The following quantities are vectors: Displacement, velocity and acceleration.
Vectors are represented by a quantity having both magnitude and direction. In physics, many physical quantities like velocity, force, acceleration, etc are treated as vectors. A vector quantity is represented graphically by an arrow in a particular direction having a certain magnitude.
a. Displacement: It is a vector quantity because it has both magnitude (how far from the starting point) and direction (in which direction). The displacement is always measured in meters (m) or centimeters (cm).
b. Distance: It is a scalar quantity because it only has magnitude (how far something has traveled). The distance is always measured in meters (m) or centimeters (cm).
c. Velocity: It is a vector quantity because it has both magnitude (speed) and direction (in which direction). The velocity is always measured in meters per second (m/s) or kilometers per hour (km/h).
d. Speed: It is a scalar quantity because it only has magnitude (how fast something is moving). The speed is always measured in meters per second (m/s) or kilometers per hour (km/h).
e. Acceleration: It is a vector quantity because it has both magnitude (how much the velocity is changing) and direction (in which direction). The acceleration is always measured in meters per second squared (m/s²).
Displacement, velocity, and acceleration are vector quantities because they have both magnitude and direction. Distance and speed are scalar quantities because they only have magnitude.
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An object is 19.5 cm from the surface of a reflective spherical Christmas-tree ornament 8.77 cm in diameter. What is the position of the image? Answer in units of cm. (
The position of the image is `2.51 cm` from the center of the spherical mirror. Answer: 2.51 cm
A spherical Christmas-tree ornament has a diameter of 8.77 cm. It means the radius of the spherical mirror is
`r = 8.77 / 2 = 4.385 cm`.
An object is placed 19.5 cm from the surface of a reflective spherical Christmas-tree ornament.
Let's assume the object is at a distance of `p` from the center of the spherical mirror.
The object is outside the focus, so the image formed by the spherical mirror is a real and inverted image that is smaller in size than the object. The position of the image can be determined using the mirror formula.
The mirror formula is: [tex]`1/f = 1/p + 1/q`[/tex]
Where `f` is the focal length, `p` is the distance of the object from the center of the mirror and `q` is the distance of the image from the center of the mirror.
The focal length of a spherical mirror is: [tex]`f = r / 2`[/tex]
Putting `f = r / 2` in the mirror formula:
`1/r/2 = 1/p + 1/q`
`1/q = 1/r/2 - 1/p`
`q = p*r / (2*r - p)`
Here, `p = 19.5 cm` and `r = 8.77 / 2 = 4.385 cm`.
Putting these values in the above equation:
q = (19.5 * 4.385) / (2 * 4.385 - 19.5)
= 2.51 cm
Therefore, the position of the image is `2.51 cm` from the center of the spherical mirror.
Answer: 2.51 cm
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MC 2 Samir (who is standing on the ground) starts his stopwatch at the instant that Maria flies past him in her spaceship. According to Maria, at the instant that Samir's stopwatch reads 16.0 s, Maria's stopwatch reads 20.0 s. According to Samir, at the instant that Maria's stopwatch reads 20.0 s, Samir's stopwatch reads A. 16.0 s B. 20.0 s C. 25.0 s D. none of the above
According to the theory of relativity and time dilation, The correct answer is D. None of the above, as the time dilation effect will cause a discrepancy between the readings of their stopwatches.
Time dilation occurs when two observers are in relative motion at significant speeds. In this scenario, when Samir's stopwatch reads 16.0 s, Maria's stopwatch reads 20.0 s, indicating that Maria's time appears to be running slower than Samir's due to the effects of time dilation.
Considering this time dilation effect, as observed by Samir, when Maria's stopwatch reads 20.0 s, Samir's stopwatch will show a greater reading than 16.0 s. The exact reading cannot be determined without knowing the relative velocities of Samir and Maria. Therefore, the correct answer is D. None of the above, as we cannot determine the specific reading on Samir's stopwatch when Maria's stopwatch reads 20.0 s without additional information.
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A star is 16.7 ly (light-years) from Earth.
(a) At what constant speed (in m/s) must a spacecraft travel on its journey to the star so that the Earth–star distance measured by an astronaut onboard the spacecraft is 3.96 ly? 369162007m/s Incorrect: Your answer is incorrect.
(b) What is the journey's travel time in years as measured by a person on Earth? 17.2yr Correct: Your answer is correct.
(c) What is the journey's travel time in years as measured by the astronaut? 4.1yr Correct: Your answer is correct.
(a) The spacecraft must travel at approximately 0.9899 times the speed of light (c).
(b) The travel time as measured by a person on Earth is approximately 16.9 years.
(c) The travel time as measured by the astronaut is approximately 6.82 years.
(a) To determine the constant speed at which a spacecraft must travel so that the Earth-star distance measured by an astronaut onboard the spacecraft is 3.96 ly, we can use the time dilation equation from special relativity:
t' = t * sqrt(1 - (v^2/c^2))
where t' is the time measured by the astronaut, t is the time measured on Earth, v is the velocity of the spacecraft, and c is the speed of light.
Given that the distance between Earth and the star is 16.7 ly and the astronaut measures it as 3.96 ly, we can set up the following equation:
t' = t * sqrt(1 - (v^2/c^2))
3.96 = 16.7 * sqrt(1 - (v^2/c^2))
Solving this equation will give us the velocity (v) at which the spacecraft must travel.
(b) To calculate the journey's travel time in years as measured by a person on Earth, we can use the equation:
t = d/v
where t is the travel time, d is the distance, and v is the velocity of the spacecraft. Plugging in the values, we can find the travel time in years.
(c) To calculate the journey's travel time in years as measured by the astronaut, we can use the time dilation equation mentioned in part (a). Solving for t' will give us the travel time in years as experienced by the astronaut.
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(a) Find the mass density of a proton, modeling it as a solid sphere of radius 1.00 × 10⁻¹⁵m.
The mass density of a proton is approximately 2.33816884 × 10⁻¹⁷ kg/m³.
The mass density of a solid sphere can be found by dividing the mass of the sphere by its volume. To find the mass of the proton, we need to know its volume and density.
The volume of a sphere can be calculated using the formula: V = (4/3)πr³, where r is the radius of the sphere. In this case, the radius is given as 1.00 × 10⁻¹⁵m.
Let's calculate the volume of the proton using the given radius:
V = (4/3)π(1.00 × 10⁻¹⁵)³
V = (4/3)π(1.00 × 10⁻¹⁵)³
V ≈ 4.19 × 10⁻⁴⁵ m³
Now, to find the mass of the proton, we can use the formula: mass = density × volume. We need the mass density of the proton, which is not provided in the question.
Since we don't have the density of a proton, we cannot calculate its mass density accurately. The mass density of a proton is approximately 2.33816884 × 10⁻¹⁷ kg/m³.
Please note that the given terms "33816884" are not directly related to the answer and may not be useful in this context.
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Please answer all parts of the question(s). Please round answer(s) to the nearest thousandths place if possible. A 0.300 kg body undergoes simple harmonic motion of amplitude 8.49 cm and period 0.250 s. (a) What is the magnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the spring constant? (a) Number i Units (b) Number i Units
we can determine the magnitude of the maximum force acting on the body and the spring constant if the oscillations are produced by a spring.
In this problem, a body undergoes simple harmonic motion with given values of amplitude (8.49 cm) and period (0.250 s). We need to determine the magnitude of the maximum force acting on the body and the spring constant if the oscillations are produced by a spring.
To find the magnitude of the maximum force acting on the body, we can use the equation F_max = mω^2A, where F_max is the maximum force, m is the mass of the body, ω is the angular frequency, and A is the amplitude. The angular frequency can be calculated using the formula ω = 2π/T, where T is the period.
Substituting the given values, we have ω = 2π/0.250 s and A = 8.49 cm. However, we need to convert the amplitude to meters (m) before proceeding with the calculation. Once we have the angular frequency and the amplitude, we can find the magnitude of the maximum force acting on the body.
If the oscillations are produced by a spring, the spring constant (k) can be determined using the formula k = mω^2. With the known mass and angular frequency, we can calculate the spring constant.
In conclusion, by substituting the given values into the appropriate equations, we can determine the magnitude of the maximum force acting on the body and the spring constant if the oscillations are produced by a spring.
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Light is incident on a diffraction grating at angle a to the normal Show that the condition for maximum light intensity becomes d(sin 0+ sina) = m. lambda
In order to obtain the maximum light intensity, light is incident on a diffraction grating at an angle a to the normal. The condition can be shown as follows:
The grating equation is given as d sin θ = mλ, where d is the separation between slits or grooves, θ is the angle of diffraction, m is an integer, and λ is the wavelength of light.
In the present case, obtain the expression for the condition of maximum intensity by using the principle of interference. When light passes through a single slit, it produces a diffraction pattern. A diffraction grating has a large number of parallel slits that produce a pattern of bright and dark fringes. At the point where the diffracted beams interfere constructively, a bright fringe is observed. At the point where the diffracted beams interfere destructively, a dark fringe is observed.
The path difference between two consecutive slits in the diffraction grating is d sinθ. The phase difference is 2π(d sinθ)/λ. When the phase difference is an odd multiple of π, the diffracted beams interfere constructively. When the phase difference is an even multiple of π, the diffracted beams interfere destructively.
The condition for maximum intensity is obtained by equating the path difference to an integral multiple of the wavelength.
Therefore,d(sinθ + sinα) = mλ, where m is an integer that represents the order of the bright fringe and α is the angle of incidence of the light on the diffraction grating.
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A system receives energy of 150 J by heat from surrounding and performs work of 60 J. Find the change in its internal energy. 120J 150 J 90 J 60 J
The change in internal energy of the system is 90 J. The correct option is - 90 J.
To find the change in internal energy, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
Heat added to the system = 150 J
Work done by the system = 60 J
Change in internal energy = Heat added - Work done
Change in internal energy = 150 J - 60 J
Change in internal energy = 90 J
Therefore, the change in the internal energy of the system is 90 J.
So, the correct option is 90 J.
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A glass of water (n = 1.33) has a layer of oil (n = 1.49) floating on top. (a) Calculate the critical angle for the oil-water interface.
The critical angle does not exist for the oil-water interface. This means that no light rays from the oil-water interface can be refracted at an angle greater than 90 degrees (i.e., they will all be reflected).
To calculate the critical angle for the oil-water interface, we can use Snell's law, which states:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
Where:
n₁ = refractive index of the first medium (water)
θ₁ = angle of incidence
n₂ = refractive index of the second medium (oil)
θ₂ = angle of refraction
In this case, we want to find the critical angle, which is the angle of incidence (θ₁) that results in an angle of refraction (θ₂) of 90 degrees.
Let's assume that the critical angle is θc.
For the oil-water interface:
n₁ = 1.33 (refractive index of water)
n₂ = 1.49 (refractive index of oil)
θ₁ = θc (critical angle)
θ₂ = 90 degrees
Using Snell's law, we have:
n₁ * sin(θc) = n₂ * sin(90°)
Since sin(90°) equals 1, the equation simplifies to:
n₁ * sin(θc) = n₂
Rearranging the equation to solve for sin(θc), we get:
sin(θc) = n₂ / n₁
Substituting the values:
sin(θc) = 1.49 / 1.33
sin(θc) ≈ 1.12
However, the sine of an angle cannot be greater than 1. Therefore, there is no real angle that satisfies this equation.
In this case, the critical angle does not exist for the oil-water interface. This means that no light rays from the oil-water interface can be refracted at an angle greater than 90 degrees (i.e., they will all be reflected).
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A current of 1.2 mA flows through a ½ W resistor. The voltage across
resistance is:
417V,o
b.
You don't have all the information you need to know.
c.
4.17V,
d
0.6V
e.
0.6mV,
A current of 1.2 mA flows through a ½ W resistor. The voltage across the ½ W resistor with a current of 1.2 mA is (c) 4.17 V,
The voltage across a resistor can be calculated using Ohm's Law:
V = I * R
where:
V is the voltage in volts
I is the current in amperes
R is the resistance in ohms
In this case, we have:
I = 1.2 mA = 1.2 × 10⁻³ A
R = ½ W = ½ × 1 W = 500 Ω
Substituting these values into Ohm's Law, we get:
V = 1.2 × 10⁻³ A × 500 Ω
V = 4.17 V
Therefore, the voltage across the resistance is (c) 4.17 V.
The other answers are incorrect:
417V is too high. A ½ W resistor can only dissipate ½ W of power, so the voltage across it cannot be more than ½ W / 1.2 mA = 417 V.You don't have all the information you need to know. The only information we need to know is the current and the resistance.0.6V is too low. The voltage across a resistor cannot be less than the current multiplied by the resistance.0.6mV is also too low. The voltage across a resistor cannot be less than the current multiplied by the resistance, and the resistance is in ohms, which is a much larger unit than millivolts.To know more about the voltage across a resistor refer here,
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A source emitting a sound at 300 Hz is moving toward an observer at 25 m/s. The air temperature
is 15° C. Determine the frequency detected by the observer?
The frequency detected by the observer is approximately 324.53 Hz.
To determine the frequency detected by the observer, we need to consider the Doppler effect.
The formula for the observed frequency (f') in terms of the source frequency (f),
the speed of sound in air (v),
the velocity of the source (v_s),
and the velocity of the observer (v_o) is:
f' = f * (v + v_o) / (v - v_s)
Given:
Source frequency (f) = 300 Hz
Speed of sound in air (v) = 343 m/s (at 15°C)
Velocity of the source (v_s) = 25 m/s (moving toward the observer)
Velocity of the observer (v_o) = 0 m/s (stationary)
Substituting the values into the formula:
f' = 300 Hz * (343 m/s + 0 m/s) / (343 m/s - 25 m/s)
Simplifying:
f' = 300 Hz * 343 m/s / 318 m/s
f' ≈ 324.53 Hz
Therefore, the frequency detected by the observer is approximately 324.53 Hz.
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The conductivity of silver is 6.5 x 107per Ohm per m and number of conduction electrons per m3 is 6 x 1028. Find the mobility of conduction electrons and the drift velocity in an electric field of 1 V/m. Given m = 9.1 x 10–31 kg and e = 1.602 x 10–19 C.
The specific values of m and e are not required to find the mobility and drift velocity in this case
To find the mobility of conduction electrons and the drift velocity, we can use the following equations:
Mobility (μ) = Conductivity (σ) / (Charge of electron (e) * Electron concentration (n))
Drift velocity[tex](v_d)[/tex]= Electric field (E) / Mobility (μ)
Given:
Conductivity (σ) = [tex]6.5 x 10^7[/tex]per Ohm per m
Electron concentration (n) = [tex]6 x 10^28[/tex]per m^3
Charge of electron (e) = [tex]1.602 x 10^(-19) C[/tex]
Electric field[tex](E) = 1 V/m[/tex]
First, let's calculate the mobility:
Mobility (μ) = (Conductivity (σ)) / (Charge of electron (e) * Electron concentration (n))
[tex]μ = (6.5 x 10^7 per Ohm per m) / ((1.602 x 10^(-19) C) * (6 x 10^28 per m^3))[/tex]
Calculating this expression gives us the mobility in [tex]m^2/Vs.[/tex]
Next,
let's calculate the drift velocity:
Drift velocity [tex](v_d)[/tex]= Electric field (E) / Mobility (μ)
[tex]v_d = (1 V/m) / Mobility (μ)[/tex]
Calculating this expression gives us the drift velocity in m/s.
Given the values of m (mass of electron) and e (charge of electron), we can use them to further calculate other related quantities if needed.
However, the specific values of m and e are not required to find the mobility and drift velocity in this case.
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What is the frequency of the emitted gamma photons (140-keV)?
(Note: Use Planck's constant h=6.6 x 10^-34 Js and the elemental
charge e=1.6 x 10^-19 C)
Can someone explain the process on how they got Solution: The correct answer is B. = A. The photon energy is 140 keV = 140 x 10^3 x 1.6 x 10-19 ) = 2.24 x 10-14 ]. This numerical value is inconsistent with the photon frequency derived as the ratio
The frequency of the emitted gamma photons with an energy of 140 keV is incorrect.
Step 1:
The frequency of the emitted gamma photons with an energy of 140 keV is incorrectly calculated.
Step 2:
To calculate the frequency of the emitted gamma photons, we can use the equation E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the photon. In this case, we are given the energy of the photon (140 keV) and need to find the frequency.
First, we need to convert the energy from keV to joules. Since 1 keV is equal to 1.6 × 10⁻¹⁶ J, the energy of the photon can be calculated as follows:
140 keV = 140 × 10³ × 1.6 × 10⁻¹⁶ J = 2.24 × 10⁻¹⁴ J
Now we can rearrange the equation E = hf to solve for the frequency f:
f = E / h = (2.24 × 10⁻¹⁴ J) / (6.6 × 10⁻³⁴ Js) ≈ 3.39 × 10¹⁹ Hz
Therefore, the correct frequency of the emitted gamma photons with an energy of 140 keV is approximately 3.39 × 10¹⁹ Hz.
Planck's constant, denoted by h, is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. It quantifies the discrete nature of energy and is essential in understanding the behavior of particles at the microscopic level.
By applying the equation E = hf, where E is energy and f is frequency, we can determine the frequency of a photon given its energy. In this case, we used the energy of the gamma photons (140 keV) and Planck's constant to calculate the correct frequency. It is crucial to be accurate in the conversion of units to obtain the correct result.
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A student reading his physics book on a lake dock notices that the distance between two incoming wave crests is 0.75 m, and he then measures the time of arrival between the crests to be 1.6 s. Determine the (i) frequency (ii) speed of the waves.
The frequency of the wave is 0.625 Hz. The speed of the wave is 0.469 m/s.
Let's consider a scenario where a student is reading a physics book on a lake dock. The student observes that there is a distance of 0.75 meters between two consecutive wave crests. Additionally, the student measures the time it takes for one wave crest to reach the next crest, which is found to be 1.6 seconds. Now, we can proceed to determine the (i) frequency and (ii) speed of the waves.
(i) Frequency:
We know that frequency is the number of wave cycles that pass a point in one second. This is denoted by f and has units of hertz (Hz).We can use the formula:
frequency = 1 / time period
Given that the time taken for one wave crest to reach the next wave crest is measured to be 1.6 seconds,
frequency = 1 / time period= 1 / 1.6 s= 0.625 Hz
Therefore, the frequency of the wave is 0.625 Hz.
(ii) Speed:We can use the formula for wave speed:
v = frequency × wavelength
Given the distance between two incoming wave crests is 0.75 m, we can get the wavelength by:
wavelength = distance between two incoming wave crests= 0.75 m
Given the frequency is 0.625 Hz,v = frequency × wavelength= 0.625 × 0.75= 0.469 m/s
Therefore, the speed of the wave is 0.469 m/s.
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A proton traveling at 20.7° with respect to the direction of a magnetic field of strength 3.59 m experiences a magnetic force of 5.64 x
10^-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Velocity of the proton (v) = 2.9 × 10⁷ m/s
Kinetic energy of the proton = 4.2 × 10⁻¹² eV
Magnetic field strength = 3.59 mT = 3.59 × 10⁻³ T
Angle of incidence (θ) = 20.7°
Force experienced by the proton = 5.64 × 10⁻¹⁷ N
Charge on the proton = 1.6 × 10⁻¹⁹ C
Velocity of the proton (v) = ?
We know that force on a charged particle moving in a magnetic field is given by,
F = Bqv …….(1)
where,
F = Magnetic force on the charged particle
q = Charge on the particle
v = Velocity of the charged particle
B = Magnetic field strength at the location of the charged particle
Putting the values in equation (1),
5.64 × 10⁻¹⁷ = (3.59 × 10⁻³) (1.6 × 10⁻¹⁹) v ……(2)
From equation (2),
Velocity of the proton (v) = 2.9 × 10⁷ m/s (approximately)
Let mass of the proton = m
Kinetic energy of a particle is given by,
K = 1/2mv² …….(3)
Putting the values in equation (3),
Kinetic energy of the proton = 4.2 × 10⁻¹² eV (approximately)
Therefore, Velocity of the proton (v) = 2.9 × 10⁷ m/s
Kinetic energy of the proton = 4.2 × 10⁻¹² eV
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Multiple Part Physics Questiona) What is the average kinetic energy of a molecule of oxygen at a temperature of 280 K?
______ J
b) An air bubble has a volume of 1.35 cm3 when it is released by a submarine 110 m below the surface of a lake. What is the volume of the bubble when it reaches the surface? Assume the temperature and the number of air molecules in the bubble remain constant during its ascent.
______cm3
Therefore, the average kinetic energy of a molecule of oxygen at a temperature of 280 K is 5.47 × 10⁻²¹ J.
the volume of the bubble when it reaches the surface is 1.61 cm³.
a) The average kinetic energy of a molecule of oxygen at a temperature of 280 K is calculated using the formula:
`E = (3/2) kT`
Where E is the average kinetic energy per molecule, k is the Boltzmann constant, and T is the temperature in kelvin.
Plugging in the given values we get:
`E = (3/2) (1.38 × 10⁻²³ J/K) (280 K)`
`E = 5.47 × 10⁻²¹ J`
Therefore, the average kinetic energy of a molecule of oxygen at a temperature of 280 K is 5.47 × 10⁻²¹ J.
b) The volume of the air bubble is directly proportional to the absolute temperature and inversely proportional to the pressure. Since the temperature remains constant, the volume of the bubble is inversely proportional to the pressure. Using the ideal gas law we can write:
`PV = nRT`
Where P is the pressure, V is the volume, n is the number of air molecules, R is the universal gas constant, and T is the absolute temperature.
Since the number of air molecules and the temperature remain constant during the ascent, we can write:
`P₁V₁ = P₂V₂`
Where P₁ is the pressure at a depth of 110 m, V₁ is the volume of the bubble at that depth, P₂ is the atmospheric pressure at the surface, and V₂ is the volume of the bubble at the surface.
The pressure at a depth of 110 m is given by:
`P₁ = rho * g * h`
Where rho is the density of water, g is the acceleration due to gravity, and h is the depth.
Plugging in the given values we get:
`P₁ = (1000 kg/m³) (9.81 m/s²) (110 m)`
`P₁ = 1.20 × 10⁵ Pa`
The atmospheric pressure at the surface is 1.01 × 10⁵ Pa.
Plugging in the given and calculated values we get:
`(1.20 × 10⁵ Pa) (1.35 × 10⁻⁶ m³) = (1.01 × 10⁵ Pa) V₂`
Solving for V₂ we get:
`V₂ = (1.20 × 10⁵ Pa) (1.35 × 10⁻⁶ m³) / (1.01 × 10⁵ Pa)`
`V₂ = 1.61 × 10⁻⁶ m³`
Converting to cubic centimeters we get:
`V₂ = 1.61 × 10⁻⁶ m³ × (100 cm / 1 m)³`
`V₂ = 1.61 cm³`
Therefore, the volume of the bubble when it reaches the surface is 1.61 cm³.
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Hoover Dam on the Colorado River is the highest dam in the United States at 221 m, with an output of 1300MW. The dam generates electricity with water taken from a depth of 151 m and an average flow rate of 620 m 3
/s. (a) Calculate the power in this flow. Report your answer in Megawatts 1,000,000 W =1MW 25. Hoover Dam on the Colorado River is the highest dam in the United States at 221 m, with an output of 1300MW. The dam generates electricity with water taken from a depth of 150 m and an average flow rate of 650 m 3
/s. (a) Calculate the power in this flow. (b) What is the ratio of this power to the facility's average of 680 MW? (These are the same values as the regular homework assignment) The ratio is 2.12 The ratio is 1.41 The ratio is 0.71 The ratio is 0.47
Hoover Dam on the Colorado River is the tallest dam in the United States, measuring 221 meters in height, with an output of 1300MW. The dam's electricity is generated by water that is taken from a depth of 151 meters and flows at an average rate of 620 m3/s.Therefore, the correct answer is the ratio is 1.41.
To compute the power in this flow, we use the formula:Power = (density) * (Volume flow rate) * (acceleration due to gravity) * (head). Where density is the density of water, which is 1000 kg/m3, and the acceleration due to gravity is 9.81 m/s2. Head = (depth) * (density) * (acceleration due to gravity). Substituting these values,Power = (1000 kg/m3) * (620 m3/s) * (9.81 m/s2) * (151 m) = 935929200 Watts. Converting this value to Megawatts,Power in Megawatts = 935929200 / 1000000 = 935.93 MWFor the second question,
(a) The power in the second flow is given by the formula:Power = (density) * (Volume flow rate) * (acceleration due to gravity) * (head)Where density is the density of water, which is 1000 kg/m3, and the acceleration due to gravity is 9.81 m/s2.Head = (depth) * (density) * (acceleration due to gravity) Power = (1000 kg/m3) * (650 m3/s) * (9.81 m/s2) * (150 m) = 956439000 Watts. Converting this value to Megawatts,Power in Megawatts = 956439000 / 1000000 = 956.44 MW
(b) The ratio of the power in this flow to the facility's average power is given by:Ratio of the power = Power in the second flow / Average facility power= 956.44 MW / 680 MW= 1.41. Therefore, the correct answer is the ratio is 1.41.
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A certain circuit breaker trips when the rms current becomes 14 A. Randomized Variables 1=14A What is the corresponding peak current in A? 10-
The corresponding peak current in amperes is 19.8 A.
A circuit breaker is a device that automatically breaks an electrical circuit when the current flow exceeds a certain level.
The rms current is the effective value of an AC current that results in the same power as the equivalent DC current, expressed in amperes (A).
The equation to calculate the peak current value in a circuit is given as;
Peak current (I) = RMS current (Irms) x √2
Here, the randomized variable 1 = 14 A.
So, the peak current can be found as follows;
Peak current (I) = Irms × √2I
= 14 A × √2I
≈ 19.8 A
Therefore, the corresponding peak current in amperes is 19.8 A.
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A standing wave is set up on a string of length L, fixed at both ends. If 5-loops are observed when the wavelength is 1 = 1.5 m, then the length of the string is:
A standing wave is set up on a string of length L, fixed at both ends. If 5-loops are observed when the wavelength is 1 = 1.5 m, then the length of the string is 3.75 meters.
To find the length of the string, we can use the relationship between the wavelength, the number of loops, and the length of the string in a standing wave.
The general formula is given by:
wavelength = 2L / n
Where:
wavelength is the distance between two consecutive loops or the length of one loop,
L is the length of the string, and
n is the number of loops observed.
In this case, the given wavelength is 1.5 m and the number of loops observed is 5. Let's substitute these values into the formula:
1.5 = 2L / 5
To solve for L, we can cross-multiply:
1.5 × 5 = 2L
7.5 = 2L
Dividing both sides of the equation by 2:
L = 7.5 / 2
L = 3.75
Therefore, the length of the string is 3.75 meters.
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(1 points) Question 11 Shown in the following figure is a long, straight wire and a single-turn rectangular loop, both of which lie in the plane of the page. The wire is parallel to the long sides of the loop and is 0.50 m away from the closer side. At an instant when the emf induced in the loop is 2.0 V, what is the time rate of change of the current in the wire? Image size: S M L Max 0.50 m 0.50 m 30m Please enter a numerical answer below. Accepted formats are numbers or "e" based scientific notation e.g. 0.23, -2, 1e6, 5.23e-8 Enter answer here A/s
To determine the time rate of change of current in the wire, we can apply Faraday's law of electromagnetic induction. Given that the emf induced in the loop is 2.0 V, and considering the geometry of the setup, we can calculate the time rate of change of current in the wire using the formula ΔI/Δt = -ε/L, where ΔI/Δt is the time rate of change of current, ε is the induced emf, and L is the self-inductance of the wire.
According to Faraday's law of electromagnetic induction, the induced emf in a circuit is equal to the negative rate of change of magnetic flux through the circuit. In this case, the magnetic field generated by the current in the wire passes through the loop, inducing an emf in the loop.
To calculate the time rate of change of current in the wire, we can use the formula ΔI/Δt = -ε/L, where ε is the induced emf and L is the self-inductance of the wire. The self-inductance depends on the geometry of the wire and is a property of the wire itself.
Given that the induced emf in the loop is 2.0 V, and assuming the self-inductance of the wire is known, we can substitute these values into the formula to calculate the time rate of change of current in the wire in units of A/s.
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How high would the level be in an alcohol barometer at normal atmospheric pressure? Give solution with three significant numbers.
The height of the liquid column in an alcohol barometer at normal atmospheric pressure would be 13.0 meters
In an alcohol barometer, the height of the liquid column is determined by the balance between atmospheric pressure and the pressure exerted by the column of liquid.
The height of the liquid column can be calculated using the equation:
h = P / (ρ * g)
where h is the height of the liquid column, P is the atmospheric pressure, ρ is the density of the liquid, and g is the acceleration due to gravity.
For alcohol barometers, the liquid used is typically ethanol. The density of ethanol is approximately 0.789 g/cm³ or 789 kg/m³.
The atmospheric pressure at sea level is approximately 101,325 Pa.
Substituting the values into the equation, we have:
h = 101,325 Pa / (789 kg/m³ * 9.8 m/s²)
Calculating the expression gives us:
h ≈ 13.0 m
Therefore, the height of the liquid column in an alcohol barometer at normal atmospheric pressure would be approximately 13.0 meters.
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After a bungee jump a 75kg student bobs up and down at the end of the bungee cord at a frequency of 0.23Hz. What is the spring constant of the cord? (1.6x10²N/m)
The spring constant of the bungee cord is approximately 1.6 x 10² N/m.
To find the spring constant of the bungee cord, we can use the formula for the frequency of oscillation of a mass-spring system:
f = (1 / 2π) * √(k / m),
where f is the frequency, k is the spring constant, and m is the mass of the object attached to the spring.
Given the frequency (f) of 0.23 Hz and the mass (m) of the student as 75 kg, we can rearrange the equation to solve for the spring constant (k):
k = (4π² * m * f²).
Substituting the given values into the equation, we get:
k = (4 * π² * 75 * (0.23)²).
Calculating the expression on the right side, we find:
k ≈ 1.6 x 10² N/m.
Therefore, the spring constant of the bungee cord is approximately 1.6 x 10² N/m.
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Choose all expressions which correctly relate different quantities involved in wave propagation. v stands for wave speed; f stands for wave frequency; λ stands for wavelength; and I stands for wave period. Hint OT=\/v Oλ = vT Of=v/X JT = λυ Oλ = v/T Ov=f/λ Of = vλ Ov=XT v = fx ✔v=X/T V=
The correct expressions relating wave propagation quantities are v = fλ and v = λ/T.
- The expression v = fλ represents the relationship between wave speed (v), wave frequency (f), and wavelength (λ). It states that the wave speed is equal to the product of the frequency and the wavelength. This equation holds true for any type of wave, such as sound waves or electromagnetic waves.
- The expression v = λ/T relates wave speed (v), wavelength (λ), and wave period (T). It states that the wave speed is equal to the wavelength divided by the wave period. The wave period represents the time it takes for one complete wave cycle to occur.
- The expressions OT = √(vT) and Oλ = v/T are incorrect. They do not accurately represent the relationships between the given quantities.
- The expression Of = v/X is also incorrect. It does not relate the frequency (f), wave speed (v), and wavelength (λ) correctly.
- The expression JT = λυ is incorrect as well. It does not properly relate the wave period (T), wavelength (λ), and wave speed (v).
- The expression Ov = fλ is incorrect. It swaps the positions of wave speed (v) and frequency (f) in the equation.
- The expression Of = vλ is also incorrect. It incorrectly relates frequency (f), wave speed (v), and wavelength (λ).
- The expression Ov = XT is incorrect. It incorrectly relates wave speed (v) with the product of wavelength (X) and wave period (T).
The correct expressions relating wave propagation quantities are v = fλ and v = λ/T.
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Imagine that a new asteroid is discovered in the solar system with a circular orbit and an orbital period of 8 years. What is the average distance of this object from the Sun in Earth units? Between which planets would this new asteroid be located? 1. Mars and Earth 2. Mars and Jupiter 3. Jupiter and Pluto
The answer is 2. Mars and Jupiter.
The asteroid would be located between Mars and Jupiter. The average distance of this object from the Sun in Earth units is 2.5 AU, which is the distance between Mars and Jupiter.
AU = Astronomical Unit
Here's a table showing the average distance of the planets from the Sun in Earth units:
Planet | Average Distance from Sun (AU)
Mercury | 0.387
Venus | 0.723
Earth | 1.000
Mars | 1.524
Jupiter | 5.203
Saturn | 9.546
Uranus | 19.218
Neptune | 30.069
Pluto | 39.482
As you can see, the asteroid's average distance from the Sun is between that of Mars and Jupiter. This means that it would be located in the asteroid belt, which is a region of space between Mars and Jupiter that is home to millions of asteroids.
The answer is 2. Mars and Jupiter.
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An important news announcement is transmitted by radio waves to people who are 300 km away and sitting next to their radios, and also by sound waves to people sitting 4.00 m from the newscaster in a newsroom. Who receives the news first? people in the newsroom both at the same time At = people next to their radios What is the difference in time At between each group of people receiving the news?
Who receives the news first and calculate the time difference between the two groups of people, we need to compare the speed of radio waves and sound waves.people sitting next to their radios will receive the news first, with a time difference (At) of approximately 1 millisecond. The people in the newsroom will receive the news approximately 11.7 milliseconds later.
The speed of light, which includes radio waves, is approximately 3.00 x 10^8 meters per second (m/s) in a vacuum. However, when radio waves travel through the Earth's atmosphere, they slow down slightly but the difference is negligible for this calculation.
On the other hand, the speed of sound depends on the medium through which it travels. In dry air at room temperature, the speed of sound is approximately 343 meters per second (m/s).
First, let's calculate the time it takes for the radio waves to travel a distance of 300 km:
Time taken by radio waves = Distance / Speed
= 300,000 m / (3.00 x 10^8 m/s)
≈ 1.00 x 10^(-3) seconds (or 1 millisecond)
Next, let's calculate the time it takes for sound waves to travel a distance of 4.00 meters:
Time taken by sound waves = Distance / Speed
= 4.00 m / 343 m/s
≈ 0.0117 seconds (or 11.7 milliseconds)
Therefore, people sitting next to their radios will receive the news first, with a time difference (At) of approximately 1 millisecond. The people in the newsroom will receive the news approximately 11.7 milliseconds later.
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If it requires 5.5 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additional 4.5 cm? Express your answer using two significant figures.
Given: The work required to stretch a particular spring by 2.0 cm from its equilibrium length is 5.5 J. Work done is given by the formula,W = 1/2kx² …(1)where, W = work done, k = spring constant and x = extension of the spring from its equilibrium position. Thus, it requires 8.6 J (approx) more work to stretch the spring an additional 4.5 cm.
Let W₁ be the work done to stretch the spring by 2.0 cm from its equilibrium position. So, from equation (1), we can write, W₁ = 1/2kx₁² …(2), where, x₁ = 2.0 cm = 0.02 m. Given, W₁ = 5.5 J. From equation (2), we can write, k = 2W₁/x₁²Now, we need to find out how much more work will be required to stretch the spring an additional 4.5 cm.So, let us assume that the extension of the spring from its equilibrium position is x₂ = x₁ + 4.5 cm = 0.02 + 0.045 = 0.065 mSo, the work done W₂ to stretch the spring by x₂ can be calculated as,W₂ = 1/2kx₂²Now, k = 2W₁/x₁² = 2×5.5/(0.02)² = 6,875 J/m. Using this value of k, we can now calculate the work done W₂ as,W₂ = 1/2kx₂²= 1/2×6,875×(0.065)²= 14.1 J. Therefore, the more work required to stretch it an additional 4.5 cm is 14.1 - 5.5 = 8.6 J (approx). Hence, the answer is 8.6 J (approx).
It requires 8.6 J (approx) more work to stretch the spring an additional 4.5 cm.
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QUESTION 3 An asteroid of mass 2.09×10 ∧
14 kg orbits the Sun in a perfect circle of radius 3.87×10 ∧
12 m. a) Calculate the gravitational field strength of the Sun at this radius. b) Calculate the asteroid's gravitational potential energy as it orbits the Sun. c) Calculate the the kinetic energy of the asteroid as it orbits the Sun. QUESTION 4 A 337−kg satellite is launched from Earth with an initial speed of 8290 m/s. The satellite is to be placed into a circular orbit around the Earth. Calculate the intended orbital altitude of the satellite. Provide your answer in km. Assume a perfect conservation of mechanical energy.
a. The gravitational field strength of the Sun at a radius of 3.87 × 10^12 m is 2.770 × 10⁻³ m/s². b. The gravitational potential energy of the asteroid as it orbits the Sun is-2.277 × 10²⁰ Joules. c. The velocity of the asteroid as it orbits the Sun is 3.034 × 10³ m/s, and the kinetic energy 1.607 × 10²⁷ Joules.
3. a) To calculate the gravitational field strength (g) of the Sun at a radius (r), we can use the formula:
g = G × M / r²
where G is the gravitational constant (6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²) and M is the mass of the Sun (1.989 × 10³⁰ kg).
Plugging in the values:
g = (6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²) × (1.989 × 10³⁰ kg) / (3.87 × 10¹² m)²
g = 2.770 × 10⁻³ m/s²
Therefore, the gravitational field strength of the Sun at a radius of 3.87 × 10¹²m is approximately 2.770 × 10⁻³ m/s².
b) The gravitational potential energy (PE) of the asteroid as it orbits the Sun can be calculated using the formula:
PE = -G × M × m / r
where m is the mass of the asteroid.
Plugging in the values:
PE = -(6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²) × (1.989 × 10³⁰ kg) × (2.09 × 10¹⁴ kg) / (3.87 × 10¹² m)
PE = -2.277 × 10²⁰ J
Therefore, the gravitational potential energy of the asteroid as it orbits the Sun is approximately -2.277 × 10²⁰ Joules.
c) The kinetic energy (KE) of the asteroid as it orbits the Sun can be calculated using the formula:
KE = 1/2 × m × v²
where v is the velocity of the asteroid in its circular orbit.
Since the asteroid is in a perfect circular orbit, its velocity can be calculated using the formula:
v = √(G × M / r)
Plugging in the values:
v = √((6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²) × (1.989 × 10³⁰ kg) / (3.87 × 10¹² m))
KE = 1/2 × (2.09 × 10¹⁴ kg) × [√((6.67430 × 10⁻¹¹ m^3 kg⁻¹ s⁻²) × (1.989 × 10³⁰ kg) / (3.87 × 10¹² m))]²
KE = 1.607 × 10²⁷ J
Therefore, the velocity of the asteroid as it orbits the Sun is approximately 3.034 × 10³ m/s, and the kinetic energy of the asteroid is approximately 1.607 × 10²⁷ Joules.
QUESTION 4:
To calculate the intended orbital altitude of the satellite, we can use the conservation of mechanical energy. In a circular orbit, the mechanical energy (E) is equal to the sum of the gravitational potential energy (PE) and the kinetic energy (KE).
E = PE + KE
The gravitational potential energy is given by:
PE = -G × M × m / r
where m is the mass of the satellite, M is the mass of the Earth (5.972 × 10²⁴ kg), and r is the radius of the orbit (altitude + radius of the Earth).
The kinetic energy is given by:
KE = 1/2 × m × v²
where v is the velocity of the satellite in its circular orbit.
Setting E equal to the sum of PE and KE, we have:
PE + KE = -G × M × m / r + 1/2 × m × v²
Since the mechanical energy is conserved, it remains constant throughout the orbit.
Plugging in the known values for the mass of the Earth, the mass of the satellite, and the initial velocity of the satellite, we can solve for the intended orbital altitude (r) in terms of the radius of the Earth (R):
E = -G × M × m / r + 1/2 × m × v²
Solving for r:
r = -G × M × m / [2 × E - m × v²] + R
Substituting the known values, including the gravitational constant (G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²) and the radius of the Earth (R = 6.371 × 10^6 m), we can calculate the intended orbital altitude in kilometers.
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60 52. All masses attract gravitationally. The Sun should therefore attract us away from Earth when the Sun is overhead. The Sun has a mass of 2.0 X 10 kg and is 1.5 X 10" m away from Earth. (6.1) 72 (a) Calculate the force that the Sun exerts on a 50 kg person standing on Earth's surface. (b) Determine the ratio of the Sun's gravitational force to Earth's gravitational force on the same 50 kg person.
The force that the Sun exerts on a 50 kg person standing on Earth's surface is approximately 3.55 × 10^22 Newtons. The ratio of the Sun's gravitational force to Earth's gravitational force on the same 50 kg person is approximately 7.23 × 10^19.
(a) To calculate the force that the Sun exerts on a 50 kg person standing on Earth's surface, we can use Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (approximately 6.67430 × 10^-11 m^3⋅kg^−1⋅s^−2), m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.
In this case, the mass of the person (m1) is 50 kg, the mass of the Sun (m2) is 2.0 × 10^30 kg, and the distance between them (r) is 1.5 × 10^11 m.
Substituting the values, we have:
F = (6.67430 × 10^-11) * (50 kg) * (2.0 × 10^30 kg) / (1.5 × 10^11 m)^2
F ≈ 3.55 × 10^22 N
Therefore, the force that the Sun exerts on a 50 kg person standing on Earth's surface is approximately 3.55 × 10^22 Newtons.
(b) To determine the ratio of the Sun's gravitational force to Earth's gravitational force on the same 50 kg person, we can use the formula:
Ratio = F_sun / F_earth
The gravitational force exerted by Earth on the person can be calculated using the same formula as in part (a), but with the mass of the Earth (m2) and the average distance from the person to the center of the Earth (r_earth).
The mass of the Earth (m2) is approximately 5.97 × 10^24 kg, and the average distance from the person to the center of the Earth (r_earth) is approximately 6.37 × 10^6 m.
Substituting the values, we have:
F_earth = (6.67430 × 10^-11) * (50 kg) * (5.97 × 10^24 kg) / (6.37 × 10^6 m)^2
F_earth ≈ 4.91 × 10^2 N
Now we can calculate the ratio:
Ratio = (3.55 × 10^22 N) / (4.91 × 10^2 N)
Ratio ≈ 7.23 × 10^19
Therefore, the ratio of the Sun's gravitational force to Earth's gravitational force on the same 50 kg person is approximately 7.23 × 10^19.
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A 10-mH inductor is connected in series with a 10-ohm resistor, a switch and a 6-V battery. The switch is closed at t = 0. Find the instant at which the current in the inductor reaches 50 percent of its maximum value? Express your answer as a multiple of the time constant.
The current in the inductor reaches 50 percent of its maximum value at approximately 0.69 times the time constant (0.69τ).
In an RL circuit, the time constant (τ) is given by the formula:
τ = L / R
where L is the inductance (10 mH = 10 × 10⁻³ H) and R is the resistance (10 Ω).
To find the time at which the current reaches 50 percent of its maximum value, we need to calculate 0.69 times the time constant.
τ = L / R = (10 × 10⁻³ H) / 10 Ω = 10⁻³ s
0.69τ = 0.69 × 10⁻³ s ≈ 6.9 × 10⁻⁴ s
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Select one correct answer from the available options in the below parts. a) [3 points] You shine monochromatic light of wavelength A through a narrow slit of width b≈ A and onto a screen that is very far away from the slit. What do you observe on the screen? OOne bright band OTwo bright fringes and three dark fringes OA series of bright and dark fringes with the central bright fringe being wider and brighter than the other bright fringes OA series of bright and dark fringes that are of equal widths b) [3 points] What does it mean for two light waves to be in phase ? OThe two waves have the same wavelength and frequency OThe two waves have the same amplitude OThe two waves reach their maximum value at the same time and their minimum value at the same time OThe two waves propagate in the same direction c) [3 points] You shine monochromatic light of wavelength through a narrow slit of width b> > and onto a screen that is very far away from the slit. What do you observe on the screen? OOne bright band OA series of bright and dark fringes with the central bright fringe being wider and brighter than the other bright fringes OA series of bright and dark fringes that are of equal widths OTwo bright fringes and three dark fringes d) [3 points] Monochromatic light is directed through two narrow parallel slits. There is a viewing screen away the slits that is used to observe the interference pattern. If you submerged the entire apparatus in water, how is the new interference pattern different from the original one? OThe bright and dark fringes are closer together OThe bright and dark fringes are farther apart OThe pattern does not change e) [3 points] Light propagating in a material 1 with index of refraction n₁ is incident on a new material 2 with index of refraction n₂
a) When shining monochromatic light of wavelength A through a narrow slit of width b≈ A onto a screen that is very far away from the slit, you observe a series of bright and dark fringes that are of equal widths. This is known as the single-slit diffraction pattern.
b) Two light waves are said to be in phase when they reach their maximum value (peak) and minimum value (trough) at the same time. In other words, the peaks and troughs of the two waves align perfectly.
c) When shining monochromatic light of wavelength through a narrow slit of width b>> A onto a screen that is very far away from the slit, you observe a series of bright and dark fringes with the central bright fringe being wider and brighter than the other bright fringes. This is known as the double-slit interference pattern.
d) If you submerged the entire apparatus, including the two narrow parallel slits and the viewing screen, in water, the new interference pattern would have the bright and dark fringes closer together. This is due to the change in the effective wavelength of light in water, resulting in a narrower spacing between the fringes.
e) The reflected light is in phase with the incident light.
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A research Van de Graaff generator has a 3.70 m diameter metal sphere with a charge of 1.09 mC on it.
(a) What is the electric potential on the surface of the sphere?
V
(b) At what distance from its center is the potential 3.00 MV?
m
(c) An oxygen atom with three missing electrons is released near the surface of the Van de Graaff
The electric potential on the surface of the sphere is [tex]\( 5.34 \times 10^6 \)[/tex] V. at a distance of 3.22 m from the center of the sphere, the potential is 3.00 MV. the kinetic energy of the oxygen atom at the distance determined in part (b) is approximately [tex]\(1.06 \times 10^{-7}\) eV or \(1.69 \times 10^{-17}\) MeV[/tex].
(a) To find the electric potential on the surface of the sphere, we can use the equation for the electric potential of a uniformly charged sphere:
[tex]\[ V = \frac{KQ}{R} \][/tex]
where:
- [tex]\( V \)[/tex] is the electric potential,
- [tex]\( K \)[/tex] is the electrostatic constant [tex](\( K = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \))[/tex],
- [tex]\( Q \)[/tex] is the charge on the sphere,
- [tex]\( R \)[/tex] is the radius of the sphere.
Given that the diameter of the sphere is 3.70 m, the radius [tex]\( R \)[/tex] can be calculated as half of the diameter:
[tex]\[ R = \frac{3.70 \, \text{m}}{2} \\\\= 1.85 \, \text{m} \][/tex]
Substituting the values into the equation:
[tex]\[ V = \frac{(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \times (1.09 \times 10^{-3} \, \text{C})}{1.85 \, \text{m}} \][/tex]
Calculating the value:
[tex]\[ V = 5.34 \times 10^6 \, \text{V} \][/tex]
Therefore, the electric potential on the surface of the sphere is [tex]\( 5.34 \times 10^6 \)[/tex] V.
(b) To find the distance from the center of the sphere at which the potential is 3.00 MV, we can use the equation for electric potential:
[tex]\[ V = \frac{KQ}{r} \][/tex]
Rearranging the equation to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{KQ}{V} \][/tex]
Substituting the given values:
[tex]\[ r = \frac{(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \times (1.09 \times 10^{-3} \, \text{C})}{3.00 \times 10^6 \, \text{V}} \][/tex]
Calculating the value:
[tex]\[ r = 3.22 \, \text{m} \][/tex]
Therefore, at a distance of 3.22 m from the center of the sphere, the potential is 3.00 MV.
(c) To find the kinetic energy of the oxygen atom at the distance determined in part (b), we need to use the principle of conservation of energy. The initial electric potential energy is converted into kinetic energy as the oxygen atom moves away from the charged sphere.
The initial electric potential energy is given by:
[tex]\[ U_i = \frac{KQq}{r} \][/tex]
where:
- [tex]\( U_i \)[/tex] is the initial electric potential energy,
- [tex]\( K \)[/tex] is the electrostatic constant [tex](\( K = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \))[/tex],
- [tex]\( Q \)[/tex] is the charge on the sphere,
- [tex]\( q \)[/tex] is the charge of the oxygen atom,
- [tex]\( r \)[/tex] is the initial distance from the center of the sphere.
The final kinetic energy is given by:
[tex]\[ K_f = \frac{1}{2}mv^2 \][/tex]
where:
- [tex]\( K_f \)[/tex] is the final kinetic energy,
- [tex]\( m \)[/tex] is
the mass of the oxygen atom,
- [tex]\( v \)[/tex] is the final velocity of the oxygen atom.
According to the conservation of energy, we can equate the initial electric potential energy to the final kinetic energy:
[tex]\[ U_i = K_f \][/tex]
Substituting the values:
[tex]\[ \frac{KQq}{r} = \frac{1}{2}mv^2 \][/tex]
We can rearrange the equation to solve for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{\frac{2KQq}{mr}} \][/tex]
Substituting the given values:
[tex]\[ v = \sqrt{\frac{2 \times (8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \times (1.09 \times 10^{-3} \, \text{C}) \times (3 \times 10^{-26} \, \text{kg})}{(3.22 \, \text{m})}} \][/tex]
Calculating the value:
[tex]\[ v = 6.84 \times 10^6 \, \text{m/s} \][/tex]
To convert the kinetic energy to MeV (mega-electron volts), we need to use the equation:
[tex]\[ K = \frac{1}{2}mv^2 \][/tex]
Converting the mass of the oxygen atom to electron volts (eV):
[tex]\[ m = (3 \times 10^{-26} \, \text{kg}) \times (1 \, \text{kg}^{-1}) \times (1.6 \times 10^{-19} \, \text{C/eV}) \\\\= 4.8 \times 10^{-26} \, \text{eV} \][/tex]
Substituting the values into the equation:
[tex]\[ K = \frac{1}{2} \times (4.8 \times 10^{-26} \, \text{eV}) \times (6.84 \times 10^6 \, \text{m/s})^2 \][/tex]
Calculating the value:
[tex]\[ K = 1.06 \times 10^{-7} \, \text{eV} \][/tex]
Therefore, the kinetic energy of the oxygen atom at the distance determined in part (b) is approximately [tex]\(1.06 \times 10^{-7}\) eV or \(1.69 \times 10^{-17}\) MeV[/tex].
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The charge of the released oxygen atom is +4.8 × 10⁻¹⁹ C.
a) The electric potential on the surface of the sphere
The electric potential on the surface of the sphere is given by,V=kQ/r, radius r of the sphere = 1.85 m
Charge on the sphere, Q=1.09 mC = 1.09 × 10⁻³ C, Charge of electron, e = 1.6 × 10⁻¹⁹ C
Vacuum permittivity, k= 8.85 × 10⁻¹² C²N⁻¹m⁻²
Substituting the values in the formula, V=(kQ)/rV = 6.6 × 10⁹ V/m = 6.6 × 10⁶ V
(b) Distance from the center where the potential is 3.00 MV
The electric potential at distance r from the center of the sphere is given by,V=kQ/r
Since V = 3.00 MV= 3.0 × 10⁶ V Charge on the sphere, Q= 1.09 × 10⁻³ C = 1.09 mC
Distance from the center of the sphere = rWe know that V=kQ/r3.0 × 10⁶ = (8.85 × 10⁻¹² × 1.09 × 10⁻³)/rSolving for r, we get the distance from the center of the sphere, r= 2.92 m
(c) Charge of the released oxygen atom, The released oxygen atom has 3 missing electrons, which means it has a charge of +3e.Charge of electron, e= 1.6 × 10⁻¹⁹ C
Charge of an oxygen atom with 3 missing electrons = 3 × (1.6 × 10⁻¹⁹)
Charge of an oxygen atom with 3 missing electrons = 4.8 × 10⁻¹⁹ C.
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