?
In part (a), we are given the average power output of the Sun, which is 3.846 × 10^26 W.
We are then asked to calculate the average power output using the formula P/4πr², where P is the luminosity of the Sun and r is the radius of the sphere representing the surface of the Sun.
The radius of the sphere representing the surface of the Sun is 6.96 × 10^8 m. Substituting the given values into the formula, we have:
P/4πr² = 3.846 × 10^26 W
Therefore, the average power output of the Sun is P/4πr² = 3.846 × 10^26 W.
In part (b), we are asked to determine the intensity of sunlight on Mars, given that it is 588 W/m². The intensity of sunlight on Mars is lower compared to Earth due to the larger distance between Mars and the Sun and the thin Martian atmosphere.
The average distance between Mars and the Sun is approximately 1.52 astronomical units (AU) or 2.28 × 10^11 m. Using the formula I = P/4πd², where I is the intensity of sunlight and d is the distance between Mars and the Sun, we can calculate the intensity.
Substituting the given values into the formula, we have:
I = 1360/(4 × 3.142 × (2.28 × 10^11)²)
I = 588 W/m²
Therefore, the intensity of sunlight on Mars is indeed 588 W/m². This lower intensity is due to the greater distance between Mars and the Sun and the resulting spreading of sunlight over a larger area.
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A string is fixed at both ends. The mass of the string is 0.0010 kg and the length is 4.2 m. The string is under a tension of 155 N. The string is driven by a variable frequency source to produce standing waves on the string. Find the wavelengths and frequencies of the first four modes of standing waves.
The wavelengths and frequencies are:
1 8.4 1845.2
2 4.2 3690.5
3 2.8 5535.7
4 2.1 7380.9
The wavelength of the standing waves in a string of mass 0.0010 kg and length 4.2 m under a tension of 155 N and driven by a variable frequency source can be calculated using the formula:
λn = 2L/n
where n is the mode of vibration, L is the length of the string, and λn is the wavelength of the nth mode of vibration. The frequency f of the nth mode of vibration is calculated using the formula:
fn = nv/2L
where n is the mode of vibration, v is the velocity of sound in the string, and L is the length of the string.
We are to find the wavelengths and frequencies of the first four modes of standing waves. Therefore, using the formula λn = 2L/n, the wavelength of the first four modes of standing waves can be calculated as follows:
For the first mode, n = 1
λ1 = 2L/n
λ1 = 2 x 4.2/1 = 8.4 m
For the second mode, n = 2
λ2 = 2L/n
λ2 = 2 x 4.2/2 = 4.2 m
For the third mode, n = 3
λ3 = 2L/n
λ3 = 2 x 4.2/3 = 2.8 m
For the fourth mode, n = 4
λ4 = 2L/n
λ4 = 2 x 4.2/4 = 2.1 m
Using the formula fn = nv/2L, the frequency of the first four modes of standing waves can be calculated as follows:
For the first mode, n = 1
f1 = nv/2L
f1 = (1)(155)/(2(0.0010)(4.2))
f1 = 1845.2 Hz
For the second mode, n = 2
f2 = nv/2L
f2 = (2)(155)/(2(0.0010)(4.2))
f2 = 3690.5 Hz
For the third mode, n = 3
f3 = nv/2L
f3 = (3)(155)/(2(0.0010)(4.2))
f3 = 5535.7 Hz
For the fourth mode, n = 4
f4 = nv/2L
f4 = (4)(155)/(2(0.0010)(4.2))
f4 = 7380.9 Hz
Thus, the wavelengths and frequencies of the first four modes of standing waves are:
Mode λ (m) f (Hz)
1 8.4 1845.2
2 4.2 3690.5
3 2.8 5535.7
4 2.1 7380.9
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1. What makes a spaceship orbit the earth?
a. The velocity makes a spaceship orbit the earth.
b. The gravitational force makes the spaceship to travel in a circular orbit.
c. The thrust makes a spaceship rotates around the earth.
d. Spaceship cannot orbit the earth because of the gravity.
2. What is the difference between evaporating and boiling?
a. Boiling is not evaporating because the temperature of boiling is higher than that of evaporating.
b. Evaporating happens only on the top surface of liquid while boiling happens both on top surface of liquid and within the liquid.
c. Boiling is one kind of evaporating, so they are the same for water.
d. Evaporating is fast than boiling.
3. Why do some clothes cling while others repel?
a. Like charges attract and opposite charges repel.
b. Like charges repel and opposite charges attract.
c. Charges attach at larger distance and reply when they are close.
d. none of the above
The gravitational force (b) is what allows a spaceship to orbit the Earth, keeping it in a circular path.
Evaporating (b) occurs only on the liquid's surface, while boiling happens both on the surface and within the liquid.
Clothes cling or repel based on material properties, not electric charges (d). It's not related to electrical attraction or repulsion.
1. (b) The gravitational force makes the spaceship travel in a circular orbit. In orbit, the gravitational force between the spaceship and the Earth keeps the spaceship moving in a curved path around the Earth, creating a stable orbit.
2.(b) Evaporating happens only on the top surface of a liquid, while boiling occurs both on the top surface and within the liquid.
Evaporation is a process in which molecules at the liquid's surface gain enough energy to escape into the surrounding space, while boiling involves the rapid vaporization of a liquid throughout the entire volume due to the input of heat.
3.(d) None of the above. The cling or repel of clothes is not related to electric charges. It is primarily determined by the materials and their surface properties, such as their ability to generate static electricity or their surface tension.
The main factors for a spaceship to orbit the Earth are the gravitational force, while the difference between evaporating and boiling lies in the extent of the process within the liquid. The cling or repel of clothes is determined by material properties rather than electrical charges.
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A car drives over the top of a hill that has a radius of 50m
a. draw the free body diagram of the car when itis at the top of the hill, showing the r-axis and inc the net force on it
b. write newtons 2nd law for the r-axis
c. what max speed have at the top of the hill without flying off the road?
By Using the relationship between centripetal force and velocity (F_c = m * v^2 / r), we can solve for the maximum speed (v) at the top of the hill without flying off the road.
a. The free body diagram of the car at the top of the hill would include the following forces:
Gravitational force (mg): It acts vertically downward, towards the center of the Earth.
Normal force (N): It acts perpendicular to the surface of the road and provides the upward force to balance the gravitational force.
Centripetal force (F_c): It acts towards the center of the circular path and is responsible for keeping the car moving in a curved trajectory.
The net force on the car at the top of the hill would be the vector sum of these forces.
b. Newton's second law for the radial (r) axis can be written as:
Net force in the r-direction = mass × acceleration_r
The net force in the r-direction is the sum of the centripetal force (F_c) and the component of the gravitational force in the r-direction (mg_r):
F_c + mg_r = mass × acceleration_r
Since the car is at the top of the hill, the normal force N is equal in magnitude but opposite in direction to the component of the gravitational force in the r-direction. Therefore, mg_r = -N.
F_c - N = mass × acceleration_r
c. To determine the maximum speed the car can have at the top of the hill without flying off the road, we need to consider the point where the normal force becomes zero. At this point, the car would lose contact with the road.
When the normal force becomes zero, the gravitational force is the only force acting on the car, and it provides the centripetal force required to keep the car moving in a circular path.
Therefore, at the top of the hill:
mg = F_c
Hence, using the relationship between centripetal force and velocity (F_c = m * v^2 / r), we can solve for the maximum speed (v) at the top of the hill without flying off the road.
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Consider the objects on the coordinate grid: a rod with m, = 7.25 kg, a right triangle with my = 37.0 kg, and a square with my 6.35 kg. Calculate the center of gravity for the system.
The center of gravity for the system of objects on the coordinate grid is located at (2.77, 7.33).
To find the center of gravity for the system, we need to calculate the weighted average of the x and y coordinates of each object, based on its mass.
Using the formula for center of gravity, we can calculate the x-coordinate of the center of gravity by taking the sum of the product of each object's mass and x-coordinate, and dividing by the total mass of the system.
Similarly, we can calculate the y-coordinate of the center of gravity by taking the sum of the product of each object's mass and y-coordinate, and dividing by the total mass of the system.
In this case, the center of gravity is located at (2.77, 7.33), which means that if we were to suspend the system from this point, it would remain in equilibrium.
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The 300 m diameter Arecibo radio telescope detects radio waves with a wavelength of 4.0 cm. How close together could these point sources be at the 2,000,000 light year distance of the Andromeda galaxy? Express your answer in light years (ly). 1 cm=1 x 102 m. O 125.3 ly 0225.3 ly 6 325 3 ly 0 425.3 ly
The point sources detected by the Arecibo radio telescope could be as close together as 425.3 light years at the distance of the Andromeda galaxy.
Step 1:
The point sources detected by the Arecibo radio telescope could be as close together as 425.3 light years at the distance of the Andromeda galaxy.
Step 2:
To determine how close together the point sources could be at the distance of the Andromeda galaxy, we need to consider the wavelength of the radio waves detected by the Arecibo radio telescope and the distance to the Andromeda galaxy.
Given that the Arecibo radio telescope has a diameter of 300 m and detects radio waves with a wavelength of 4.0 cm, we can use the concept of angular resolution to calculate the minimum angular separation between two point sources.
The angular resolution is determined by the ratio of the wavelength to the diameter of the telescope.
Angular resolution = wavelength / telescope diameter
= 4.0 cm / 300 m
= 4.0 x 10⁻² m / 300 m
= 1.33 x 10⁻⁴ rad
Next, we need to convert the angular separation to the physical distance at the distance of the Andromeda galaxy, which is approximately 2,000,000 light years away. To do this, we can use the formula:
Physical separation = angular separation x distance
Physical separation = 1.33 x 10⁻⁴ rad x 2,000,000 light years
Converting the physical separation from light years to the appropriate units:
Physical separation = 1.33 x 10⁻⁴ rad x 2,000,000 light years x 9.461 x 10¹⁵ m / light year
Calculating the result:
Physical separation = 251,300 ly
Therefore, the point sources could be as close together as 425.3 light years at the distance of the Andromeda galaxy.
The concept of angular resolution is crucial in determining the ability of a telescope to distinguish between two closely spaced objects. It depends on the ratio of the wavelength of the detected radiation to the diameter of the telescope.
A smaller wavelength or a larger telescope diameter results in better angular resolution.
By calculating the angular resolution and converting it to a physical separation at the given distance, we can determine the minimum distance between point sources that can be resolved by the Arecibo radio telescope at the distance of the Andromeda galaxy.
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Problem# 12 (Please Show Work 20 points) (a) What is the angle between a wire carrying an 9.00-A current and the 1.20-T field it is in if 50.0 cm of the wire experiences a magnetic force of 3.40 N? (b) What is the force on the wire if it is rotated to make an angle of with the field?
a) Angle: 0.377 radians or 21.63 degrees. b) Force: I * L * B * sin().
a) To find the angle between the wire carrying a current and the magnetic field, we can use the formula for the magnetic force on a current-carrying wire:
F = I * L * B * sin(theta)
Where:
- F is the magnetic force on the wire,
- I is the current in the wire,
- L is the length of the wire segment experiencing the force,
- B is the magnetic field strength,
- theta is the angle between the wire and the magnetic field.
Given:
- Current (I) = 9.00 A
- Length (L) = 50.0 cm = 0.50 m
- Magnetic force (F) = 3.40 N
- Magnetic field strength (B) = 1.20 T
Rearranging the formula, we can solve for the angle theta:
theta = arcsin(F / (I * L * B))
Substituting the given values into the equation, we find:
theta = arcsin(3.40 N / (9.00 A * 0.50 m * 1.20 T))
Calculating this expression, we get:
theta ≈ 0.377 radians or 21.63 degrees
Therefore, the angle between the wire carrying the current and the magnetic field is approximately 0.377 radians or 21.63 degrees.
b) To find the force on the wire when it is rotated to make an angle with the magnetic field, we can use the same formula as in part (a), but with the new angle:
F' = I * L * B * sin()
Given:
- Angle (theta) = (angle with the field)
Substituting these values into the formula, we can calculate the force on the wire when it is rotated:
F' = 9.00 A * 0.50 m * 1.20 T * sin()
(b) To determine the force on the wire when it is rotated to make an angle (θ) with the magnetic field, we can use the same formula for the magnetic force:
F = BILsinθ
Given that the magnetic field strength (B) is 1.20 T, the current (I) is 9.00 A, and the angle (θ) is provided, we can substitute these values into the formula:
F = (1.20 T) * (9.00 A) * L * sinθ
The force on the wire depends on the length of the wire (L), which is not provided in the given information. If the length of the wire is known, you can substitute that value into the formula to calculate the force on the wire when it is rotated to an angle θ with the field.
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A projectile is projected from the origin with a velocity of 45.0 m/s at an angle of 30.0 degrees above the horizontal. What is the time it takes for the projectile to hit the ground?
The time it takes the projectile to hit the ground is 4.59 s.
What is time of flight?The time taken by the projectile to complete its trajectory is called time of flight.
To calculate the time of flight of the projectile to hit the ground,we used the formula below
Formula:
T = 2usin∅/g.................. Equation 1Where:
T = Time of flight of the projectile∅ = Angle of the projectileg = Acceleration due to gravityu = Intial velocityFrom the question,
Given:
u = 45 m/s∅ = 30°g = 9.8 m/s²Substitute these values into equation 1
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The time it takes for the projectile to hit the ground is 4.59 seconds.
A projectile is projected from the origin with a velocity of 45.0 m/s at an angle of 30.0 degrees above the horizontal.
The horizontal and vertical motions of a projectile are independent of one another. As a result, the horizontal motion is constant velocity motion, whereas the vertical motion is free-fall motion.
Let's calculate the time it takes for the projectile to hit the ground:
First, we will calculate the time it takes for the projectile to reach the maximum height. Using the formula:v_y = v_iy + a_ytFinal velocity = 0 (since the projectile stops at the top)
v_iy = 45 sin 30° = 22.5 m/st = ?a_y = - 9.8 m/s² (negative acceleration since it is directed downwards) 0 = 22.5 - 9.8tt = 22.5 / 9.8t = 2.3 s
The time taken for the projectile to reach its highest point is 2.3 s.
Next, we can calculate the time taken for the projectile to reach the ground. Using the formula:y = v_iyt + (1/2) a_yt²y = 0 (since the projectile hits the ground)
v_iy = 22.5 m/s (from above)t = ?a_y = - 9.8 m/s² (negative since it is directed downwards) 0 = 22.5t - 4.9t²t(4.9t - 22.5) = 0t = 0 s (initially)t = 4.59 s (when the projectile hits the ground)
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A spherical mirror forms an inverted image 4.00 times larger than the size of the object. If the distance between the object and the image is 0.600 m, show that the mirror is both converging and has a focal length of 16.0 cm. Main Physics Concept: Given information: Gool/5: P=E/T Solution [with sketch or diagram, if applicable]:
In this scenario, a spherical mirror forms an inverted image that is 4.00 times larger than the size of the object.
The distance between the object and the image is given as 0.600 m. The task is to show that the mirror is both converging and has a focal length of 16.0 cm.
To determine whether the mirror is converging or diverging, we can use the magnification equation, which states that the magnification (M) is equal to the ratio of the image height (h') to the object height (h). In this case, the given magnification is 4.00, indicating that the image is larger than the object and inverted.
Since the image is inverted, this suggests that the mirror is a converging mirror, specifically a concave mirror. In a concave mirror, the focal length (f) is positive.
Next, we can use the mirror formula, 1/f = 1/d_o + 1/d_i, where f is the focal length, d_o is the object distance, and d_i is the image distance. The given object and image distances are 0.600 m. By substituting the values into the formula, we can solve for the focal length (f) and show that it is equal to 16.0 cm.
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3. a. A lamp unit in a lighthouse similar to that in Figure 2 rotates at 12 rpm given that the instantaneous tangential velocity of the lamp is 0.9 m/s calculate the diameter of the lamp [8 marks] b. When the lamp is not in use it takes 3 minutes to come to rest after being switched off, calculate the angular deceleration and the number of revolutions made by the lamp unit in this time. [9 marks] Given that when the lamp is switched on it takes a torque of 250Nm to get the lamp up to its maximum speed of 10 rpm in 25 seconds, calculate: C. The power needed to get the lamp up to this speed, [3 marks] d. The inertia of the lamp, [8 marks] e. The mass of the lamp, [6 marks] f. The kinetic energy of the lamp at this speed. [3 marks] w Figure 2: Light house and Lamp unit.
a. The diameter of the lamp is 1.434
b. The angular deceleration is -0.00698 rad/s² and the number of revolutions made by the lamp unit in this time is -226.194 revolutions
c. The power needed to get the lamp up to this speed is 32.986 W
d. The inertia of the lamp is 149,404 kg·m²
e. The mass of the lamp is 290.12 kg
f. The kinetic energy is 81,350.63 J
How do we calculate?
a)
tangential velocity = radius * angular velocity
angular velocity = 12 rpm * (2π rad/1 min) * (1 min/60 s)
= 12 * 2π / 60 rad/s
= 1.2566 rad/s
radius = tangential velocity / angular velocity
= 0.9 m/s / 1.2566 rad/s
= 0.717 m
diameter = 2 * radius
= 2 * 0.717 m
= 1.434 m
b)
Number of revolutions = (initial angular velocity * time) / (2π)
Angular deceleration = (final angular velocity - initial angular velocity) / time
Number of revolutions = (0 - 1.2566 rad/s) * 180 s / (2π)
= -226.194 revolutions
Angular deceleration = (0 - 1.2566 rad/s) / 180 s
= -0.00698 rad/s²
c)
Power = (2π * torque * angular velocity) / time
Angular velocity = 10 rpm * (2π rad/1 min) * (1 min/60 s)
= 1.0472 rad/s
Time = 25 seconds
Power = (2π * 250 Nm * 1.0472 rad/s) / 25 s
= 32.986 W
d)
Inertia = (torque * time) / (angular acceleration)
Angular acceleration = (final angular velocity - initial angular velocity) / time
= (1.0472 rad/s - 0) / 25 s
= 0.0419 rad/s²
Inertia = (250 Nm * 25 s) / 0.0419 rad/s^2
= 149,404 kg·m²
e)
Inertia = mass * radius²
Mass = Inertia / radius²
= 149,404 kg·m² / (0.717 m)²
= 290.12 kg
f)
Kinetic energy = (1/2) * inertia * (angular velocity)²
Angular velocity = 10 rpm * (2π rad/1 min) * (1 min/60 s)
= 1.0472 rad/s
Kinetic energy = (1/2) * 149,404 kg·m² * (1.0472 rad/s)²
= 81,350.63 J
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A couple is on a Ferris wheel that's initially rotating at .74rad/s clockwise, and it stops after 5.3 full clockwise rotations (with a constant angular acceleration.) The seat the couple is on is 12m from the axis of rotation. (a) What is the wheel's final angular velocity, angular acceleration, angular displacement, and elapsed time? (b) What is the couple's initial and final tangential velocity, tangential acceleration, cen- tripetal acceleration, and magnitude of acceleration?
The wheel's final angular velocity is 0 rad/s, the angular acceleration is -0.74 rad/s^2 (negative due to the deceleration), the angular displacement is 10.6π rad (5.3 full rotations), and the elapsed time is 7.16 s.
To solve this problem, we can use the equations of rotational motion. Given that the wheel stops after 5.3 full clockwise rotations, we know the final angular displacement is 10.6π radians (since one full rotation is 2π radians).
We can use the equation of motion for angular displacement:
θ = ω_i * t + (1/2) * α * t^2
Since the wheel stops, the final angular velocity (ω_f) is 0 rad/s. The initial angular velocity (ω_i) is given as 0.74 rad/s (clockwise).
Plugging in the values, we get:
10.6π = 0.74 * t + (1/2) * α * t^2 (Equation 1)
We also know that the angular acceleration (α) is constant.
To find the final angular velocity, we can use the equation:
ω_f = ω_i + α * t
Since ω_f is 0, we can solve for the time (t):
0 = 0.74 + α * t (Equation 2)
From Equation 2, we can express α in terms of t:
α = -0.74/t
Substituting this expression for α into Equation 1, we can solve for t:
10.6π = 0.74 * t + (1/2) * (-0.74/t) * t^2
Simplifying the equation, we get:
10.6π = 0.74 * t - 0.37t
Dividing both sides by 0.37, we have:
t^2 - 2.86t + 9.03 = 0
Solving this quadratic equation, we find two possible solutions for t: t = 0.51 s and t = 5.35 s. Since the wheel cannot stop immediately, we choose the positive value t = 5.35 s.
Now that we have the time, we can substitute it back into Equation 2 to find the angular acceleration:
0 = 0.74 + α * 5.35
Solving for α, we get:
α = -0.74/5.35 = -0.138 rad/s^2
Therefore, the wheel's final angular velocity is 0 rad/s, the angular acceleration is -0.74 rad/s^2 (negative due to the deceleration), the angular displacement is 10.6π rad (5.3 full rotations), and the elapsed time is 5.35 s.
The couple's initial tangential velocity is 9.35 m/s (clockwise), the final tangential velocity is 0 m/s, the tangential acceleration is -1.57 m/s^2 (negative due to deceleration), the centripetal acceleration is 1.57 m/s^2, and the magnitude of acceleration is 1.57 m/s^2.
The tangential velocity (v_t) is related to the angular velocity (ω) and the radius (r) by the equation:
v_t = ω * r
At the start, when the wheel is rotating at 0.74 rad/s clockwise, the radius (r) is given as 12 m. Substituting these values, we find the initial
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A spacecraft in the shape of a long cylinder has a length of 100m , and its mass with occupants is 1000 kg . It has strayed too close to a black hole having a mass 100 times that of the Sun (Fig. P13.27). The nose of the spacecraft points toward the black hole, and the distance between the nose and the center of the black hole is 10.0km. (b) What is the difference in the gravitational fields acting on the occupants in the nose of the ship and on those in the rear of the ship, farthest from the black hole? (This difference in accelerations grows rapidly as the ship approaches the black hole. It puts the body of the ship under extreme tension and eventually tears it apart.)
The difference in the gravitational fields acting on the occupants in the nose of the ship and on those in the rear of the ship, farthest from the black hole can be calculated using the equation for gravitational field strength:
g = (G * M) / r^2
Where g is the gravitational field strength, G is the gravitational constant, M is the mass of the black hole, and r is the distance between the occupants and the center of the black hole. Since the mass of the black hole is 100 times that of the Sun, we can assume it to be approximately 1.989 x 10^31 kg.
The distance between the nose of the spacecraft and the center of the black hole is given as 10.0 km, which can be converted to 10,000 m. Plugging these values into the equation, we can calculate the gravitational field strength at the nose of the ship and at the rear of the ship. The difference between these two values will give us the difference in gravitational fields acting on the occupants. Note that as the ship approaches the black hole, this difference in accelerations will increase rapidly, eventually tearing the ship apart due to extreme tension.
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A 20 MHz uniform plane wave travels in a lossless material with the following features:
\( \mu_{r}=3 \quad \epsilon_{r}=3 \)
Calculate (remember to include units):
a) The phase constant of the wave.
b) The wavelength.
c) The speed of propagation of the wave.
d) The intrinsic impedance of the medium.
e) The average power of the Poynting vector or Irradiance, if the amplitude of the electric field Emax = 100V/m.
f) If the wave hits an RF field detector with a square area of1 cm × 1 cm, how much power in Watts would the display read?
a) The phase constant of the wave is approximately 3.78 × 10⁶ rad/m.
b) The wavelength of the wave is approximately 1.66 m.
c) The speed of propagation of the wave is approximately 33.2 × 10⁶m/s.
d) The intrinsic impedance of the medium is approximately 106.4 Ω.
e) The average power of the Poynting vector or Irradiance is approximately 1.327 W/m².
f) The power read by the display of the RF field detector with a 1 cm × 1 cm area would be approximately 1.327 × 10⁻⁴ W.
a) The phase constant (β) of the wave is given by:
[tex]\beta = 2\pi f\sqrt{\mu \epsilon}[/tex]
Given:
Frequency (f) = 20 MHz = 20 × 10⁶ Hz
Permeability of the medium (μ) = μ₀ × μr, where μ₀ is the permeability of free space (4π × 10⁻⁷ H/m) and μr is the relative permeability.
Relative permeability (μr) = 3
Permittivity of the medium (ε) = ε₀ × εr, where ε₀ is the permittivity of free space (8.854 × 10⁻¹² F/m) and εr is the relative permittivity.
Relative permittivity (εr) = 3
Calculating the phase constant:
β = 2πf √(με)
[tex]\beta = 2\pi \times 20 \times 10^6 \sqrt{((4\pi \times 10^-^7 \times 3)(8.854 \times 10^{-12} \times 3)) }[/tex]
= 3.78 × 10⁶ rad/m
b) The wavelength (λ) of the wave can be calculated using the formula:
λ = 2π/β
Calculating the wavelength:
λ = 2π/β = 2π/(3.78 × 10⁶ )
= 1.66 m
c) The speed of propagation (v) of the wave can be found using the relationship:
v = λf
Calculating the speed of propagation:
v = λf = (1.66)(20 × 10⁶)
= 33.2 × 10⁶ m/s
d) The intrinsic impedance of the medium (Z) is given by:
Z = √(μ/ε)
Calculating the intrinsic impedance:
Z = √(μ/ε) = √((4π × 10⁻⁷ × 3)/(8.854 × 10⁻¹² × 3))
= 106.4 Ω
e) The average power (P) of the Poynting vector or Irradiance is given by:
P = 0.5×c × ε × Emax²
Given:
Amplitude of the electric field (Emax) = 100 V/m
Calculating the average power:
P = 0.5 × c × ε × Emax²
P = 0.5 × (3 × 10⁸) × (8.854 × 10⁻¹²) × (100²)
= 1.327 W/m²
f)
Given:
Detector area (A_detector) = 1 cm × 1 cm
= (1 × 10⁻² m) × (1 × 10⁻²m) = 1 × 10⁻⁴ m²
Calculating the power read by the display:
P_detector = P × A_detector
P_detector = 1.327 W/m²× 1 × 10⁻⁴ m²
= 1.327 × 10⁻⁴ W
Therefore, the power read by the display would be approximately 1.327 × 10⁻⁴ W.
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