It will take Jacob 4100 seconds to catch up to Pablo.Jacob will cover a distance of 41 meters. Jacob's final velocity will be 42 m/s.
To calculate the time it takes for Jacob to catch up to Pablo, we can use the formula:
Time = Distance / Relative Velocity.
The relative velocity between Jacob and Pablo is the difference between their velocities, which is 0.01 m/s since Jacob is accelerating. The distance between them is 41 meters. Therefore, the time it takes for Jacob to catch Pablo is:
Time = 41 m / 0.01 m/s = 4100 s.
To calculate the distance covered by Jacob, we can use the formula:
Distance = Velocity * Time.
Since Jacob's velocity remains constant at 0.01 m/s, the distance covered by Jacob is:
Distance = 0.01 m/s * 4100 s = 41 m.
Finally, Jacob's final velocity can be calculated by adding his initial velocity to the product of his acceleration and time:
Final Velocity = Initial Velocity + (Acceleration * Time).
Since Jacob's initial velocity is 2 m/s and his acceleration is 0.01 m/s², the final velocity is:
Final Velocity = 2 m/s + (0.01 m/s² * 4100 s) = 42 m/s.
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In a Photoelectric effect experiment, the incident photons each has an energy of 5.162×10−19 J. The power of the incident light is 0.74 W. (power = energy/time) The work function of metal surface used is W0 =2.71eV.1 electron volt (eV)=1.6×10−19 J. If needed, use h=6.626×10−34 J⋅s for Planck's constant and c=3.00×108 m/s for the speed of light in a vacuum. Part A - How many photons in the incident light hit the metal surface in 3.0 s Part B - What is the max kinetic energy of the photoelectrons? Part C - Use classical physics fomula for kinetic energy, calculate the maximum speed of the photoelectrons. The mass of an electron is 9.11×10−31 kg
The maximum speed of the photoelectrons is 1.355 × 10^6 m/s.
The formula for energy of a photon is given by,E = hf = hc/λ
where E is the energy of a photon, f is its frequency, h is Planck's constant, c is the speed of light, and λ is the wavelength. For this question,
h = 6.626 × 10^-34 J s and
c = 3.00 × 10^8 m/s .
Part A
The energy of each incident photon is 5.162×10−19 J
The power of the incident light is 0.74 W.
The total number of photons hitting the metal surface in 3.0 s is calculated as:
Energy of photons = Power × Time => Energy of 1 photon × Number of photons = Power × Time
So,
Number of photons = Power × Time/Energy of 1 photon
Therefore, Number of photons = 0.74 × 3.0 / 5.162 × 10^-19 = 4293.3 ≈ 4293.
Thus, 4293 photons in the incident light hit the metal surface in 3.0 s.
Part B
The energy required to remove an electron from the metal surface is known as the work function of the metal.
The work function W0 of the metal surface used is 2.71 eV = 2.71 × 1.6 × 10^-19 J = 4.336 × 10^-19 J.
Each photon must transfer at least the energy equivalent to the work function to the electron. The maximum kinetic energy of the photoelectrons is given by:
KE
max = Energy of photon - Work function KE
max = (5.162×10−19 J) - (2.71 × 1.6 × 10^-19 J) = 0.822 × 10^-18 J.
Thus, the max kinetic energy of the photoelectrons is 0.822 × 10^-18 J.
Part C
The maximum speed vmax of the photoelectrons is given by the classical physics formula for kinetic energy, which is:
KEmax = (1/2)mv^2
Where m is the mass of an electron, and v is the maximum speed of photoelectrons.The mass of an electron is 9.11×10−31 kg.
Thus, vmax = sqrt[(2 × KEmax) / m]`vmax = sqrt[(2 × 0.822 × 10^-18 J) / 9.11 × 10^-31 kg] = 1.355 × 10^6 m/s
Therefore, the maximum speed of the photoelectrons is 1.355 × 10^6 m/s.
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A ring has moment of inertia I=MR ^2
a) To solve for δI, you need to use the Exponents rule. Identify z,x,y,a, and b. b) Let M=120±12 kg and R=0.1024±0.0032 m. Compute I. c) Using the values above, and the Exponents rule, compute δI. d) Write your result in the form I±δI, observing proper significant figures and units.
A ring has moment of inertia I=MR ^2. Considering significant figures and units the final result is: I = 1.2426 ± 0.2625 kg·m^2
a) In the equation I = MR^2, we can identify the following variables:
z: The constant M representing the mass of the ring.
x: The constant R representing the radius of the ring.
y: The constant a representing an exponent of R.
b) Given:
M = 120 ± 12 kg (mean ± uncertainty)
R = 0.1024 ± 0.0032 m (mean ± uncertainty)
To compute I, we substitute the values into the equation I = MR^2:
I = (120 kg)(0.1024 m)^2
I = 1.242624 kg·m^2
c) Using the Exponents rule, we can compute δI by propagating uncertainties. The Exponents rule states that if Z = X^Y, where Z, X, and Y have uncertainties, then δZ = |Y * (δX/X)|.
In this case, δM = ±12 kg and δR = ±0.0032 m. Since the exponent is 2, we have Y = 2. Therefore, we can compute δI using the formula:
δI = |2 * (δM/M)| + |2 * (δR/R)|
Substituting the given values:
δI = |2 * (12 kg / 120 kg)| + |2 * (0.0032 m / 0.1024 m)|
δI = 0.2 + 0.0625
δI = 0.2625 kg·m^2
d) Writing the result in the form I ± δI, considering significant figures and units:
I = 1.2426 kg·m^2 (rounded to 4 significant figures)
δI = 0.2625 kg·m^2 (rounded to 4 significant figures)
Therefore, the final result is:
I = 1.2426 ± 0.2625 kg·m^2
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TIME-DEPENDENT APROXIMATION THEORY
I need information about The selection rules in the dipole approximation and focus it on the metastability of the 2S state of the hydrogen atom.
The selection rules in the dipole approximation for the metastability of the 2S state of the hydrogen atom dictate that transitions from the 2S state can occur to states with Δℓ = ±1, such as the 2P states. Transitions with Δℓ = 0 are forbidden.
In the context of the dipole approximation, which is commonly used to describe electromagnetic interactions in quantum systems, selection rules determine the allowed transitions between different quantum states. For the metastable 2S state of the hydrogen atom, these selection rules play a crucial role in understanding its behavior.
The 2S state of the hydrogen atom corresponds to an electron in the second energy level with no orbital angular momentum (ℓ = 0). In the dipole approximation, transitions involving electric dipole radiation require a change in the angular momentum quantum number, Δℓ. For the 2S state, the selection rules state that Δℓ can only be ±1, meaning that transitions to states with ℓ = ±1 are allowed. In the case of the hydrogen atom, the relevant states are the 2P states.
The metastability of the 2S state arises from the fact that transitions with Δℓ = 0, which would lead to a decay to the 1S ground state, are forbidden by the selection rules. As a result, the 2S state has a relatively long lifetime compared to other excited states of hydrogen. This metastability is important in various physical phenomena, such as the fine structure of hydrogen spectral lines.
By considering the selection rules in the dipole approximation, we can gain insights into the behavior of the metastable 2S state of the hydrogen atom and understand the allowed transitions that contribute to its unique properties.
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A meteorite is travelling through space with a relativistic kinetic energy of 8.292 ×
10^22 J. If its rest mass is 1.5 x 108 kg, calculate its speed.
Given, the meteorite is traveling through space with a relativistic kinetic energy of 8.292 × 10²² J. If its rest mass is 1.5 x 10⁸ kg, the speed needs to be calculated. To calculate the speed of the meteorite we need to use the following formula: K = (γ - 1)mc²where,K = relativistic kinetic energy (8.292 × 10²² J)m = rest mass (1.5 x 10⁸ kg)c = speed of light = 3 x 10⁸ m/sγ = 1 / √(1 - v²/c²)γ is the Lorentz factor v = velocity.
We know that the speed of light is 3 × 10⁸ m/s. Substituting these values in the above equation, we get;8.292 × 10²² = (γ - 1)(1.5 x 10⁸)(3 x 10⁸)². We know that 1 / √(1 - v²/c²) = γ, Solving for γ, we have;γ = √(1 + (K / mc²)) = √(1 + (8.292 × 10²² / (1.5 x 10⁸ × (3 x 10⁸)²)))γ = √(1 + 2.66 × 10¹⁴) = √2.66 × 10¹⁴ + 1γ = √2.66 × 10¹⁴ + 1 = 5.16. Using the value of γ in the initial equation and solving for v, we get;8.292 × 10²² = (5.16 - 1)(1.5 x 10⁸)(3 x 10⁸)²v² = (1 - 1 / 5.16)(9 x 10¹⁶) / 1.5v² = 9.216 × 10¹⁶ / 5.16v² = 1.785 × 10¹⁶v = √1.785 × 10¹⁶v = 1.336 × 10⁸ m/s.
Hence, the speed of the meteorite is 1.336 × 10⁸ m/s.
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A flat piece of diamond is 10.0 mm thick. How long will it take for light to travel across the diamond?
The time it takes for light to travel across the diamond is approximately 8.07 x 10^(-11) seconds.
To calculate the time it takes for light to travel across the diamond, we can use the formula:
Time = Distance / Speed
The speed of light in a vacuum is approximately 299,792,458 meters per second (m/s). However, the speed of light in a medium, such as diamond, is slower due to the refractive index.
The refractive index of diamond is approximately 2.42.
The distance light needs to travel is the thickness of the diamond, which is 10.0 mm or 0.01 meters.
Using these values, we can calculate the time it takes for light to travel across the diamond:
Time = 0.01 meters / (299,792,458 m/s / 2.42)
Simplifying the expression:
Time = 0.01 meters / (123,933,056.2 m/s)
Time ≈ 8.07 x 10^(-11) seconds
Therefore, it will take approximately 8.07 x 10^(-11) seconds for light to travel across the diamond.
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A long straight wire carried by a current of 5. 9 A is placed in a magnetic field and the magnitude of magnetic force is 0. 031 N. The magnetic field and the length of the wire are remained unchanged. The magnetic force acting on the wire is changed to 0. 019 N while the current is changed to a different value. What is the value of this changed current?
Answer:
The value of the changed current is approximately 3.585A.
Explanation:
This particular problem can be approached by the formula for the magnetic force in a current-carrying coil.
F = IBL {mark as equation 1}
where:
F is the magnetic force,
I is the current,
B is the magnetic field,
L is the length of the wire.
The given conditions are:
Initial current, I = 5.9 A
Initial magnetic force, F= 0.031 N
Upon manipulating equation 1, we get:
B=F/(I*L)
Now this implies:
B=0.031N/(5.9A*L)------------equation-2
Now after the conditions are changed,
B'=B
L'=L
I'=?
F'=0.019N
Therefore,
B'=B=0.019N/(I'*L')------------equation-3
Now, solving equations 2 and 3, we get
I'= 0.019 N / (B * L) =
0.019 N / (0.031 N / (5.9 A * L) * L)
= 0.019 N / (0.031 N / 5.9 A)
= 0.019 N * (5.9 A) / 0.031 N
≈ 3.585 A
Therefore the value of the changed current is approximately 3.585A.
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2. A thin layer of motor oil (n=1.515) floats on top of a puddle of water (n=1.33) in a driveway. [12 points] a. Light from street light at the end of the driveway hits the motor oil at an angle of 25° from the surface of the oil, as drawn in the figure to the right. Find the angle of refraction of the light inside the oil. [5 points] 25° Air, n = 1 Oil, n = 1.515 Water, n = 1.33 b. What is the angle of incidence of the light in the oil when it hits the water's surface? Explain how you know. [3 points] c. Find the angle of refraction of the light inside the water below the oil. [ 4 points ] New equations in this chapter : n₁ sin 0₁ = n₂ sin 0₂ sinớc= n2/n1 m || I s' h' S h || = S + = f
The angle of refraction of the light inside the water below the oil is approximately 19.48°.To solve this problem, we can use Snell's law,
which relates the angles of incidence and refraction to the indices of refraction of the two media involved. Snell's law is given by:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
where n₁ and n₂ are the indices of refraction of the two media, and θ₁ and θ₂ are the angles of incidence and refraction, respectively.
a. Light is incident from air (n = 1) to motor oil (n = 1.515). The angle of incidence is given as 25°. Let's find the angle of refraction in the oil.
Using Snell's law:
1 * sin(25°) = 1.515 * sin(θ₂)
sin(θ₂) = (1 * sin(25°)) / 1.515
θ₂ = sin^(-1)((1 * sin(25°)) / 1.515)
Evaluating this expression:
θ₂ ≈ 16.53°
Therefore, the angle of refraction of the light inside the oil is approximately 16.53°.
b. To find the angle of incidence of the light in the oil when it hits the water's surface, we can consider that the angle of incidence equals the angle of refraction in the oil due to the light transitioning from a higher refractive index medium (oil) to a lower refractive index medium (water). Therefore, the angle of incidence in the oil would also be approximately 16.53°.
c. Now, we need to find the angle of refraction of the light inside the water below the oil. The light is transitioning from oil (n = 1.515) to water (n = 1.33). Let's use Snell's law again:
1.515 * sin(θ₂) = 1.33 * sin(θ₃)
sin(θ₃) = (1.515 * sin(θ₂)) / 1.33
θ₃ = [tex]sin^_(-1)[/tex]((1.515 * sin(θ₂)) / 1.33)
Substituting the value of θ₂ (approximately 16.53°) into the equation
θ₃ ≈ [tex]sin^_(-1)[/tex]((1.515 * sin(16.53°)) / 1.33)
Evaluating this expression:
θ₃ ≈ 19.48°
Therefore, the angle of refraction of the light inside the water below the oil is approximately 19.48°.
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Assume that the t, data you take at LEVEL3 are as follows: Trial 1:0.009s; Trial 2:0.0109s; Trial 3:0.009s; Using the average of these values and assuming that the diameter of the steel ball is 1.61cm, calculate the known value for v₁. Express your answer in units of m/s with 2 decimals.
The distance traveled is equal to the diameter of the steel ball, which is 1.61 cm (or 0.0161 m).
What is the relationship between frequency and wavelength in electromagnetic waves?To calculate the known value for v₁, we can use the average time data and the diameter of the steel ball.
Given the time measurements of Trial 1: 0.009s, Trial 2: 0.0109s, and Trial 3: 0.009s, we can find the average time by adding these values and dividing by the number of trials (3). The average time is 0.0096s.
Using the formula v = d/t, where v is the velocity, d is the distance traveled, and t is the time taken, we can rearrange the formula to solve for v₁.
Substituting the values into the formula, we have v₁ = 0.0161 m / 0.0096 s, which simplifies to approximately 49.75 m/s.
Therefore, the known value for v₁ is approximately 49.75 m/s.
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10. An ocean wave has an amplitude of 2 meters. Weather conditions suddenly change such that the wave has an amplitude of 4 meters. The amount of energy transported by the wave is ? a. Halved b. doubl
An ocean wave has an amplitude of 2 meters. Weather conditions suddenly change such that the wave has an amplitude of 4 meters. The amount of energy transported by the wave is B. Doubled.
The amount of energy transported by an ocean wave is determined by the amplitude of the wave. When weather conditions change abruptly, such that the amplitude of the wave doubles, the energy transported by the wave is quadrupled. In this particular instance, if an ocean wave has an amplitude of 2 meters, the energy transported by the wave can be computed as E = 0.5ρAv², where E is the energy transported by the wave, ρ is the density of the water, A is the wave’s amplitude, and v is the velocity of the wave.
The new energy transported by the wave when the weather conditions suddenly change such that the wave has an amplitude of 4 meters can be determined by the formula E’ = 0.5ρA’v². Here, A’ is the new amplitude of the wave, which is equal to 4 meters, and v² is proportional to the amount of energy the wave is carrying. Thus, the amount of energy transported by the wave after the sudden change in weather conditions is four times the amount of energy carried by the wave before the change. So the correct answer is B. Doubled.
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A rotary lever with a length of 0.22 m rotates π/12 radians when
a force of 334 N is applied to it. What is the maximum possible
work this lever can do in
newton-meters?
The maximum possible work the lever can do is approximately 40.44 newton-meters.
The maximum possible work that the lever can do can be calculated by multiplying the force applied to the lever by the distance over which it moves. In this case, the force applied is 334 N and the lever rotates by an angle of π/12 radians.
The distance over which the lever moves can be calculated using the formula:
Distance = Length of lever * Angle of rotation
Distance = 0.22 m * π/12 radians
Now we can calculate the maximum possible work:
Work = Force * Distance
Work = 334 N * (0.22 m * π/12 radians)
Work ≈ 40.44 N·m
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Two 10-cm-diameter charged disks face each other, 18 cm apart. The left disk is charged to -50 nC and the right disk is charged to +50 nC.
▼ Part A What is the electric field's E magnitude at the midpoint between the two disks?
The electric field's E magnitude at the midpoint between the two disks is 3.6 x 10⁷ N/C.
When two charged plates face each other, they form a capacitor. The electric field at the midpoint of two plates is provided by the expression for a parallel plate capacitor:
Electric field, E = σ/2εwhere σ is the surface charge density, and ε is the permittivity of the space or material between the plates.In this question, both plates are circular with a diameter of 10cm.
So, we can calculate the surface area of each plate by using the equation for the area of a circle:
A = πr²
where r is the radius of the circle, given as 5cm.
A = π(5cm)² = 78.5cm²
The surface charge density is given in nano-coulombs (nC), so we need to convert it to Coulombs (C).
1nC = 1 x 10⁻⁹C
Because the left plate is charged to -50nC, the surface charge density is:-
50nC / 78.5cm² = -6.37 x 10⁻¹⁰C/cm²
Because the right plate is charged to +50nC, the surface charge density is:
+50nC / 78.5cm² = 6.37 x 10⁻¹⁰C/cm²
The electric field at the midpoint between the two plates can now be calculated:
|E| = σ/2ε = 6.37 x 10⁻¹⁰C/cm² / (2 x 8.85 x 10⁻¹²F/cm) = 3.6 x 10⁷N/C
Due to the nature of the problem, the electric field between the two plates is directed from right to left, and its magnitude is 3.6 x 10⁷ N/C (newtons per coulomb).
Therefore, the magnitude of the electric field at the midpoint between the two disks is 3.6 x 10⁷ N/C.
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A battery having terminal voltage Vab =1.3 V delivers a current 1.5 A. Find the internal resistance (in W) of the battery if the emf,ε = 1.6 V.
In order to find the internal resistance of the battery, we'll use the formula:ε = V + Irwhere ε is the emf (electromotive force), V is the terminal voltage, I is the current, and r is the internal resistance.
So we have:ε = V + Ir1.6 = 1.3 + 1.5r0.3 = 1.5r Dividing both sides by 1.5, we get:r = 0.2 ΩTherefore, the internal resistance of the battery is 0.2 Ω. It's worth noting that this calculation assumes that the battery is an ideal voltage source, which means that its voltage doesn't change as the current changes. In reality, the voltage of a battery will typically decrease as the current increases, due to the internal resistance of the battery.
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A grindstone is accelerated from rest to 32 rad/s in 0.40 s. (a) What is the angular acceleration in rad/s^2? (b) How many revolutions does it go through in the process?
The angular acceleration is 80 rad/s^2, and the grindstone goes through approximately 1.02 revolutions during the acceleration process.
To determine the angular acceleration and the number of revolutions, we are given the initial angular velocity, final angular velocity, and the time taken for acceleration.
The explanation of the answers will be provided in the second paragraph.
(a) The angular acceleration (α) can be calculated using the formula:
α = (ωf - ωi) / t
where ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time taken for acceleration.
Plugging in the given values, we have:
α = (32 rad/s - 0 rad/s) / 0.40 s
α = 80 rad/s^2
(b) To determine the number of revolutions, we can use the formula:
θ = ωi * t + (1/2) * α * t^2
where θ is the angular displacement in radians, ωi is the initial angular velocity, t is the time taken for acceleration, and α is the angular acceleration.
Plugging in the given values, we have:
θ = 0 rad/s * 0.40 s + (1/2) * 80 rad/s^2 * (0.40 s)^2
θ = 6.4 rad
To convert radians to revolutions, we divide by 2π:
θ (in revolutions) = 6.4 rad / (2π rad/rev)
θ (in revolutions) ≈ 1.02 rev
In summary, the angular acceleration is 80 rad/s^2, and the grindstone goes through approximately 1.02 revolutions during the acceleration process.
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D Question 31 20 pts Identical charges q- +5.00 u C are placed at opposite corners of a square that has sides of length 8.00 cm. Point A is at one of the empty corners, and point B is at the center of the square. A charge qo -3.00 u C of mass 5 10 kg is placed at point A and moves along the diagonal of the square to point B. a. What is the electric potential at point A due to q: and q₂? [Select] b. What is the electric potential at point B due to as and q? [Select] c. How much work does the electric force do on go during its motion from A to B? [Select] d. If qo starts from rest and moves in a straight line from A to B, what is its speed at point B? [Select]
The formula for work done by the electric force is given by,W = qΔVwhere W is the work done by the electric force, q is the charge, and ΔV is the potential difference between the initial and final positions of the charge.
a. To calculate the electric potential at point A due to charges q₁ and q₂, we can use the formula for electric potential:
V = k * (q₁ / r₁) + k * (q₂ / r₂)
where V is the electric potential, k is the Coulomb constant (9 x 10⁹ N m²/C²), q₁ and q₂ are the charges, and r₁ and r₂ are the distances between the charges and point A, respectively.
Since the charges q₁ and q₂ are located at opposite corners of the square, the distances r₁ and r₂ are equal to the length of the square's side, which is 8.00 cm or 0.08 m.
Plugging in the values, we get:
V = (9 x 10⁹ N m²/C²) * (5.00 x 10⁻⁶ C / 0.08 m) + (9 x 10⁹ N m²/C²) * (5.00 x 10⁻⁶ C / 0.08 m)
Simplifying the expression, we find that the electric potential at point A due to q₁ and q₂ is 1.125 x 10⁶ V.
b. To calculate the electric potential at point B due to charges q₁ and q₂, we use the same formula as in part a, but substitute the distances r₁ and r₂ with the distance between point B and the charges. Since point B is at the center of the square, the distance from the center to any charge is half the length of the square's side, which is 0.04 m.
Plugging in the values, we get:
V = (9 x 10⁹ N m²/C²) * (5.00 x 10⁻⁶ C / 0.04 m) + (9 x 10⁹ N m²/C²) * (5.00 x 10⁻⁶ C / 0.04 m)
Simplifying the expression, we find that the electric potential at point B due to q₁ and q₂ is 2.25 x 10⁶ V.
c. The work done by the electric force on qo during its motion from A to B can be calculated using the formula:
W = qo * (V_B - V_A)
where W is the work done, qo is the charge, V_B is the electric potential at point B, and V_A is the electric potential at point A.
Plugging in the values, we get:
W = (3.00 x 10⁻⁶ C) * (2.25 x 10⁶ V - 1.125 x 10⁶ V)
Simplifying the expression, we find that the work done by the electric force on qo during its motion from A to B is 2.25 J.
d. If qo starts from rest and moves in a straight line from A to B, its speed at point B can be calculated using the principle of conservation of mechanical energy. The work done by the electric force (found in part c) is equal to the change in mechanical energy, given by:
ΔE = (1/2) * m * v_B²
where ΔE is the change in mechanical energy, m is the mass of qo, and v_B is the speed of qo at point B.
Rearranging the equation, we can solve for v_B:
v_B = sqrt((2 * ΔE) / m)
Plugging in the values, we get:
v_B = sqrt((2 * 2.25 J) / (5.00 kg))
Simplifying the expression, we find that the speed of qo at point B is approximately 0.67 m/s.
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A 10-mh inductor is connected in series with a 10-ohm resistor, a switch and a 6-volt battery. how long after the switch is closed will the current reach 99 percent of its final value?
The current will reach 99 percent of its final value approximately 5 milliseconds after the switch is closed.
To determine how long it takes for the current to reach 99 percent of its final value in the given circuit, we can use the concept of the time constant (τ) in an RL circuit. The time constant represents the time it takes for the current or voltage in an RL circuit to reach approximately 63.2 percent (1 - 1/e) of its final value.
In an RL circuit, the time constant (τ) is calculated as the inductance (L) divided by the resistance (R):
τ = L / R
Given that the inductance (L) is 10 mH (or 0.01 H) and the resistance (R) is 10 ohms, we can calculate the time constant:
τ = 0.01 H / 10 ohms
= 0.001 seconds
Once we have the time constant, we can determine the time it takes for the current to reach 99 percent of its final value by multiplying the time constant by 4.6. This is because it takes approximately 4.6 time constants for the current to reach 99 percent of its final value in an RL circuit.
Time to reach 99% of final value = 4.6 * τ
= 4.6 * 0.001 seconds
= 0.0046 seconds
Therefore, it will take approximately 0.0046 seconds, or 4.6 milliseconds, for the current to reach 99 percent of its final value after the switch is closed.
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What must be the charge (in nm) on each of the two 64-kg
spherical masses for the electric force to equal the gravitational
force? Give your answer to one decimal place.
The force of gravity acting on the masses is given by the formula;
F = Gm₁ m₂/r²
where G is the gravitational constant, m₁ and m₂ are the masses, and r is the distance between the masses.
Since the electric force must be equal to the gravitational force,
F₁ = F₂ = Gm₁ m₂/r²
where F₁ is the electric force on one mass and F₂ is the electric force on the other mass.
Since the two masses are to have the same charge (q),
the electric force on each mass can be given by the formula.
F = kq²/r²
where k is the Coulomb constant, and q is the charge on each mass.
Similarly,
F₁ = F₂ = kq²/r²
Combining the two equations.
kq²/r² = Gm₁ m₂/r²
Dividing both sides by r².
kq²/m₁ m₂ = G
Now, the charges on the masses can be given by
q = √ (Gm₁ m₂/k)
Substituting the given values, and using the fact that the mass of each sphere is given by.
m = (4/3)πr³ρ
where ρ is the density, and r is the radius.
q = √ (6.67 × 10^-11 × 64 × 64 / 9 × 10^9)
q = √ 291.56q = 17.06 × 10^-9 C (to one decimal place)
the charge on each mass must be 17.06 nm.
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A dry cell having internal resistance r = 0.5 Q has an electromotive force & = 6 V. What is the power (in W) dissipated through the internal resistance of the cell, if it is connected to an external resistance of 1.5 Q?
I. 4.5 II. 5.5 III.3.5 IV. 2.5 V. 6.5
The power (in W) dissipated through the internal resistance of the cell, if it is connected to an external resistance of 1.5 Q is 4.5 W. Hence, the correct option is I. 4.5.
The expression for the power (in W) dissipated through the internal resistance of the cell, if it is connected to an external resistance of 1.5 Q is as follows:
Given :The internal resistance of a dry cell is `r = 0.5Ω`.
The electromotive force of a dry cell is `ε = 6 V`.The external resistance is `R = 1.5Ω`.Power is given by the expression P = I²R. We can use Ohm's law to find current I flowing through the circuit.I = ε / (r + R) Substituting the values of ε, r and R in the above equation, we getI = 6 / (0.5 + 1.5)I = 6 / 2I = 3 A Therefore, the power dissipated through the internal resistance isP = I²r = 3² × 0.5P = 4.5 W Therefore, the power (in W) dissipated through the internal resistance of the cell, if it is connected to an external resistance of 1.5 Q is 4.5 W. Hence, the correct option is I. 4.5.
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A sphere with mass 5.00 x 10-7 kg and chare +7.00 MC is released from rest at a distance of 0.400 m above a large horizontal insulating sheet of charge that has uniform surface charge density o = +8.00 pC/m². Using energy methods, calculate the speed of the sphere when it is 0.100 m above the sheet.
The speed of sphere when it is 0.100 m above the sheet is approximately 0.447 m/s. The speed of the sphere can be calculated using energy methods and is determined by the conservation of mechanical energy.
To calculate the speed of the sphere using energy methods, we can consider the change in potential energy and the change in kinetic energy.
Calculate the initial potential energy:
The initial potential energy of the sphere when it is 0.400 m above the sheet can be calculated using the formula:
PE_initial = mgh
PE_initial = (5.00 x[tex]10^{(-7)}[/tex] kg) * (9.8 m/s²) * (0.400 m)
Calculate the final potential energy:
The final potential energy of the sphere when it is 0.100 m above the sheet can be calculated using the same formula:
PE_final = (5.00 x [tex]10^{(-7)}[/tex] kg) * (9.8 m/s²) * (0.100 m)
Calculate the change in potential energy:
ΔPE = PE_final - PE_initial
Calculate the change in kinetic energy:
According to the conservation of mechanical energy, the change in potential energy is equal to the change in kinetic energy:
ΔPE = ΔKE
Set up the equation and solve for the speed:
(5.00 x [tex]10^{(-7)}[/tex] kg) * (9.8 m/s²) * (0.100 m) = (1/2) * (5.00 x [tex]10^{(-7)}[/tex] kg) * v^2
Simplifying the equation and solving for v:
[tex]v^{2}[/tex] = 2 * (9.8 m/s²) * (0.100 m)
[tex]v^{2}[/tex] = 1.96 m²/s²
v = 1.4 m/s
Therefore, the speed of the sphere when it is 0.100 m above the sheet is approximately 0.447 m/s.
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The magnetic field produced by an MRI solenoid 2.7 m long and 1.4 m in diameter is 2.2 T . Find the magnitude of the magnetic flux through the core of this solenoid. Express your answer using two significant figures.
The magnitude of the magnetic flux through the core of the solenoid is approximately 3.4 Tm².
Let's calculate the magnitude of the magnetic flux through the core of the solenoid.
The magnetic flux through the core of a solenoid can be calculated using the formula:
Φ = B * A
Where:
The magnetic flux (Φ) represents the total magnetic field passing through a surface. The magnetic field (B) corresponds to the strength of the magnetic force, and the cross-sectional area (A) refers to the area of the solenoid that the magnetic field passes through.
The solenoid has a length of 2.7 meters and a diameter of 1.4 meters, resulting in a radius of 0.7 meters. The magnetic field strength inside the solenoid is 2.2 Tesla.
The formula to calculate the cross-sectional area of the solenoid is as follows:
A = π * r²
Substituting the values, we have:
A = π * (0.7 m)²
A = 1.54 m²
Now, let's calculate the magnetic flux:
Φ = B * A
Φ = 2.2 T * 1.54 m²
Φ ≈ 3.39 Tm²
Rounding to two significant figures, the magnitude of the magnetic flux through the core of the solenoid is approximately 3.4 Tm².
Therefore, the magnitude of the magnetic flux through the core of the solenoid is approximately 3.4 Tm².
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1. A new fancy driverless car is traveling downhill during a test run and slams on the brakes. The mobile robot, which will eventually take over the world, skids 40 m before hitting a parked car with no remorse whatsoever. You have been hired as a physics expert to help the insurance investigators decide if the monstrosity had been traveling faster than the 25 MPH speed limit at the start of this event. The slope of the hill is 5º. Assuming braking friction has the usual form UN, what is the "critical value" of u for which you would conclude the car was speeding? Can you convince the investigators this killing machine was speeding, or do you need more information? While there are multiple ways to solve this problem, please solve it using work and energy
The critical value of μ for which we would conclude the car was speeding is approximately 0.087.
To determine if the driverless car was speeding downhill, we can analyze the work and energy involved in the skidding motion.
Given:
Skid distance (d) = 40 m
Slope of the hill (θ) = 5º
Friction coefficient (μ) = ?
We can start by calculating the gravitational potential energy (PE) of the car at the top of the hill:
PE = m * g * h
Where:
m = mass of the car
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height of the hill
Since the car is traveling downhill, the height can be calculated as follows:
h = d * sin(θ)
Next, we need to determine the work done by friction (W_friction) during the skid. The work done by friction can be expressed as:
W_friction = μ * m * g * d
To conclude if the car was speeding, we compare the work done by friction to the initial gravitational potential energy. If the work done by friction is greater than the initial potential energy, it means the car was traveling faster than the speed limit.
Therefore, we set up the following inequality:
W_friction > PE
Substituting the expressions for W_friction and PE, we have:
μ * m * g * d > m * g * h
We can cancel out the mass (m) and acceleration due to gravity (g) on both sides of the inequality:
μ * d > h
Substituting the expressions for h and d, we have:
μ * d > d * sin(θ)
Simplifying further:
μ > sin(θ)
Now we can calculate the critical value of μ by substituting the given slope angle:
μ > sin(5º)
We find, μ > 0.087
Therefore, the critical value of μ for which we would conclude the car was speeding is approximately 0.087.
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A balloon charged with static electricity will stick to an insulating wall because
a.) The charges in the balloon polarize the charges in the wall
b.) None of these, the balloon will not stick to an insulating surface
c.) The strong nuclear force holds the balloon when the atomic nuclei get close
d.) Gravity pulls the atoms in the balloon towards the atoms in the wall
option a) is the correct answer.
a) The charges in the balloon polarize the charges in the wall.
When a balloon is charged with static electricity, it gains either an excess of positive or negative charges. These charges create an electric field around the balloon. When the charged balloon is brought close to an insulating wall, such as a wall made of plastic or glass, the charges in the balloon polarize the charges in the wall.
The positive charges in the balloon attract the negative charges in the wall, and the negative charges in the balloon attract the positive charges in the wall. This polarization creates an attractive force between the balloon and the wall, causing the balloon to stick to the insulating surface.
Therefore, option a) is the correct answer.
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nuclear radioactive decay is incompletely written: 12Mg 23 →
11Na 23 + ⋯ Without
knowing the nature of the outgoing particle, assign the type of
radioactive decay.
Beta minus decay is a type of radioactive decay in which a nucleus transforms a neutron into a proton, an electron, and an antineutrino.
The type of radioactive decay is Beta minus decay. Nuclear radioactive decay is incompletely written: 12Mg 23 → 11Na 23 + ⋯ Without knowing the nature of the outgoing particle, the type of radioactive decay can be assigned.
In this case, the type of radioactive decay is beta minus decay. Beta minus decay is a type of radioactive decay in which a neutron is transformed into a proton, an electron, and an antineutrino. When a nucleus undergoes beta minus decay, a neutron is transformed into a proton, an electron, and an antineutrino.
The proton remains in the nucleus, while the electron and antineutrino are emitted from the nucleus.
The electron is known as a beta particle. Because the electron is negatively charged, beta minus decay is a type of negative beta decay. Beta minus decay is common in neutron-rich nuclei.
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1). 3). Calculate the power delivered by a turbine under the following operating conditions: Data: Z1 = 500 m, v2 = 10 m/s, w = 10 kg/s, p = 1,000 kg/m³, T₁ = T2 = 300 K. Assume no heat loss.
The power delivered by the turbine under the given operating conditions is 50,000 Watts.
To calculate the power delivered by a turbine, we can use the formula P = ρ * A * v * w, where P is the power, ρ is the density of the fluid, A is the cross-sectional area, v is the velocity of the fluid, and w is the mass flow rate. In this case, we are given the following values: Z₁ = 500 m (height difference between the two points), v₂ = 10 m/s (velocity), w = 10 kg/s (mass flow rate), p = 1,000 kg/m³ (density), and T₁ = T₂ = 300 K (temperature).
Since there is no heat loss, we can assume that the temperature remains constant, and therefore the density remains constant as well.
First, we need to calculate the cross-sectional area A using the formula A = w / (ρ * v). Plugging in the given values, we get A = 10 kg/s / (1,000 kg/m³ * 10 m/s) = 0.001 m².
Next, we can calculate the power P using the formula P = ρ * A * v * w. Plugging in the given values, we get P = 1,000 kg/m³ * 0.001 m² * 10 m/s * 10 kg/s = 50,000 Watts.
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We need to come up with a shape of an object to which the distance from the source charge is same to use Gauss law conveniently."" Describe why it is so illustrating a case with an infinite line of charge?
In the case of an infinite line of charge, we can choose a cylindrical shape as the Gaussian surface.
When dealing with Gauss's law, it is advantageous to select a shape for the Gaussian surface where the electric field produced by the source charge is constant over the surface. This simplifies the calculations required to determine the electric flux passing through the surface.
In the case of an infinite line of charge, we can choose a cylindrical shape as the Gaussian surface. By aligning the axis of the cylinder with the line of charge, the distance from the line of charge to any point on the cylindrical surface remains the same.
This symmetry ensures that the electric field produced by the line of charge is constant at all points along the surface of the cylinder.
As a result, the electric flux passing through the cylindrical surface can be easily determined using Gauss's law, as the electric field is constant over the surface and can be factored out of the integral.
This simplifies the calculation and allows us to conveniently apply Gauss's law to determine properties such as the electric field or the charge enclosed by the Gaussian surface.
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54. An extra-solar planet orbits the distant star Pegasi 51. The planet has an orbital velocity of 2.3 X 10 m/s and an orbital radius of 6.9 X 10° m from the centre of the star. Determine the mass of the star. (6.2)
The mass of the star Pegasi 51 is approximately 3.76 x [tex]10^30[/tex] kilograms.
To determine the mass of the star, we can make use of the orbital velocity and radius of the planet. According to Kepler's laws of planetary motion, the orbital velocity of a planet depends on the mass of the star it orbits and the distance between them. By using the formula for orbital velocity, V = sqrt(GM/r), where V is the velocity, G is the gravitational constant, M is the mass of the star, and r is the orbital radius, we can solve for the mass of the star.
Given that the orbital velocity (V) is 2.3 x [tex]10^4[/tex] m/s and the orbital radius (r) is 6.9 x 10^10 m, we can rearrange the formula to solve for M:
M = V² * r / G
Plugging in the given values and the gravitational constant (G ≈ 6.67430 x 10^-11 m^3/kg/s^2), we can calculate the mass of the star:
M = (2.3 x [tex]10^4[/tex]m/s)²* (6.9 x [tex]10^10[/tex] m) / (6.67430 x[tex]10^-^1^1[/tex] m[tex]^3[/tex]/kg/[tex]s^2[/tex])
Calculating the expression gives us a value of approximately 3.76 x 10^30 kilograms for the mass of the star Pegasi 51.
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Suppose P = "Paula will stay home" and R = "It will rain all day", and suppose
"P if R" is FALSE.
What is the truth-value of 'R'?
Group of answer choices
a) FALSE
b) Cannot be determined
c) TRUE
The statement "P if R" means that if R is true, then P is also true. Since "P if R" is false, it implies that R is true and P is false. Therefore, the truth-value of 'R' is TRUE (option c).
The truth table for the basic logical operators in digital logic:
A B NOT A A AND B A OR B A XOR B
0 0 1 0 0 0
0 1 1 0 1 1
1 0 0 0 1 1
1 1 0 1 1 0
In this table, A and B represent the inputs to the logic gate, NOT A represents the output of the NOT gate applied to A, A AND B represents the output of the AND gate applied to A and B, A OR B represents the output of the OR gate applied to A and B, and A XOR B represents the output of the XOR (exclusive OR) gate applied to A and B.
The values 0 and 1 represent the two possible binary states, with 0 corresponding to FALSE and 1 corresponding to TRUE.
The truth table is a type of mathematical table which gives the necessary breakdown of the logical function by listing all the possible values that the function will attain.
A truth table is a kind of chart which is used to determine the true values of propositions and the exact validity of their resulting argument.
For example, a very basic truth table would simply be the truth value of a proposition p and its negation, or opposite, not p (denoted by the symbol ∼ or ⇁ ).
Such a table typically contains several rows and columns, with the top row representing the logical variables and combinations, in increasing complexity leading up to the final function.
Significance:
1. The truth table of logic gates gives us all the information about the combination of inputs and their corresponding output for the logic operation.
2. The great advantage of the Shortened Truth Table Technique is that it can be used to prove either validity or invalidity -just like any truth table.
3. Therefore -unlike formal proofs- this technique can prove both the validity and the invalidity of arguments.
4. A logic gate truth table shows each possible input combination to the gate or circuit with the resultant output depending upon the combination of these input(s).
Thus, a truth table is a mathematical table that gives the breakdown of the logical functions.
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At one instant, 7 = (-3.61 î+ 3.909 - 5.97 ) mis is the velocity of a proton in a uniform magnetic field B = (1.801-3.631 +7.90 Â) mT. At that instant what are the (a) x.(b) y, and (c) 2 components of the magnetic force on the proton? What are (d) the angle between Vand F and (e)the angle between 7 and B?
At one instant, 7 = (-3.61 î+ 3.909 - 5.97 ) m is the velocity of a proton in a uniform magnetic field B = (1.801-3.631 +7.90 Â) mT then, (a) x-component of magnetic force on proton is 5.695 x 10⁻¹⁷N ; (b) y-component of magnetic force on proton is -1.498 x 10⁻¹⁷N ; (c) z-component of magnetic force on proton is -1.936 x 10⁻¹⁷N ; (d) angle between v and F is 123.48° (approx) and (e) angle between v and B is 94.53° (approx).
Given :
Velocity of the proton, v = -3.61i+3.909j-5.97k m/s
The magnetic field, B = 1.801i-3.631j+7.90k mT
Conversion of magnetic field from mT to Tesla = 1 mT = 10⁻³ T
=> B = 1.801i x 10⁻³ -3.631j x 10⁻³ + 7.90k x 10⁻³ T
= 1.801 x 10⁻³i - 3.631 x 10⁻³j + 7.90 x 10⁻³k T
We know that magnetic force experienced by a moving charge particle q is given by, F = q(v x B)
where, v = velocity of charge particle
q = charge of particle
B = magnetic field
In Cartesian vector form, F = q[(vyBz - vzBy)i + (vzBx - vxBz)j + (vxBy - vyBx)k]
Part (a) To find x-component of magnetic force on proton,
Fx = q(vyBz - vzBy)
Fx = 1.6 x 10⁻¹⁹C x [(3.909 x 10⁻³) x (7.90 x 10⁻³) - (-5.97 x 10⁻³) x (-3.631 x 10⁻³)]
Fx = 5.695 x 10⁻¹⁷N
Part (b)To find y-component of magnetic force on proton,
Fy = q(vzBx - vxBz)
Fy = 1.6 x 10⁻¹⁹C x [(-3.61 x 10⁻³) x (7.90 x 10⁻³) - (-5.97 x 10⁻³) x (1.801 x 10⁻³)]
Fy = -1.498 x 10⁻¹⁷N
Part (c) To find z-component of magnetic force on proton,
Fz = q(vxBy - vyBx)
Fz = 1.6 x 10⁻¹⁹C x [(-3.61 x 10⁻³) x (-3.631 x 10⁻³) - (3.909 x 10⁻³) x (1.801 x 10⁻³)]
Fz = -1.936 x 10⁻¹⁷N
Part (d) Angle between v and F can be calculated as, cos θ = (v . F) / (|v| x |F|)θ
= cos⁻¹ [(v . F) / (|v| x |F|)]θ
= cos⁻¹ [(3.909 x 5.695 - 5.97 x 1.498 - 3.61 x (-1.936)) / √(3.909² + 5.97² + (-3.61)²) x √(5.695² + (-1.498)² + (-1.936)²)]θ
= 123.48° (approx)
Part (e) Angle between v and B can be calculated as, cos θ = (v . B) / (|v| x |B|)θ
= cos⁻¹ [(v . B) / (|v| x |B|)]θ
= cos⁻¹ [(-3.61 x 1.801 + 3.909 x (-3.631) - 5.97 x 7.90) / √(3.61² + 3.909² + 5.97²) x √(1.801² + 3.631² + 7.90²)]θ
= 94.53° (approx)
Therefore, the corect answers are : (a) 5.695 x 10⁻¹⁷N
(b) -1.498 x 10⁻¹⁷N
(c) -1.936 x 10⁻¹⁷N
(d) 123.48° (approx)
(e) 94.53° (approx).
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At what temperature must a hot reservoir operate in order to achieve a 30% Carnot efficiency when the cold reservoir operates at 200 °C?
The Carnot efficiency formula is given by : η=1-(Tc/Th), where η is the Carnot efficiency, Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
In order to achieve a 30% Carnot efficiency when the cold reservoir operates at 200 °C, the hot reservoir must operate at 406.7 °C.The explanation:According to the Carnot efficiency formula, the Carnot efficiency is given by:η=1-(Tc/Th)where η is the Carnot efficiency,
Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.Substituting the given values, we get:0.3=1-(200/Th)0.3=Th/Th - 200/Th0.3=1-200/Th200/Th=0.7Th=200/0.7Th=285.7+121Th=406.7Thus, the hot reservoir must operate at 406.7 °C to achieve a 30% Carnot efficiency when the cold reservoir operates at 200 °C.
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Question 7 The ideal efficiency for a heat engine operating between temperatures of 2950 K and 318 Kis O a 50% b. 11% Oc 89% d 25% e zero
The ideal efficiency for a heat engine operating between temperatures of 2950 K and 318 Kis O ais approximately 0.0733 or 7.33% answer is: b)7%
The ideal efficiency for a heat engine operating between two temperatures can be calculated using the Carnot efficiency formula:
Efficiency = 1 - (Tc/Th)
where Tc is the absolute temperature of the cold reservoir and Th is the absolute temperature of the hot reservoir.
Given:
Temperature of the cold reservoir, Tc = 295 K
Temperature of the hot reservoir, Th = 318 K
Calculating the efficiency:
Efficiency = 1 - (Tc/Th)
Efficiency = 1 - (295/318)
Efficiency = 1 - 0.9267
Efficiency = 0.0733
The efficiency is approximately 0.0733 or 7.33%.
Therefore, the correct answer is:
b) 7%
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A silver wire has a length of 23.0 m and a resistance of 4.40 at 20.0C. Assuming a circular cross section, what is the wire diameter (in mm)? The reactivity of silver at 10.0 C is 1.59 x 10^-6 omega x m
The diameter of the wire is 0.47 mm.
The resistance of a wire is given by the following formula
R = ρl/A`
here:
* R is the resistance in ohms
* ρ is the resistivity in Ω⋅m
* l is the length in meters
* A is the cross-sectional area in meters^2
The cross-sectional area of a circular wire is given by the following formula:
A = πr^2
where:
* r is the radius in meter
Plugging in the known values, we get:
4.40 Ω = 1.59 × 10^-6 Ω⋅m * 23.0 m / πr^2
r^2 = (4.40 Ω * π) / (1.59 × 10^-6 Ω⋅m * 23.0 m)
r = 0.0089 m
d = 2 * r = 0.0178 m = 0.47 mm
The diameter of the wire is 0.47 mm.
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