What is the maximum kinetic energy and speed of an electron ejected from a Na surface, in a photo-electric effect apparatus, when the surface is illuminated by light of wavelength 410 nm ? The work function for sodium is 2.28eV. b) What is the critical frequency below which no electrons are ejected from sodium? c) What is the kinetic energy of electrons emitted when yellow light of λ=600 nm is incident on Na ? d) Sketch a graph of kinetic energy of the ejected electron vs. frequency of incident light for the photoelectric effect in sodium. Indicate the work function and critical frequency on your graph. What is the slope of the graph?
Answers
The maximum kinetic energy of the ejected electron is approximately [tex]\(4.51 \times 10^{-19}\, \text{J}\)[/tex], and its maximum speed is approximately [tex]\(5.79 \times 10^5\, \text{m/s}\)[/tex]. the critical frequency below which no electrons are ejected from sodium is approximately [tex]\(9.27 \times 10^{14}\, \text{Hz}\)[/tex].
The kinetic energy of electrons emitted when yellow light of [tex]\(\lambda = 600\)[/tex] nm is incident on sodium is approximately [tex]\(2.15 \times 10^{-19}\, \text{J}\)[/tex].
a) To calculate the maximum kinetic energy [tex](\(E_{\text{max}}\))[/tex] and speed [tex](\(v_{\text{max}}\))[/tex] of an electron ejected from a sodium surface in the photoelectric effect, we can use the following formulas:
[tex]\[E_{\text{max}} = h \cdot \nu - \phi\]\\\\\v_{\text{max}} = \sqrt{\frac{{2E_{\text{max}}}}{{m_e}}}\][/tex]
where:
[tex]\(h\) is Planck's constant (\(6.62607015 \times 10^{-34}\, \text{J}\cdot\text{s}\))[/tex],
[tex]\(\nu\)[/tex] is the frequency of the incident light [tex](\(\frac{c}{\lambda}\), where \(c\)[/tex] is the speed of light and [tex]\(\lambda\)[/tex] is the wavelength of the light),
[tex]\(\phi\)[/tex] is the work function of sodium (in electron volts, eV),
[tex]\(m_e\)[/tex] is the mass of an electron [tex](\(9.10938356 \times 10^{-31}\, \text{kg}\))[/tex].
Given:
Wavelength of the incident light [tex](\(\lambda\))[/tex] = 410 nm [tex](\(410 \times 10^{-9}\, \text{m}\))[/tex],
Work function of sodium [tex](\(\phi\))[/tex] = 2.28 eV [tex](\(2.28 \times 1.602176634 \times 10^{-19}\, \text{J}\))[/tex].
First, calculate the frequency of the incident light:
[tex]\[\nu = \frac{c}{\lambda} = \frac{3 \times 10^8\, \text{m/s}}{410 \times 10^{-9}\, \text{m}} \\\\= 7.317 \times 10^{14}\, \text{Hz}\][/tex]
Now substitute the values into the equations to calculate [tex]\(E_{\text{max}}\) and \(v_{\text{max}}\)[/tex]:
[tex]\[E_{\text{max}} = (6.62607015 \times 10^{-34}\, \text{J}\cdot\text{s}) \cdot (7.317 \times 10^{14}\, \text{Hz}) - (2.28 \times 1.602176634 \times 10^{-19}\, \text{J})\]\\\v_{\text{max}} = \sqrt{\frac{{2 \cdot E_{\text{max}}}}{{9.10938356 \times 10^{-31}\, \text{kg}}}}\][/tex]
After evaluating the equations, we find:
[tex]\(E_{\text{max}} \approx 4.51 \times 10^{-19}\, \text{J}\)[/tex]
[tex]\(v_{\text{max}} \approx 5.79 \times 10^5\, \text{m/s}\)[/tex]
Therefore, the maximum kinetic energy of the ejected electron is approximately [tex]\(4.51 \times 10^{-19}\, \text{J}\)[/tex], and its maximum speed is approximately [tex]\(5.79 \times 10^5\, \text{m/s}\)[/tex].
b) The critical frequency [tex](\(\nu_{\text{c}}\))[/tex] is the threshold frequency below which no electrons are ejected. It can be calculated using the formula:
[tex]\[\nu_{\text{c}} = \frac{{\phi}}{{h}}\][/tex]
Substituting the values into the formula:
[tex]\[\nu_{\text{c}} = \frac{{2.28 \times 1.602176634 \times 10^{-19}\, \text{J}}}{{6.62607015 \times 10^{-34}\, \text{J}\cdot\text{s}}}\][/tex]
After evaluating the equation, we find:
[tex]\(\nu_{\text{c}} \approx 9.27 \times 10^{14}\, \text{Hz}\)[/tex]
Therefore, the critical frequency below which no electrons are ejected from sodium is approximately [tex]\(9.27 \times 10^{14}\, \text{Hz}\)[/tex].
c) To calculate the kinetic energy of electrons emitted when yellow light of [tex]\(\lambda = 600\)[/tex] nm is incident on sodium, we can use the same formula as in part (a):
[tex]\[E_{\text{max}} = h \cdot \nu - \phi\][/tex]
Given:
Wavelength of the yellow light [tex](\(\lambda\))[/tex] = 600 nm [tex](\(600 \times 10^{-9}\, \text{m}\))[/tex]
Calculate the frequency of the yellow light:
[tex]\[\nu = \frac{c}{\lambda} = \frac{3 \times 10^8\, \text{m/s}}{600 \times 10^{-9}\, \text{m}} \\\\= 5 \times 10^{14}\, \text{Hz}\][/tex]
Substitute the values into the equation to calculate [tex]\(E_{\text{max}}\)[/tex]:
[tex]\[E_{\text{max}} = (6.62607015 \times 10^{-34}\, \text{J}\cdot\text{s}) \cdot (5 \times 10^{14}\, \text{Hz}) - (2.28 \times 1.602176634 \times 10^{-19}\, \text{J})\][/tex]
After evaluating the equation, we find:
[tex]\(E_{\text{max}} \approx 2.15 \times 10^{-19}\, \text{J}\)\\[/tex]
Therefore, the kinetic energy of electrons emitted when yellow light of [tex]\(\lambda = 600\)[/tex] nm is incident on sodium is approximately [tex]\(2.15 \times 10^{-19}\, \text{J}\)[/tex].
d).
graph is in the image attached
KE is kinetic energy
f is frequency
Wo is work function and h is slope of the graph
fo is critical frequency
slope of the graph will represent plank's constant
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Related Questions
How much heat is necessary to change 20g of ice at 0 degree C into water at 0 degree C? (Lf = 80kcal/kg)
Answers
To change 20g of ice at 0 degree C into water at 0 degree C, 1600 calories of heat energy is required.Latent heat of fusion (Lf) is the energy released or absorbed by a substance during a change in state (from solid to liquid or liquid to solid) without any change in temperature.
Latent heat of fusion (Lf) is the energy released or absorbed by a substance during a change in state (from solid to liquid or liquid to solid) without any change in temperature.In this case, we are required to calculate the amount of heat energy required to change 20g of ice at 0 degree C into water at 0 degree C.Using the given formula:Heat energy = mass × latent heat of fusion= 20g × 80 kcal/kg= 1600 calories. Therefore, 1600 calories of heat energy is required to change 20g of ice at 0 degree C into water at 0 degree C.
When heat is applied to a substance, its temperature rises as the molecules in the substance vibrate more and move apart from each other. Eventually, the heat supplied is used up in breaking the intermolecular bonds between the molecules and overcoming the forces of attraction holding them together.At this point, the substance begins to change its state (e.g. from solid to liquid). During the state change, the temperature of the substance remains constant as the heat energy is being used to break the bonds between the molecules and not to increase their kinetic energy (i.e. temperature).This energy required to change the state of a substance without any change in temperature is called the latent heat of fusion. The value of latent heat of fusion for ice is 80 kcal/kg.To change 20g of ice at 0 degree C into water at 0 degree C, 1600 calories of heat energy is required. This is calculated using the formula:Heat energy = mass × latent heat of fusion= 20g × 80 kcal/kg= 1600 calories.Therefore, 1600 calories of heat energy is required to change 20g of ice at 0 degree C into water at 0 degree C.
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The total power of a sound wave remains constant as it travels away from a source, but the intensity changes. We perceive the loudness of sound relative to the intensity. The intensity of sound can be measured in W/m². (a) Why are noises quieter when they are farther away? (b) How much quieter is a train when you are 200 m away compared to when you are 100 m away?
Answers
(a) Because Sound waves spread out in three-dimensional space, and their intensity decreases as they move further away from the source. b) if you are 200m away from the source of sound compared to 100m, the sound will be four times quieter or 6 decibels quieter.
This results in a reduction in the amount of sound energy reaching our ears, making the sound seem quieter.The further a sound wave travels, the more it will spread out. As a result, the intensity of the sound wave decreases, resulting in a decrease in the amount of sound energy that reaches the ear. This leads to a reduction in the perceived loudness of the sound.
When the distance between the train and the listener increases, the intensity of the sound wave decreases as the sound wave spreads out. Thus, the train would be four times quieter (i.e., 6 decibels quieter) at a distance of 200m compared to 100m.Explanation:We have already discussed that the loudness of sound is a measure of how much sound energy is detected by the ear, while the intensity of sound can be measured in watts per square meter (W/m²).
The intensity of sound waves decreases as they travel further away from the source because the sound waves spread out in three-dimensional space. This leads to a reduction in the amount of sound energy reaching our ears, making the sound seem quieter. According to the inverse square law, if the distance is doubled, the intensity decreases to one-fourth of its original value. This means that if you are 200m away from the source of sound compared to 100m, the sound will be four times quieter or 6 decibels quieter.
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use thermal expansion to find the difference in length between an object that is heated and when it is cooled.
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Thermal expansion is the phenomenon in which the length of a material changes as a result of temperature changes. When heated, most materials expand, whereas when cooled, they shrink. As a result, we can determine the difference in length between a heated and a cooled object using thermal expansion.
Thermal expansion is a phenomenon in which the length of a material changes as a result of temperature changes. When a substance is heated, its constituent atoms vibrate more quickly and energetically, causing them to spread out and take up more space. The difference in length between an object that is heated and when it is cooled can be calculated using the following formula:ΔL = αL₀ΔTWhere:ΔL is the change in length of the object, α is the coefficient of linear expansion, L₀ is the original length of the object, and ΔT is the temperature difference experienced by the object. The coefficient of linear expansion is a measure of how much a material's length changes in response to temperature changes. It is represented by the symbol α and has units of per degree Celsius (°C⁻¹) or per kelvin (K⁻¹).
Thermal expansion is a phenomenon in which the length of a material changes as a result of temperature changes. When a substance is heated, its constituent atoms vibrate more quickly and energetically, causing them to spread out and take up more space. As a result, the substance's volume expands as well as its length. Similarly, when a substance is cooled, its atoms vibrate less quickly and energetically, causing them to pack together more tightly and take up less space. The temperature difference is the difference between the object's temperature when it is heated and when it is cooled.To illustrate this, consider an iron rod that is 1.0 meter long at a temperature of 20°C. The coefficient of linear expansion for iron is 1.2 x 10⁻⁵ K⁻¹. Suppose the rod is heated to a temperature of 200°C and then cooled back to its original temperature. The temperature difference experienced by the rod is therefore ΔT = 200 - 20 = 180°C. Using the formula above, we can calculate the difference in length of the rod as follows:ΔL = αL₀ΔT = (1.2 x 10⁻⁵ K⁻¹) (1.0 m) (180°C) = 0.00216 m = 2.16 mmTherefore, the difference in length between the rod when it is heated and when it is cooled is 2.16 mm.
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(0)
1) What is the effect of crossing over in the gametes? What is its effect in the next generation?
2) How would nondisjunction be different if it occurred in anaphase II? Draw the results of nondisjunction in anaphase II for a cell that starts meiosis with four chromosomes.
3) Using the 40X lens, focus on the cells within the pollen sacs in late prophase. Draw a cell in late prophase. Describe any differences from the early prophase slide.
4) Use the 5X lens to observe cells in late prophase. Note that some of our slides show sections of single pollen sacs, not whole anthers.
Answers
It is not possible to observe the details of the chromosomes and the spindle fibers. This is because the 5X lens has a lower magnification than the 40X lens, which is needed to observe these details. Nonetheless, the 5X lens can be useful for observing the general organization of the cells and the structure of the pollen sacs.
1) Crossing over during gamete formation results in the exchange of genetic information between homologous chromosomes. This increases genetic variation within the gametes, as well as in the resulting offspring. The effect of crossing over can be observed in the next generation as it leads to new combinations of traits and may increase diversity within a population.
2) Two gametes would have two chromosomes each (normal), andTwo gametes would have only one chromosome or three chromosomes (aneuploidy).3) During late prophase, the chromosomes are fully condensed, and the nuclear envelope has broken down. This allows for the chromosomes to interact with the spindle fibers, which will pull them apart during cell division.
In addition, the nucleolus is no longer visible, and the chromosomes are arranged in the center of the cell. In the case of pollen sacs, this is the stage where the cells are preparing to divide into haploid gametes through meiosis. A cell in late prophase would have more condensed and visible chromosomes than in early prophase.
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swell is more regular than waves in the wind-generated area because of what?
Answers
Swell is more regular than waves in the wind-generated area because of the properties of the waves. Swell is a set of ocean waves that are formed by far-off storms or winds.
They move away from the storm that generated them, traveling over vast expanses of water. Their wavelength, amplitude, and speed depend on their distance from the storm and the duration and strength of the storm that created them.
In general, swells have longer wavelengths, larger amplitudes, and higher speeds than wind-generated waves. As a result, they travel faster and maintain their shape over greater distances. This makes them more regular than wind-generated waves because they are less influenced by local wind conditions and topography.
Wind-generated waves, on the other hand, are created by local winds that blow over the surface of the ocean. They have shorter wavelengths and lower speeds than swells and are more affected by the local wind conditions and topography. Because of this, wind-generated waves are more irregular and variable than swells.
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Q11. A 5 cm long solenoid of 12 turns carriers a current of 4 A. What is the magnetic field at its center? a) 301.6 G b) 6.03 G (C) 12.1 G d) 0.121 G
Answers
The magnetic field at the center of a 5 cm long solenoid carrying a current of 4 A with 12 turns is 301.6 G.
A solenoid refers to a coil with several turns of wire. Magnetic field is generated within the solenoid when electric current flows through it. The formula to determine the magnetic field at the center of the solenoid is given as;B = (μ * n * I) / L Where B is the magnetic fieldμ is the permeability of free s pace is the current is the number of turns L is the length of the solenoid Substituting the given values; B = (μ * n * I) / L = (4π * 10⁻⁷ T*m/A * 12 * 4A) / 0.05m = 301.6 G
therefore, the magnetic field at the center of the solenoid is 301.6 G.
Attractive Field is the district around an attractive material or a moving electric charge inside which the power of attraction acts. A pictorial portrayal of the attractive field which depicts how an attractive power is circulated inside and around an attractive material.
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Question 9 A U-tube contains some mercury. 12 cm of water is added to one side of the U-tube. Part A Find how high the mercury rises on the other side from its original level. Use 13.5 g/cm³ as the d
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When 12 cm of water is added to one side of a U-tube, the mercury on the other side rises to a height of approximately 0.889 cm. This is based on the given density of mercury (13.5 g/cm³) and the principle of Pascal's law.
To find how high the mercury rises on the other side of the U-tube, we need to consider the principle of Pascal's law, which states that pressure is transmitted equally in all directions within an incompressible fluid.
Here's how we can approach the problem:
Density of mercury = 13.5 g/cm³
Height of water column = 12 cm
First, we need to calculate the pressure exerted by the water column. Pressure can be calculated using the formula P = ρgh, where P is the pressure, ρ is the density, g is the acceleration due to gravity, and h is the height of the column.
For the water column, the pressure exerted is P_water = ρ_water * g * h_water.
Next, we need to find the height to which the mercury rises on the other side.
Since the pressure is transmitted equally, the pressure exerted by the mercury column should balance the pressure exerted by the water column.
Let h_mercury be the height to which the mercury rises. The pressure exerted by the mercury column is P_mercury = ρ_mercury * g * h_mercury.
Since the pressures are equal, we have P_water = P_mercury.
Therefore, ρ_water * g * h_water = ρ_mercury * g * h_mercury.
Simplifying the equation, we find h_mercury = (ρ_water * h_water) / ρ_mercury.
Substituting the given values, we have h_mercury = (1 g/cm³ * 12 cm) / 13.5 g/cm³.
Simplifying further, h_mercury ≈ 0.889 cm.
Therefore, the mercury rises to a height of approximately 0.889 cm on the other side from its original level.
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The position function X (t) of a particle moving along an X axis is X = 4,0 - 6,0t^(2), with X in in meters and t in seconds. At what negative time and positive time does the particle pass through the origin?
Answers
The given position function of a particle moving along an X-axis is X = 4,0 - 6,0t^2with X in meters and t in seconds. To find the negative time and positive time at which the particle passes through the origin, we will set X = 0.
Given, position function of a particle X (t) = 4.0 - 6.0t^2.
For the particle to pass through the origin, X (t) must be 0.X (t) = 4.0 - 6.0t^2 = 0Solving for t, we get; t^2 = 4/3t = ± √(4/3)t = (2/√3) and t = -(2/√3).
Since time cannot be negative, negative time is not possible.
The particle passes through the origin at t = (2/√3) seconds or approximately 1.1547 seconds.
The negative time is not possible because time cannot be negative.
So, the particle passes through the origin at t = (2/√3) seconds or approximately 1.1547 seconds.
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Two 1.0 kg masses are 1.0 m apart on a frictionless table. Each has +1.5 μ
C of charge.
a) What is the magnitude of the electric force on one of the masses?
b) What is the initial acceleration of each mass if they are released and allowed to move?
Answers
The magnitude of the electric force acting on one of the masses is [tex]5.67 * 10^{-8}[/tex] N. The magnitude of the initial acceleration of each mass is 9.8 m/s², and the direction is towards the other mass.
Given that two 1.0 kg masses are 1.0 m apart on a frictionless table. Each has +1.5 μC of charge. We need to find the magnitude of the electric force on one of the masses and the initial acceleration of each mass if they are released and allowed to move. a) Magnitude of the electric force on one of the masses
The electric force between two charged particles can be calculated using Coulomb's law. The electric force acting on one of the masses is given by the following formula:
[tex]F = \frac{1}{4πε₀} \frac{q1q2}{r^{2}}[/tex]
Where, q₁ = Charge on first particle
q₂ = Charge on the second particle r = Distance between the charges
ε₀ = Permittivity of free space
Mass of each object, m = 1.0 kg, Charge on each object, q = +1.5 μC = [tex]1.5 * 10^{-6}[/tex] C Distance between the objects, r = 1.0 m.
Permittivity of free space, [tex]ε₀ = 8.85 * 10^{-12}[/tex]C²/N·m²
Substitute the given values in the Coulomb's law,
[tex]F = \frac{1}{4πε₀} \frac{q1q2}{r^{2}}[/tex]
[tex]F = \frac{1}{4πε₀} \frac{(1.5 * 10^{-6}) *(1.5* 10^{-6})}{1^{2}}[/tex]
[tex]F = \frac{1}{4πε₀} \frac{(2.25 * 10^{-12})}{1}[/tex]/1
F = [tex]5.67 * 10^{-8}[/tex] N.
Thus, the magnitude of the electric force acting on one of the masses is [tex]5.67 * 10^{-8}[/tex] N. b) Initial acceleration of each mass if they are released and allowed to move.
The gravitational force acting on each of the masses is given by,
Fg = mgFg
= 1.0 × 9.8
Fg = 9.8 N. The net force acting on each of the masses is given by,
Fnet = ma From Newton's second law of motion,
Net force = (mass) × (acceleration)
Fnet = Fe - Fg.
Where, Fe = Electric force, Fg = Gravitational force. Substitute the given values,
Fnet = Fe - FgFnet
= [tex]5.67 * 10^{-8}[/tex] - 9.8
Fnet = -9.8 N
As the net force is acting in the opposite direction to the gravitational force, the initial acceleration of the mass will be negative. The magnitude of the initial acceleration of each mass is 9.8 m/s², and the direction is towards the other mass.
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A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 18.0 degrees below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes sre defective. The car rolls from rest down the incline with a constant acceleration of 2.90 m/s2 for a distance of 60.0 m to the endge of the cliff, which is 30.0m above the ocean. (a) Find the cars position relative to the base of the cliff when the car lands in the ocean. (b) Find the length of the time the car is in the air.
Answers
(a) The car lands in the ocean at a horizontal distance of approximately 65.53 meters from the base of the cliff.
(b) The car is in the air for approximately 2.70 seconds.
To solve part (a), we can use the kinematic equation for displacement in the x-direction:
x = x₀ + v₀x * t + 0.5 * a * t²
where x is the final position, x₀ is the initial position (0 in this case), v₀x is the initial velocity in the x-direction (0 m/s since the car starts from rest), a is the constant acceleration (2.90 m/s²), and t is the time taken.
Plugging in the values, we have:
x = 0 + 0 * t + 0.5 * 2.90 m/s² * t²
Since we want to find the position when the car lands in the ocean, we set x to be equal to 60.0 m (the distance traveled by the car). Solving for t, we get:
60.0 m = 0.5 * 2.90 m/s² * t²
t² = 41.38 s²
t ≈ 6.43 s
Now, to find the horizontal distance from the base of the cliff, we can use trigonometry. The horizontal distance traveled by the car is given by:
x_horizontal = x * cos(18.0°)
Plugging in the values, we have:
x_horizontal = 60.0 m * cos(18.0°)
x_horizontal ≈ 65.53 m
Therefore, the car lands in the ocean at a horizontal distance of approximately 65.53 meters from the base of the cliff.
To solve part (b), we can use the equation for time of flight of a projectile:
t_flight = 2 * t
Plugging in the value of t we obtained earlier, we have:
t_flight = 2 * 6.43 s
t_flight ≈ 12.86 s
Therefore, the car is in the air for approximately 12.86 seconds.
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A speedboat accelerates uniformly from rest at 3.0 m/s. The distance the speedboat will travel between 4.0 s and 6.0 s is m. (Record your two-digit answer in the answer space.)
Answers
The distance the speedboat will travel between 4.0 s and 6.0 s is 12 m.
The speedboat starts from rest and undergoes uniform acceleration at a rate of 3.0 m/s². We need to find the distance traveled by the speedboat between 4.0 s and 6.0 s.
To calculate the distance traveled, we can use the kinematic equation:
d = v₀t + (1/2)at²
where d is the distance, v₀ is the initial velocity, t is the time, and a is the acceleration.
Given that the initial velocity v₀ is 0 m/s, the time interval is 2.0 s (6.0 s - 4.0 s), and the acceleration a is 3.0 m/s², we can substitute these values into the equation:
d = 0 + (1/2)(3.0 m/s²)(2.0 s)²
d = (1/2)(3.0 m/s²)(4.0 s²)
d = 6.0 m
Therefore, the distance the speedboat will travel between 4.0 s and 6.0 s is 12 m. The speedboat undergoes constant acceleration, and by using the kinematic equation, we find that the distance traveled is 6.0 m. This represents the total displacement of the speedboat within the given time interval.
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If the net torque applied to an object is constant and the rotational inertia is doubled, what happens to the angular acceleration?
a) doubles
b) half is reduced
c) it does not change
d) a quarter is reduced
Answers
If the net torque applied to an object is constant and the rotational inertia is doubled, the angular acceleration will be reduced by half.
This can be explained using Newton's second law for rotation, which states that the net torque applied to an object is equal to the product of its rotational inertia (I) and its angular acceleration (α): τ = I * α If the net torque remains constant, but the rotational inertia (I) is doubled, the equation becomes: τ = 2I * α. Since τ is constant, if I is doubled, α must be halved in order to maintain the equality. Therefore, the angular acceleration is reduced by half. So, the correct answer is: b) Half is reduced.
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The potential energy of a +6.00 x 10-6 C charge
decreases from 0.06 J to 0.02 J when it is moved from point A to
point B. What is the change in electric potential between these two
points?
Answers
The change in electric potential between points A and B, as a +6.00 x 10^-6 C charge moves, is approximately 6.67 x 10^3 volts. Electric potential is the work done per unit charge in an electric field.
The electric potential difference, also known as voltage, is defined as the change in electric potential energy per unit charge between two points. Mathematically, it can be calculated using the formula:
ΔV = ΔPE / q
Where ΔV represents the change in electric potential, ΔPE represents the change in potential energy, and q represents the charge.
In this case, the charge q is given as +6.00 x 10^-6 C.
The change in potential energy ΔPE is the difference between the initial potential energy (0.06 J) and the final potential energy (0.02 J), which is 0.06 J - 0.02 J = 0.04 J.
Now, we can calculate the change in electric potential:
ΔV = ΔPE / q
ΔV = 0.04 J / (+6.00 x 10^-6 C)
ΔV ≈ 6.67 x 10^3 V
Therefore, the change in electric potential between points A and B is approximately 6.67 x 10^3 volts.
It's important to note that electric potential is a scalar quantity and is measured in volts (V). It represents the amount of work done to move a unit positive charge from one point to another in an electric field.
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1550-kg car rounds a circular turn of radius 145 m, toward the left, on a horizontal road. Its angular momentum about the center of the turn has magnitude 2.46 x 106 kg. m/s. Part A What is the direction of the car's angular momentum? away from the center of the turn vertically downward toward the center of the turn vertically upward Submit Previous Answers Correct Part B What is the speed of the car? Express your answer with the appropriate units. μΑ ? V= Value Units Submit Request Answer Part C What is the magnitude of the car's angular momentum about the center of the turn, once it's exited the turn and is on a straight stretch of the road? Express your answer with the appropriate units. μΑ ? Lfinal = Value Units
Answers
The direction of the car's angular momentum is vertically downward towards the center of the turn. The speed of the car is 23.6 m/s. The magnitude of the car's angular momentum about the center of the turn, once it's exited the turn and is on a straight stretch of the road is 5.07 x 10⁶ kg·m/s.
Part A:What is the direction of the car's angular momentum
The direction of the car's angular momentum is vertically downward towards the center of the turn.Let the direction of the angular momentum be the z-axis (perpendicular to the horizontal plane) since the car is turning leftward, its velocity is directed towards the center of the turn, and so its angular momentum is directed towards the bottom.
Part B:What is the speed of the car?
The formula for angular momentum is given by:L = I ωwhereL is the angular momentum, I is the moment of inertia, and ω is the angular velocity.Since the car is a point particle, its moment of inertia is given by:
I = MR2where M is the mass of the car and R is the radius of the turn.
The angular velocity can be determined using the speed and the radius of the turn.ω = v/R
We can write the expression for angular momentum as:L = MR2(v/R)R = MvR
The magnitude of angular momentum is given as:L = 2.46 x 10⁶ kg·m/sMass of the car, M = 1550 kg
Radius of the turn, R = 145 m
Substituting the values in the above equation:2.46 x 10⁶ kg·m/s = 1550 kg × v × 145 mv = 23.6 m/s
Part C:What is the magnitude of the car's angular momentum about the center of the turn, once it's exited the turn and is on a straight stretch of the road Once the car is on the straight stretch of the road, there is no centripetal force acting on it. Therefore, it will move in a straight line with a constant speed. Since there is no force acting on the car perpendicular to its motion, the angular momentum will remain constant since torque is zero.
After the car exits the turn, the magnitude of the angular momentum remains constant:
Lfinal = MRv where M is the mass of the car, R is the radius of the turn, and v is the speed of the car.Lfinal = (1550 kg)(145 m)(23.6 m/s)Lfinal = 5.07 x 10⁶ kg·m/s
Therefore, the magnitude of the car's angular momentum about the center of the turn, once it's exited the turn and is on a straight stretch of the road is 5.07 x 10⁶ kg·m/s.
In conclusion, the direction of the car's angular momentum is vertically downward towards the center of the turn. The speed of the car is 23.6 m/s. The magnitude of the car's angular momentum about the center of the turn, once it's exited the turn and is on a straight stretch of the road is 5.07 x 10⁶ kg·m/s.
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A particale's velocity function is given by V=4t² + 3t - 2 with V in meter/second and t in second Find the acceleration at t-2 s 19m/s2.a 11m/s2.b 20m/s2.c 8m/s2 d 1507 SIMET 150 N 10 PEET 1655 PELLE
Answers
The acceleration at t = 2 seconds is 20 m/s2.The velocity function for a particle is V = 4t² + 3t - 2, with V in meters/second and t in seconds.
The acceleration is the time derivative of velocity. It is denoted as a(t) or V'(t). The acceleration at a specific point in time t can be found by differentiating the velocity function with respect to time t. Thus, the acceleration function a(t) = dV(t)/dt. Differentiating the velocity function V(t) = 4t² + 3t - 2 with respect to t gives the acceleration function a(t) = 8t + 3. When t = 2 seconds, the acceleration is a(2) = 8(2) + 3 = 16 + 3 = 19 m/s2. Therefore, the acceleration at t = 2 seconds is 19 m/s2.
Speed increase is characterized as. The pace of progress of speed as for time. Because it has both magnitude and direction, acceleration is a vector quantity. It is additionally the second subordinate of position concerning time or it is the primary subsidiary of speed regarding time
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A bag containing 0°C ice is much more effective in absorbing energy than one containing the same amount of 0°C water.
a) What heat transfer, in joules, is necessary to raise the temperature of 0.75 kg of water (c = 4186 J/(kg⋅°C)) from 0°C to 30.0°C?
Qw = J
b) How much heat transfer, in joules, is required to first melt 0.75 kg of 0°C ice (Lf = 334 kJ/kg) and then raise its temperature from 0°C to 30°C?
Qtot = J
Answers
a) Qw = 94335 J
b) Qtot = 344835 J
What is the required heat transfer in joules to raise the temperature of 0.75 kg of water from 0°C to 30.0°C, and how much heat transfer in joules is needed to melt 0.75 kg of 0°C ice and raise its temperature to 30°C? To calculate the heat transfer necessary to raise the temperature of water, we can use the formula:Qw = mcΔT
Qw is the heat transfer in joules,
m is the mass of water in kilograms (0.75 kg in this case),
c is the specific heat capacity of water (4186 J/(kg⋅°C)),
ΔT is the change in temperature (30.0°C - 0°C = 30.0°C).
Substituting the values into the formula, we have:
Qw = (0.75 kg) × (4186 J/(kg⋅°C)) × (30.0°C) = 94335 J
Therefore, the heat transfer necessary to raise the temperature of 0.75 kg of water from 0°C to 30.0°C is 94335 joules.
To calculate the total heat transfer required to melt ice and then raise its temperature, we need to consider two steps:First, we calculate the heat transfer required to melt the ice:
Qmelt = mLf
Where:
Qmelt is the heat transfer for melting,
m is the mass of ice in kilograms (0.75 kg in this case),
Lf is the latent heat of fusion for ice (334 kJ/kg = 334000 J/kg).
Substituting the values into the formula, we have:
Qmelt = (0.75 kg) × (334000 J/kg) = 250500 J
Next, we calculate the heat transfer required to raise the temperature of water from 0°C to 30.0°C, as calculated in part a:
Qraise = Qw = 94335 J
The total heat transfer is the sum of the heat transfers for melting and raising the temperature
Qtot = Qmelt + Qraise = 250500 J + 94335 J = 344835 J
Therefore, the total heat transfer required to first melt 0.75 kg of 0°C ice and then raise its temperature from 0°C to 30.0°C is 344835 joules.
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1 T² 2 money T₂2 Application: We know that the moon orbits the earth. The orbital period of the moon is 27.32 days, and the distance from the moon to the earth is 384,000 km, We wish to use this to compute the size of the orbit of geosynchronous orbits. (A geosynchronous orbit is one for which the orbital period is 1 day.) So we know r1 - = 384, 000 km, T₁ - 27.32 days, and T₂ 1 day Plug in those values in the formula from Kepler's 3rd law of planetary motion, and solve for №2. Note that you will need to watch out for units: you can use whatever unit you want, so long as you are consistent. Do work on paper, and upload here.
Answers
It should be noted that based on the information, the size of the orbit for geosynchronous orbits is approximately 6929.38 km.
How to calculate the valueThe formula can be written as:
(T₁ / T₂)² = (r₁ / r₂)³
Substituting the known values:
(27.32 days / 1 day)² = (384,000 km / r₂)³
Simplifying:
27.32² = (384,000³) / r₂³
To solve for r₂, we can rearrange the equation:
r₂³ = (384,000³) / 27.32²
Taking the cube root on both sides:
r₂ = ∛[(384,000³) / 27.32²]
r₂ ≈ ∛(231,449,856,000,000 / 747.5024)
r₂ ≈ ∛(309,579,898,531.2)
r₂ ≈ 6929.38 km
Therefore, the size of the orbit for geosynchronous orbits is approximately 6929.38 km.
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You are standing 0.50 m in front of a lens that projects an image of you onto a wall 3.0 m on the other side of the lens. What is the focal length of the lens? What is the magnification?
Answers
The image distance, the object distance, and the focal length are related by the lens equation which is given as 1/f = 1/o + 1/i. The focal length, object distance, and image distance of a lens system can be determined using this equation.
The magnification of the lens can also be determined. The problem requires determining the focal length of a lens given the object distance and image distance. Given values include; the object distance, u = 0.50 m and the image distance, v = 3.0 m. Then we can substitute the given values into the lens formula as follows;
1/f = 1/o + 1/i
= 1/0.5 + 1/3
= 1.67/f
= 1.67f
= 1/1.67
= 0.6 m
The focal length of the lens is 0.6 m. To determine the magnification, we can use the magnification equation, m = -v/u. Therefore, the magnification is;
m = -v/u
= -3.0/0.5
= -6.
The magnification is -6. Therefore, the focal length of the lens is 0.6 m and the magnification is -6.
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what power does the resistor consume if it is connected to a 12.6- vv car battery? assume that rr remains constant when the power consumption changes.
Answers
If a resistor is connected to a 12.6 V car battery, then the power consumed by the resistor can be calculated using the formula: Power = V²/R where V is the voltage across the resistor and R is the resistance of the resistor.
Given, the voltage across the resistor is 12.6 V. If the resistance of the resistor is not provided, then it is impossible to determine the power consumed by the resistor. However, it is given that the resistance of the resistor remains constant when the power consumption changes.
Therefore, let's assume that the resistance of the resistor is 10 ohms.
Using the formula,
Power = V²/R = 12.6²/10 = 15.876 W
Thus, the power consumed by the resistor is 15.876 W when it is connected to a 12.6 V car battery and has a resistance of 10 ohms.
Therefore, the power consumed by a resistor connected to a 12.6 V car battery can be calculated using the formula Power = V²/R, where V is the voltage across the resistor and R is the resistance of the resistor. If the resistance of the resistor is not provided, then it is impossible to determine the power consumed by the resistor.
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Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress. Specify the orientation of the element in each case 10 MP 20 MPa
Answers
(a) The principal stresses are 10 MPa and 20 MPa.
(b) The maximum in-plane shear stress and average normal stress need further information to be determined.
To determine the principal stresses, we need to know the orientation of the element in question. The principal stresses are the maximum and minimum normal stresses experienced by the element. However, without knowing the orientation, we cannot calculate the in-plane shear stress or average normal stress.
These values depend on the orientation of the element's surface relative to the applied forces. Therefore, additional information is required to calculate the maximum in-plane shear stress and average normal stress.
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what is the angular resolution limit (degrees) set by diffraction for the 279- cm mirror diameter telescope ( λ =560 nm )?
Answers
To calculate the angular resolution limit set by diffraction for a telescope, we can use the formula: θ = 1.22 * (λ / D)
where θ is the angular resolution limit, λ is the wavelength of light, and D is the diameter of the telescope's mirror. Given that the mirror diameter is 279 cm (or 2.79 meters) and the wavelength of light is 560 nm (or 5.6 × 10^-7 meters), we can plug in these values into the formula to find the angular resolution limit: θ = 1.22 * (5.6 × 10^-7 / 2.79) Calculating the value: θ = 1.22 * 2.00716845878136200716845878136e-7. θ ≈ 2.45 × 10^-7 degrees. Therefore, the angular resolution limit set by diffraction for the 279-cm mirror diameter telescope with a wavelength of 560 nm is approximately 2.45 × 10^-7 degrees.
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"
Which of the following statements are TRUE about a body moving in
circular motion?
A. For a body moving in a circular motion at constant speed,
the direction of the velocity vector is the same as the
10 1 point A Which of the following statements are TRUE about a body moving in circular motion? A. For a body moving in a circular motion at constant speed, the direction of the velocity vector is the same as the direction of
the acceleration
B. At constant speed and radius, increasing the mass of an object moving in a circular path will increase the net force.
C. If an object moves in a circle at a constant speed, its velocity vector will be constant in magnitude but changing in direction
a.) A and B
b.) A, B and C
c.) A and C
d.) B and C
Answers
Option c) A and C statements are TRUE about a body moving in circular motion.
a) For a body moving in circular motion at a constant speed, the direction of the velocity vector is the same as the direction of the acceleration. This is true because in circular motion, the velocity vector is always tangential to the circular path, and the acceleration vector is directed towards the center of the circle, perpendicular to the velocity vector.
b) Increasing the mass of an object moving in a circular path will not directly affect the net force. The net force is determined by the centripetal force required to keep the object in circular motion, which is determined by the object's mass, speed, and radius of the circular path. Increasing the mass alone does not change the net force.
c) If an object moves in a circle at a constant speed, its velocity vector will be constant in magnitude but changing in direction. This is because the object is constantly changing its direction while maintaining the same speed. Velocity is a vector quantity that includes both magnitude (speed) and direction, so if the direction is changing, the velocity vector is also changing.
Therefore, the correct statements are A and C.
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A company decides to sell the cherries stacked on top of eachother in a cylindrical cardboard container. What would be the smallest possible diameter, height, and volume of the containe
Answers
The smallest possible diameter, height, and volume of the cylindrical cardboard container for stacking cherries on top of each other would depend on the size and quantity of cherries, but a smaller diameter would generally lead to a smaller height and volume.
The size of the container needed to stack cherries in a cylindrical shape can vary depending on the size and quantity of the cherries. However, to minimize the container's dimensions, we can consider a few factors.
First, let's focus on the diameter. A smaller diameter would allow for a tighter stack of cherries and reduce the empty space between them. This means more cherries can be accommodated in a smaller area. By minimizing the diameter, we optimize the container's capacity for the given quantity of cherries.
Next, let's consider the height of the container. The height should be sufficient to accommodate the desired quantity of cherries while ensuring they are stacked securely. However, a smaller diameter can compensate for a taller height, as it allows for a more compact arrangement of cherries within the container. Thus, a smaller diameter would generally lead to a smaller height.
Finally, the volume of the container can be calculated using the formula for the volume of a cylinder: V = πr²h, where V represents volume, r represents the radius (half the diameter), and h represents the height. By minimizing the diameter and height, we can achieve the smallest possible volume for the given quantity of cherries.
In conclusion, the smallest possible diameter, height, and volume of the cylindrical cardboard container for stacking cherries depend on the size and quantity of cherries, but a smaller diameter would generally lead to a smaller height and volume. By optimizing these dimensions, we can ensure an efficient use of space while maintaining the structural integrity of the cherry stack.
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A car and its suspension system can be simply modelled as a
large mass (the mass of the car) on a spring.
Calculate the effective spring constant in this model if the
suspension is adjusted so the 130
Answers
Answer:
Explanation:
To calculate the effective spring constant of a car's suspension system, we need to know the mass of the car and the displacement of the suspension system under a certain load.
Let's assume that the car's mass is represented by m and the displacement of the suspension system under a load of 130 N is represented by x.
In this model, the force exerted by the spring is given by Hooke's Law:
F = -k * x
where F is the force, k is the spring constant, and x is the displacement.
The weight of the car is given by:
Weight = m * g
where m is the mass of the car and g is the acceleration due to gravity.
Since the suspension system is adjusted so that the force exerted by the spring is equal to the weight of the car, we have:
k * x = m * g
Rearranging the equation, we can solve for the spring constant k:
k = (m * g) / x
Therefore, to calculate the effective spring constant in this model, we need to know the mass of the car (m), the acceleration due to gravity (g), and the displacement of the suspension system under a load of 130 N (x).
In a simplified model of a car and its suspension system, the effective spring constant can be calculated by adjusting the suspension to achieve a specific frequency. By using Hooke's law and the equation for natural frequency, the effective spring constant is found to be (4π²m)/130², where m is the car's mass.
In the simplified model of a car and its suspension system, where the car's mass is represented as a large mass on a spring, the effective spring constant can be calculated using Hooke's law and the equation for the natural frequency of a mass-spring system.
Hooke's law states that the force exerted by a spring is proportional to the displacement from its equilibrium position. Mathematically, it can be expressed as F = -kx, where F is the force, k is the spring constant, and x is the displacement.
The natural frequency of a mass-spring system is given by the equation f = 1/(2π√(m/k)), where f is the frequency, m is the mass, and k is the spring constant.
In this case, the suspension is adjusted so the 130
By equating the two equations above, we have:
130 = 1/(2π√(m/k))
Rearranging the equation, we get:
k = (4π²m)/130²
The effective spring constant (k) is therefore given by (4π²m)/130², where m is the mass of the car.
It's important to note that this simplified model neglects many factors and complexities of a real suspension system, and actual suspensions are more intricate with multiple components and nonlinear behaviors.
This model serves as a basic approximation for understanding the behavior of the system under simplified conditions.
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A dog walks at 6 m/s on the deck of a boat that is travelling
at 7.6 m/s with respect to the water.
What is the velocity of the dog with respect to the water if it
walks towards the bow (front)?
What
Answers
Velocity when the When the dog walks towards the
Bow: 13.6 m/s forward.
Stern: 1.6 m/s backward.
Starboard: 6 m/s to the right.
When the dog walks towards the bow (front) of the boat, its velocity with respect to the water is the vector sum of its velocity on the deck and the velocity of the boat. Since the boat is traveling forward at 7.6 m/s and the dog is walking at 6 m/s in the same direction, the resulting velocity is the sum of these two velocities: 7.6 m/s + 6 m/s = 13.6 m/s forward.
When the dog walks towards the stern (back) of the boat, its velocity with respect to the water is the difference between the velocity of the boat and its own velocity on the deck: 7.6 m/s - 6 m/s = 1.6 m/s backward.
When the dog walks towards the starboard (right) side of the boat, its velocity with respect to the water remains unchanged at 6 m/s to the right, as it is not affected by the boat's motion in that direction.
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the complete question is:
A dog walks at 6 m/s on the deck of a boat that is travelling at 7.6 m/s with respect to the water.
What is the velocity of the dog with respect to the water if it walks towards the bow (front)?
What is the velocity of the dog with respect to the water if it walks towards the stern (back)?
What is the velocity of the dog with respect to the water if it walks towards the starboard (right)?
A turntable slows from an initial rate of 28.0 rad/s at a rate of 0.580 rad/s2. The turntable is a disk with a diameter of 40.0 cm and mass of 2.00 kg: The slowing of the turntable is due to a frictional force exerted 1.00 cm from the axis of rotation (a) Determine the magnitude of the tangential acceleration of a point on the edge of the turntable: m/s2 Determine the time it takes the turntable to come to rest_ (c) Determine the number of revolutions the turntable makes before stopping: revolutions (d) Determine the magnitude of the torque exerted on the turntable Nm (e) Determine the magnitude of the frictional force. Determine the magnitude of the initial angular momentum of the turntable_ kg m?/s
Answers
(a) The magnitude of the tangential acceleration of a point on the edge of the turntable is 0.290 m/s².
(b) The time it takes the turntable to come to rest is 48.3 s.
(c) The number of revolutions the turntable makes before stopping is 4.79 revolutions.
(d) The magnitude of the torque exerted on the turntable is 0.116 Nm.
(e) The magnitude of the frictional force is 0.116 N.
(f) The magnitude of the initial angular momentum of the turntable is 0.056 kg m²/s.
(a) The tangential acceleration can be calculated using the formula a = α × r, where α is the angular acceleration and r is the radius of the turntable. Given α = -0.580 rad/s² and r = 0.20 m (half of the diameter), we find a = 0.290 m/s².
(b) The time it takes for the turntable to come to rest can be determined using the equation vf = vi + at, where vf is the final velocity (zero in this case), vi is the initial velocity (28.0 rad/s), a is the acceleration (-0.580 rad/s²), and t is the time. Rearranging the equation, we have t = (vf - vi) / a = -28.0 rad/s / (-0.580 rad/s²) = 48.3 s.
(c) The number of revolutions the turntable makes before stopping can be found using the equation θ = ωi × t + 0.5 × α × t², where θ is the angle in radians, ωi is the initial angular velocity, t is the time, and α is the angular acceleration. Since ωi = 28.0 rad/s, α = -0.580 rad/s², and t = 48.3 s, we can calculate θ = 4.79 revolutions.
(d) The magnitude of the torque exerted on the turntable can be determined using the equation τ = I × α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. The moment of inertia for a disk rotating about its axis is given by I = (1/2) × m × r², where m is the mass of the disk and r is its radius. Substituting the given values, we find τ = 0.116 Nm.
(e) The magnitude of the frictional force can be calculated using the equation f = m × a, where f is the force, m is the mass, and a is the acceleration. Substituting the given values, we find f = 0.116 N.
(f) The magnitude of the initial angular momentum of the turntable can be calculated using the equation L = I × ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. Substituting the given values, we find L = 0.056 kg m²/s.
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what is the y component of the force on the particle at y=0.5m? what is the y-component of the force on the particle at y=4m?
Answers
The y-component of the force on the particle at y = 4 m is -4.9 N.
To calculate the y-component of the force on the particle at y = 0.5 m and at y = 4 m, we need to first understand the given information. In physics, force is the push or pull on an object due to its interaction with another object. The formula for calculating force is F = ma, where F is the force, m is the mass of the object and a is the acceleration due to force. In this case, we have a gravitational force acting on the particle and the formula for gravitational force is Fg = mg, where Fg is the gravitational force, m is the mass of the particle and g is the acceleration due to gravity. We know that the force is acting on a particle and the force is the gravitational force which is acting due to the interaction between the particle and the Earth. The y-component of the force on the particle at y = 0.5 m:
From the given information, we know that the particle is located at a height of 0.5 m above the ground.
Hence, we can calculate the gravitational force acting on the particle by the following formula: Fg = mg, where m is the mass of the particle and g is the acceleration due to gravity. Fg = 0.5 kg * 9.8 m/s^2 = 4.9 N
Now, we know that the gravitational force is acting downwards on the particle, hence the y-component of the force will be negative. Thus, the y-component of the force on the particle at y = 0.5 m is -4.9 N. The y-component of the force on the particle at y = 4 m: From the given information, we know that the particle is located at a height of 4 m above the ground. Hence, we can calculate the gravitational force acting on the particle by the following formula:Fg = mg, where m is the mass of the particle and g is the acceleration due to gravity. Fg = 0.5 kg * 9.8 m/s^2 = 4.9 N
Now, we know that the gravitational force is acting downwards on the particle, hence the y-component of the force will be negative. Thus, the y-component of the force on the particle at y = 4 m is -4.9 N.
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The 2-in.-diameter drive shaft AB on the helicopter is subjected to an axial tension of 10 000 lb and a torque of 300 lb. ft. Determine the principal stresses and the maximum in-plane shear stress that act at a point on the surface of the shaft AB. Full credit will be given for problems that list assumptions, show an appropriate free body diagram of the pipe, and show work to solve equilibrium equations.
Answers
The maximum in-plane shear stress acting at the surface of the drive shaft is 67.64 x 10⁴ lb/ft².
The ratio of the tangential force to the cross-sectional area of the surface it is acting on is known as the shear stress.
Diameter of the drive shaft, d = 2 in = 0.167 ft
The tension acting on the drive shaft, T = 10⁴ lb
The torque acting on the drive shaft, τ = 300 lb.ft
The expression for the stress acting on the drive shaft is given by,
σ = T/A
σ = 10⁴/(πd²/4)
σ = 4 x 10⁴/3.14 x (0.167)²
σ = 45.67 x 10⁴ lb/ft²
The expression for the shear stress is given by,
τ' = T/J
τ' = 10⁴x 1/(π/2 x 1)
τ' = 10⁴/1.57
τ' = 63.67 x 10⁴ lb/ft²
The expression for the principal shear stress is given by,
σ₍₁,₂₎ = σₓ + σy ± √{[(σₓ - σy)/2]² + (τ')²}
σ₍₁,₂₎ = 45.67 x 10⁴+ 0 ± √[(45.67 x 10⁴/2)² + (63.67 x 10⁴)²]
σ₍₁,₂₎ = 45.67 x 10⁴ ± 67.64 x 10⁴
So,
σ₁ = -21.97 x 10⁴ lb/ft²
σ₂ = 113.31 x 10⁴ lb/ft²
Therefore, the maximum in-plane shear stress acting at the surface of the drive shaft is given by,
τ(max) = √{[(σₓ - σy)/2]² + (τ')²}
τ(max) = √[(45.67 x 10⁴/2)² + (63.67 x 10⁴)²]
τ(max) = 67.64 x 10⁴ lb/ft²
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"A charged particle of charge q moves perpendicular to a uniform
magnetic field whose magnetic flux density has magnitude b Find the
magnitude of q if b is 0.4T The speed of the particle is 8m/s and
the magnitude of the force on it is 96mN (milli newton)
Answers
The magnitude of the charge (q) of the particle is approximately 0.3 Coulombs.
To find the magnitude of the charge (q) of a charged particle moving perpendicular to a uniform magnetic field, we can use the formula for the magnetic force on a charged particle:
F = qvB,
where F is the force, q is the charge of the particle, v is the speed of the particle, and B is the magnetic flux density.
Magnetic flux density (B): 0.4 T
Speed of the particle (v): 8 m/s
Magnitude of the force (F): 96 mN (converted to N)
We can rearrange the formula to solve for q:
q = F / (vB).
Substituting the given values:
q = (96 x 10^-3 N) / ((8 m/s)(0.4 T)).
Simplifying the expression, we find:
q ≈ 0.3 C.
Therefore, the magnitude of the charge (q) of the particle is approximately 0.3 Coulombs.
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A "7805" is a (select best answer) o power IC. super capacitor that dramatically reduces the ripple voltage in a power supply, linear voltage regulator. switching voltage regulator that reduces the ripple voltage in a power supply, low drop-out voltage regulator (LDO) that reduces the ripple voltage in a power supply A power supply has a no-load output voltage of 5 V. A load is connected to the power supply and the output voltage drops to 4.9 V. What is the magnitude of the change in output voltage in percent? 2.041% 2% Need additional information Save Consider a linear power supply consisting of a transformer, 4-diode bridge rectifier, smoothing capacitor, and linear voltage regulator. The regulator has a 65 dB ripple rejection ratio. The ripple voltage at the smoothing capacitor is 350 mV. The output ripple voltage is Need more information. about 0.11 muV. about 200,uV. about 5.4 mV
Answers
Output ripple voltage is about 5.4 mV. A "7805" is a linear voltage regulator. Linear voltage regulators are DC to DC voltage converters that reduce the DC voltage. They accept an input voltage and regulate it to a fixed output voltage. This output voltage is highly regulated, and it's referred to as a clean voltage.
Linear voltage regulators function by reducing the voltage between the input and output of the regulator to keep the output voltage constant.
A power supply has a no-load output voltage of 5 V. A load is connected to the power supply and the output voltage drops to 4.9 V.
The magnitude of the change in output voltage in percent is:2%.
Given, no-load output voltage of 5 V and output voltage drops to 4.9 V when load is connected.
Output voltage ripple can be calculated using the formula;
Output voltage ripple = (Ripple factor × VDC) / (Load resistance).
Ripple factor = √(VRMS / VDC).
Ripple rejection ratio is given by,RRR = 20 log (Vout with ripple / Vout without ripple)
dB is the unit of measurement used for the ripple rejection ratio.
Rearranging the equation we get,
Vout with ripple / Vout without ripple = antilog (RRR/20).
We can calculate the Vout with ripple using the given ripple voltage at the smoothing capacitor.
Output voltage with ripple = VDC ± Vripple
where, VDC = DC voltage, V ripple = Ripple voltage
Vripple = 350 mV
RRR = 65 dB
Vout with ripple / Vout without ripple = antilog (65/20)= 1778.27941
Vout with ripple = 1778.27941 × Vout without ripple
= 1778.27941 × 5 V
= 8891.39706 mV
= 8.89139706 V
Output voltage with ripple = VDC ± V ripple
= 8.89139706 mV ± 350 mV
= 5.2 mV (approx)
Output ripple voltage is about 5.4 mV.
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