The pressure of the gas in the cell decreased.
The mercury manometer shown in Figure 1 is attached to a gas cell. The mercury height h is 120 mm when the cell is placed in an ice-water mixture.
The mercury height drops to 30 mm when the device is carried into an industrial freezer. The right tube of the manometer is much narrower than the left tube.
The assumption that can be made about the gas volume is that it remains constant. The volume of a gas in a closed container is constant unless the pressure, temperature, or number of particles in the gas changes. The device is carried from an ice-water mixture (which is about 0°C) to an industrial freezer.
It is assumed that the freezer is set to a lower temperature than the ice-water mixture. We'll need to determine the freezer temperature. The pressure exerted by the mercury is equal to the pressure exerted by the gas in the cell.
We may use the atmospheric pressure at sea level to calculate the gas pressure: Pa = 101,325 Pa Using the data provided in the problem, we can now determine the freezer temperature:
[tex]Δh = h1 − h2 Δh = 120 mm − 30 mm = 90 mm[/tex]
We'll use the difference in height of the mercury column, which is Δh, to determine the pressure change between the ice-water mixture and the freezer:
[tex]P2 = P1 − ρgh ΔP = P2 − P1 ΔP = −ρgh[/tex]
The pressure difference is expressed as a negative value because the pressure in the freezer is lower than the pressure in the ice-water mixture.
[tex]ΔP = −ρgh = −(13,600 kg/m3)(9.8 m/s2)(0.09 m) = −11,956.8 PaP2 = P1 + ΔP = 101,325 Pa − 11,956.8 Pa = 89,368.2 Pa[/tex]
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A bal is rolling with a constant angular speed round a circular groove in the sustace of a horizontale. If the word is 3.7 rad in the counteedoch reco, herause the circular groove is 0.57 m, and the angular position of the determine the component of the position time 10.40s and 55
To determine the component of the position of the ball, we need the values of the angular speed, time, and radius. Using the formulas θ = ω * t and s = r * θ, we can calculate the angular position and linear position of the ball, respectively. Once the values are known, the positions can be determined accordingly.
To determine the component of the position of the ball at a given time, we need to consider the angular displacement and radius of the circular groove.
The ball has a constant angular speed and completes an angular displacement of 3.7 rad in the counterclockwise direction, we can calculate the angular position (θ) using the formula:
θ = ω * t
where ω is the angular speed and t is the time. Plugging in the values, we can find the angular position.
Next, we can calculate the linear position (s) of the ball using the formula:
s = r * θ
where r is the radius of the circular groove. Substituting the given values, we can calculate the linear position of the ball.
It's important to note that the linear position will depend on the reference point chosen on the circular groove. If a specific reference point is mentioned or if further clarification is provided, the exact position of the ball can be determined.
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Mercury is poured into a U-tube as shown in Figure a. The left arm of the tube has cross-sectional area A1 of 10.9 cm2, and the right arm has a cross-sectional area A2 of 5.90 cm2. Three hundred grams of water are then poured into the right arm as shown in Figure b.
Figure (a) shows a U-shaped tube filled with mercury. Both arms of the U-shaped tube are vertical. The left arm with cross-sectional area A1 is wider than the right arm with cross-sectional area A2. The height of the mercury is the same in both arms. Figure (b) shows the same U-shaped tube, but now most of the right arm is filled with water. The height of the column of water in the right arm is much greater than the height of the column of mercury in the left arm. The height of the mercury in the left arm is greater than the height of the mercury in the arms in Figure (a), and the difference in height is labeled h.
(a) Determine the length of the water column in the right arm of the U-tube.
cm
(b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?
cm
The mercury rises by 0.53 cm in the left arm of the U-tube. The length of the water column in the right arm of the U-tube can be calculated as follows:
Water Column Height = Total Height of Right Arm - Mercury Column Height in Right Arm
Water Column Height = 20.0 cm - 0.424 cm = 19.576 cm
The mercury rises in the left arm of the U-tube because of the difference in pressure between the left arm and the right arm. The pressure difference arises because the height of the water column is much greater than the height of the mercury column. The difference in height h can be calculated using Bernoulli's equation, which states that the total energy of a fluid is constant along a streamline.
Given,
A1 = 10.9 cm²
A2 = 5.90 cm²
Density of Mercury, ρ = 13.6 g/cm³
Mass of water, m = 300 g
Now, let's determine the length of the water column in the right arm of the U-tube.
Based on the law of continuity, the volume flow rate of mercury is equal to the volume flow rate of water.A1V1 = A2V2 ... (1)Where V1 and V2 are the velocities of mercury and water in the left and right arms, respectively.
The mass flow rate of mercury is given as:
m1 = ρV1A1
The mass flow rate of water is given as:
m2 = m= 300g
We can express the volume flow rate of water in terms of its mass flow rate and density as follows:
ρ2V2A2 = m2ρ2V2 = m2/A2
Substituting the above expression and m1 = m2 in equation (1), we get:
V1 = (A2/A1) × (m2/ρA2)
So, the volume flow rate of mercury is given as:
V1 = (5.90 cm²/10.9 cm²) × (300 g)/(13.6 g/cm³ × 5.90 cm²) = 0.00891 cm/s
The volume flow rate of water is given as:
V2 = (A1/A2) × V1
= (10.9 cm²/5.90 cm²) × 0.00891 cm/s
= 0.0164 cm/s
Now, let's determine the height of the mercury column in the left arm of the U-tube.
Based on the law of conservation of energy, the pressure energy and kinetic energy of the fluid at any point along a streamline is constant. We can express this relationship as:
ρgh + (1/2)ρv² = constant
Where ρ is the density of the fluid, g is the acceleration due to gravity, h is the height of the fluid column, and v is the velocity of the fluid.
Substituting the values, we get:
ρgh1 + (1/2)ρv1² = ρgh2 + (1/2)ρv2²
Since h1 = h2 + h, v1 = 0, and v2 = V2, we can simplify the above equation as follows:
ρgh = (1/2)ρV2²
h = (1/2) × (V2/V1)² × h₁
h = (1/2) × (0.0164 cm/s / 0.00891 cm/s)² × 0.424 cm
h = 0.530 cm = 0.53 cm (rounded to two decimal places)
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As an object moves away from any kind of spherical mirror, its
image
1. goes out of focus
2. gets closer to the focus
3. becomes virtual
4. flips between inverted and erect
As an object moves away from any kind of spherical mirror, the characteristics of its image becomes virtual.
1. The image goes out of focus: This is not necessarily true. The focus of a spherical mirror remains fixed, regardless of the position of the object. If the object moves away from the mirror, the image may become blurred or less sharp, but it doesn't necessarily go out of focus.
2. The image gets closer to the focus: This statement is incorrect. The position of the image formed by a spherical mirror depends on the position of the object and the focal length of the mirror. As the object moves away from the mirror, the image generally moves farther away from the mirror as well.
3. The image becomes virtual: This is generally true. A virtual image is formed when the reflected rays do not actually converge at a physical point. In the case of a convex (outwardly curved) mirror, the image formed is always virtual, regardless of the position of the object. As the object moves away from the mirror, the virtual image remains behind the mirror and appears smaller.
4. The image flips between inverted and erect: This statement is incorrect. The nature of the image formed by a spherical mirror (inverted or erect) depends on whether the mirror is concave (inwardly curved) or convex (outwardly curved) and the position of the object relative to the focal point. However, as the object moves away from either type of spherical mirror, the image formed remains inverted.
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1A) Applying Gauss’s Law to the charged spherical shell shows us that on the surface of the shell and beyond we can compute the electric field with what the formula for the electric field of what type of charge? Write that formula below, using the following symbols: for the charge, for Coulomb’s constant, and for the distance from the center of the sphere. Show your work.
1B) According to the answers above, where will the electric field be the largest? Explain.
1C) Enter the dielectric strength of air for the electric field and the answer to (4) for the radius and calculate a value for the maximum charge that can build up before Carona discharge. Show your work.
It's one question with 3 parts.
When applying Gauss's Law to a charged spherical shell, the formula for the electric field can be used to compute the electric field for a type of charge known as "surface charge density" (σ).
The formula for the electric field due to a charged spherical shell is given by
E = σ / (ε₀),
where
E represents the electric field,
σ is the surface charge density, and
ε₀ is Coulomb's constant.
The electric field is largest on the surface of the charged shell due to the distribution of the charges. The dielectric strength of air can be used to calculate the maximum charge that can build up before Corona discharge occurs.
1B) The electric field is largest on the surface of the charged shell. This is because the surface charge density is concentrated on the outer surface of the shell, leading to a higher electric field intensity. Inside the shell, the electric field cancels out due to the charge distribution, resulting in a lower electric field magnitude.
1C) The dielectric strength of air refers to the maximum electric field that air can withstand before it breaks down and leads to a discharge. The dielectric strength of air is approximately 3 x 10^6 V/m.
To calculate the maximum charge that can build up before Corona discharge, we can use the formula for electric field E = σ / (ε₀) and the given value for the radius. By rearranging the formula, we can solve for the surface charge density σ:
σ = E * (ε₀)
Substituting the value for the electric field (3 x 10^6 V/m) and the value for ε₀, we can calculate the maximum charge that can build up before Corona discharge occurs.
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In a photoelectric effect experiment, a metal with a work function of 1.4 eV is used.
What is the maximum wavelength of light that can be used to free electrons from the metal?
Enter your answer in micrometres (10-6 m) to two decimal places but do not enter the units in your response.
The energy of a photon of light is given by
E = hc/λ,
where
h is Planck's constant,
c is the speed of light and
λ is the wavelength of the light.
The photoelectric effect can occur only if the energy of the photon is greater than or equal to the work function (φ) of the metal.
Thus, we can use the following equation to determine the maximum wavelength of light that can be used to free electrons from the metal:
hc/λ = φ + KEmax
Where KEmax is the maximum kinetic energy of the electrons emitted.
For the photoelectric effect,
KEmax = hf - φ
= hc/λ - φ
We can substitute this expression for KEmax into the first equation to get:
hc/λ = φ + hc/λ - φ
Solving for λ, we get:
λmax = hc/φ
where φ is the work function of the metal.
Substituting the given values:
Work function,
φ = 1.4 e
V = 1.4 × 1.6 × 10⁻¹⁹ J
= 2.24 × 10⁻¹⁸ J
Speed of light, c = 3 × 10⁸ m/s
Planck's constant,
h = 6.626 × 10⁻³⁴ J s
We get:
λmax = hc/φ
= (6.626 × 10⁻³⁴ J s)(3 × 10⁸ m/s)/(2.24 × 10⁻¹⁸ J)
= 8.84 × 10⁻⁷ m
= 0.884 µm (to two decimal places)
Therefore, the maximum wavelength of light that can be used to free electrons from the metal is 0.884 µm.
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The rms speed of the molecules of a gas at 143 °C is 217 m/s. Calculate the mass m of a single molecule in the gas.
The mass of a single molecule in the gas is approximately 4.54 x 10^(-26) kg.
The root mean square (rms) speed of gas molecules can be related to the temperature and the molar mass of the gas using the following equation:
v(rms) = √(3kT / m)
Where v(rms) is the rms speed, k is the Boltzmann constant (1.38 x 10^(-23) J/K), T is the temperature in Kelvin, and m is the molar mass of the gas in kilograms.
To solve for the mass of a single molecule, we need to convert the temperature from Celsius to Kelvin:
T(K) = 143°C + 273.15
Substituting the given values into the equation, we can solve for m:
217 m/s = √(3 * 1.38 x 10^(-23) J/K * (143 + 273.15) K / m)
Squaring both sides of the equation:
(217 m/s)^2 = 3 * 1.38 x 10^(-23) J/K * (143 + 273.15) K / m
Simplifying and rearranging the equation to solve for m:
m = 3 * 1.38 x 10^(-23) J/K * (143 + 273.15) K / (217 m/s)^2
Calculating the right-hand side of the equation:
m ≈ 4.54 x 10^(-26) kg
Therefore, the mass of a single molecule in the gas is approximately 4.54 x 10^(-26) kg.
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Which graphs could represent the Position versus Time for CONSTANT VELOCITY MOTION
The graph of position versus time would also be a straight line in constant velocity motion.
In constant velocity motion, the distance travelled by an object increases at a constant rate over time. The object has a constant speed in this situation. As a result, the graph of distance versus time is a straight line.
The reason for this is that velocity is constant, and the slope of the position versus time graph is equal to velocity. As a result, the slope is constant, and the graph is a straight line.
The following graphs could represent the position versus time for constant velocity motion:
A straight line with a positive slope
The graph of the line is determined by the position of the object and the time elapsed. The slope of the line indicates the velocity of the object. When the slope of the line is constant, the object is travelling at a constant velocity.
A horizontal line
If the object is stationary, the position versus time graph would show a horizontal line because the position of the object would remain constant over time. The velocity would be zero in this situation.
When an object is moving with constant velocity, the position versus time graph is linear with a positive slope. The reason for this is that the velocity is constant, meaning that the object covers equal distances in equal time intervals. The graph of the position versus time would thus show a straight line. Similarly, the slope of the line will indicate the velocity of the object. As a result, when the object has a constant velocity, the slope of the position versus time graph would be constant. The velocity can be calculated as the ratio of the displacement over time, which is equal to the slope of the position versus time graph.
Alternatively, if an object is stationary, then the position versus time graph would display a horizontal line at the point where the object is located. This is because the object would remain in the same position over time.
In constant velocity motion, the position versus time graph would show a straight line with a positive slope. The slope of the line indicates the velocity of the object. Additionally, if the object is stationary, then the position versus time graph would display a horizontal line.
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a 601nm light and a 605nm light are to be resolved using a
diffraction grating. How many lines must be illuminated to resolve
the light in the 2nd order?
When a 601nm light and a 605nm light are to be resolved using a diffraction grating, the number of lines that must be illuminated to resolve the light in the 2nd order is approximately 9589.
When diffraction grating is illuminated with light, it diffracts the light into several beams in various angles. In this process, the distance between lines on a diffraction grating should be less than the wavelength of the light to diffract light into a pattern of bright and dark fringes.
Diffracted order is said to be second when the light bends twice, from the line of the diffraction grating and from the screen.
Here, the difference between the two wavelengths is : 605 nm - 601 nm = 4 nm
To resolve the difference between these two wavelengths, there should be a difference of at least one fringe (or one period).
The formula to calculate the number of fringes or lines illuminated is given as : d sin(θ) = mλ
where,
d is the distance between two lines on the diffraction grating
sin(θ) is the angle at which the light bends
m is the order of diffraction, here m = 2
λ is the wavelength of the light
To resolve the light in the 2nd order, we will substitute the given values in the formula above :
4 × 10⁻⁹ m = d sin(θ) × 2 × 10⁻⁶ m
601 nm and 605 nm light are to be resolved using a diffraction grating.
The number of lines that must be illuminated to resolve the light in the 2nd order is approximately 9589.
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What is the binding energy per nucleon of 302Hg that has an atomic mass of 201.9706177? Note: Use the following atomic masses in your calculation: H = 1.007825 u and in = 1.008665 u. (a) 8.647 Mev. (b
The binding energy per nucleon of 302Hg is approximately 1.17220976 × 10¹⁶MeV/ nucleon.
To calculate the binding energy per nucleon of a nucleus, we need to determine the total binding energy of the nucleus and then divide it by the total number of nucleons.
The total binding energy of a nucleus can be calculated using the formula:
Binding Energy = (Z × mp + N × mn - M) × c²
Where
Z is the number of protons,
mp is the mass of a proton,
N is the number of neutrons,
mn is the mass of a neutron,
M is the atomic mass of the nucleus, and
c is the speed of light.
For the nucleus of 302Hg, we have:
No. of protons, Z = 30
No. of neutrons, N = 200
Total Number of Nucleons = Z + N
= 30 +200
= 230
The mass of a proton (mp) ≈ 1.007825 u,
The mass of a neutron (mn) ≈ 1.008665 u.
The atomic mass of 302Hg ≈201.9706177 u.
The speed of light (c) ≈ 2.998 × 10^8 m/s.
Substituting these values into the formula, we can calculate the binding energy:
Binding Energy = (30 × 1.007825 + 200 ×1.008665 - 201.9706177) × (2.998 × 10⁸)²
Binding energy = 2.69614345 × 10¹⁸ MeV
To find the binding energy per nucleon, we divide the binding energy by the total number of nucleons:
Binding Energy per Nucleon = Binding Energy / Total Number of Nucleons
= 2.69614345 × 10¹⁸ MeV / 230
≈ 1.17220976 × 10¹⁶MeV/ nucleon
Therefore, the binding energy per nucleon of 302Hg is approximately 1.17220976 × 10¹⁶MeV/ nucleon.
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(a) The current in a wire is 2.0 mA. In 2.0 ms. how much charge flows through a point in a wire, and how many electrons pass the point?
2.5 × 10¹³ electrons pass through the point in the wire in 2.0 ms.
Current is the rate of flow of charge, typically measured in amperes (A). One ampere is equivalent to one coulomb of charge flowing per second. For a current of 2.0 mA, which is 2.0 × 10⁻³ A, the charge that flows through a point in the wire in 2.0 ms can be calculated using the formula Q = I × t, where Q represents the charge in coulombs, I is the current in amperes, and t is the time in seconds.
By substituting the given values into the formula, we can calculate the resulting value.
Q = (2.0 × 10⁻³ A) × (2.0 × 10⁻³ s)
Q = 4.0 × 10⁻⁶ C
Therefore, 4.0 × 10⁻⁶ C of charge flows through the point in the wire in 2.0 ms. To determine the number of electrons that pass the point, we can use the formula n = Q/e, where n represents the number of electrons, Q is the charge in coulombs, and e is the charge on an electron.
Substituting the values into the formula:
n = (4.0 × 10⁻⁶ C) / (1.6 × 10⁻¹⁹ C)
n = 2.5 × 10¹³
Hence, 2.5 × 10¹³ electrons pass through the point in the wire in 2.0 ms.
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1.Find the force on a particle of mass m=1.70×10-27kg and charge q=1.60×10-19C if it enters a field B=5 mT with an initial speed of v=83.5 km/s. Assume the velocity is in the x direction and the magnetic field enters perpendicular to the screen. Also make a schematic drawing of these vectors. Don't forget to place your reference system.
2.Find the force on a straight conductor of length 0.3 m, which carries a current of 5 A in the negative z-direction. In that space there is a magnetic field given by the vector B=3.5×10-3Ti-3.5×10-3Tj . Make a schematic drawing of the situation. (We do not require precision in your drawing for the direction of the magnetic field, only approximate).
3.A conductor of length 2.5 m is located at z=0, x=4m with a current of 12 A in the -y direction. Find the magnetic field that exists in that region if the force on the conductor is F=-1.20×10-2N(-12i-12j).
4.A long thin wire carries a current I. A metal bar of length L is moving with a constant speed v as shown in the figure. Point a is a distance d from the wire a) Calculate the electromotive force induced in the bar. b) If the bar is replaced by a rectangular circuit of resistance R, what is the magnitude of the induced current in the circuit?
1. The force on the particle is 1.36 x 10^-14 N, schematic drawing shows velocity in x-direction, magnetic field entering perpendicular to the screen, and force perpendicular to both.
2. The force on the straight conductor is 5.25 x 10^-3 N, schematic drawing shows conductor in negative z-direction and magnetic field vectors approximately orthogonal to the conductor.
3. The magnetic field is approximately -0.01 T in the x-direction and -0.01 T in the y-direction.
4. a) The electromotive force induced in the bar is BLv. b) The magnitude of the induced current in the rectangular circuit is V/R.
1. The force on the particle can be calculated using the equation F = qvB, where q is the charge, v is the velocity, and B is the magnetic field. Plugging in the given values, the force is 1.36 x 10^-14 N. A schematic drawing would show the velocity vector in the x-direction, the magnetic field vector entering perpendicular to the screen, and the force vector perpendicular to both.
2. The force on the straight conductor can be calculated using the equation F = IL x B, where I is the current, L is the length of the conductor, and B is the magnetic field. Plugging in the given values, the force is 5.25 x 10^-3 N. A schematic drawing would show the conductor in the negative z-direction, with the magnetic field vectors shown approximately orthogonal to the conductor.
3. The magnetic field can be determined using the equation F = IL x B. Since the force is given as F = -1.20 x 10^-2 N (-12i - 12j), we can equate the force components to the corresponding components of the equation and solve for B. The resulting magnetic field is approximately -0.01 T in the x-direction and -0.01 T in the y-direction.
4. To calculate the electromotive force induced in the bar, we can use the equation emf = BLv, where B is the magnetic field, L is the length of the bar, and v is the speed of the bar. The magnitude of the induced current in the rectangular circuit can be calculated using Ohm's Law, I = V/R, where V is the electromotive force and R is the resistance of the circuit.
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Light of wavelength 553.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 91.5 cm from the slit. The distance on the screen between the fourth order minimum and the central maximum is 1.19 cm. What is the width of the slit in micrometers (μm)?
The width of the slit is approximately 21.1 μm, calculated using the diffraction pattern and given parameters.
The width of the slit can be calculated using the formula for the diffraction pattern:
d * sin(θ) = m * λ
where d is the width of the slit, θ is the angle of diffraction, m is the order of the minimum, and λ is the wavelength of light.
In this case, we have the following information:
λ = 553.0 nm = 553.0 × 10^(-9) m
m = 4 (for the fourth-order minimum)
d = ? (to be determined)
To find the angle of diffraction θ, we can use the small angle approximation:
θ ≈ tan(θ) = (x/L)
where x is the distance between the central maximum and the fourth-order minimum on the screen (1.19 cm = 1.19 × 10^(-2) m), and L is the distance from the slit to the screen (91.5 cm = 91.5 × 10^(-2) m).
θ = (1.19 × 10^(-2) m) / (91.5 × 10^(-2) m) = 0.013
Now, we can rearrange the formula to solve for the slit width d:
d = (m * λ) / sin(θ)
= (4 * 553.0 × 10^(-9) m) / sin(0.013)
Calculating the value of sin(0.013), we find:
sin(0.013) ≈ 0.013
Substituting the values into the formula, we get:
d = (4 * 553.0 × 10^(-9) m) / 0.013 ≈ 0.0211 m = 21.1 μm
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You push a 25-kg block 10 m along a horizontal floor at constant speed. Your force F is directed 30
degrees below the horizontal. The coefficient of kinetic friction between the block and floor is 0.1.
a. How much work did you do on the block? (Hint: first you need to calculate your applied force
F.)
b. How much thermal (i.e. wasted) energy was dissipated in the process?
c. Are there any non-conservative forces at work in this problem?
The force of friction is a non-conservative force, since it depends on the path taken by the block.
The given values are the mass of the block m = 25-kg, the distance it was pushed along the floor d = 10 m, the coefficient of kinetic friction between the block and the floor μk = 0.1 and the angle that the force was directed below the horizontal θ = 30 degrees.
We are to find (a) the amount of work done on the block, (b) the amount of thermal energy that was dissipated in the process, and (c) whether there are any non-conservative forces at work in this problem. (a) The work done by the force F on the block is given by W = Fd cos θ,
where F is the applied force, d is the distance moved, and θ is the angle between the force and the direction of motion.
The force F can be calculated as follows: F = ma + mg sin θ - μk mg cos θ
where a is the acceleration of the block and g is the acceleration due to gravity. Since the block is moving at constant speed, its acceleration is zero.
Thus, we have F = mg sin θ - μk mg cos θ
= (25 kg)(9.8 m/s^2)(sin 30°) - (0.1)(25 kg)(9.8 m/s^2)(cos 30°)
= 122.5 N
The work done on the block is then W = (122.5 N)(10 m)(cos 30°) = 1060 J (b)
The amount of thermal energy that was dissipated in the process is equal to the work done by the force of friction, which is given by Wf = μk mgd
= (0.1)(25 kg)(9.8 m/s^2)(10 m) = 245 J (c)
The force of friction is a non-conservative force, since it depends on the path taken by the block.
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"Two converging lenses with the same focal length of 10 cm are 40
cm apart. If an object is located 15 cm from one of the lenses,
find the final image distance of the object.
a. 0 cm
b. 5 cm
c. 10 cm
d 15 cm
The final image distance of the object, if the object is located 15 cm from one of the lenses is 6 cm. So none of the options are correct.
To determine the final image distance of the object in the given setup of two converging lenses, we can use the lens formula:
1/f = 1/di - 1/do
Where: f is the focal length of the lens, di is the image distance, do is the object distance.
Given that both lenses have the same focal length of 10 cm, we can consider them as a single lens with an effective focal length of 10 cm. The lenses are 40 cm apart, and the object distance (do) is 15 cm.
Using the lens formula, we can rearrange it to solve for di:
1/di = 1/f + 1/do
1/di = 1/10 cm + 1/15 cm
= (15 + 10) / (10 * 15) cm⁻¹
= 25 / 150 cm⁻¹
= 1 / 6 cm⁻¹
di = 1 / (1 / 6 cm⁻¹) = 6 cm
Therefore, the final image distance of the object is 6 cm. So, none of the options are correct.
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An object 1.50 cm high is held 3.20 cm from a person's cornea, and its reflected image is measured to be 0.175 cm high. (a) What is the magnification? Х (b) Where is the image (in cm)? cm (from the corneal "mirror") (C) Find the radius of curvature (in cm) of the convex mirror formed by the cornea.
The magnification of the object is -0.1167. The image is 1.28 cm from the corneal "mirror". The radius of curvature of the convex mirror formed by the cornea is -0.1067 cm.
It is given that, Height of object, h = 1.50 cm, Distance of object from cornea, u = -3.20 cm, Height of image, h' = -0.175 cm
(a) Magnification:
Magnification is defined as the ratio of height of the image to the height of the object.
So, Magnification, m = h'/h m = -0.175/1.50 m = -0.1167
(b)
Using the mirror formula, we can find the position of the image.
The mirror formula is given as :1/v + 1/u = 1/f Where,
v is the distance of the image from the mirror.
f is the focal length of the mirror.
Since we are considering a mirror of the cornea, which is a convex mirror, the focal length will be negative.
Therefore, we can write the formula as:
1/v - 1/|u| = -1/f
1/v = -1/|u| - 1/f
v = -|u| / (|u|/f - 1)
On substituting the given values, we have:
v = 1.28 cm
So, the image is 1.28 cm from the corneal "mirror".
(c)
The radius of curvature, R of a convex mirror is related to its focal length, f as follows:R = 2f
By lens formula,
1/v + 1/u = 1/f
1/f = 1/v + 1/u
We already have the value of v and u.
So,1/f = 1/1.28 - 1/-3.20
1/f = -0.0533cmS
o, the focal length of the convex mirror is -0.0533cm.
Now, using the relation,R = 2f
R = 2 × (-0.0533)
R = -0.1067 cm
Therefore, the radius of curvature of the convex mirror formed by the cornea is -0.1067 cm.
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A spider’s web can undergo SHM when a fly lands on it and displaces the web. For simplicity, assume that a web is described by Hooke’s law (even though really it deforms permanently when displaced). If the web is initially horizontal and a fly landing on the web is in equilibrium when it displaces the web by 0.0430 mm, what is the frequency of oscillation when the fly lands? Hz
the frequency of oscillation when the fly lands on the web cannot be determined without additional details about the spring constant and mass of the web.
To determine the frequency of oscillation when the fly lands on the spider's web, we can use Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement from equilibrium.
The equation for the frequency of simple harmonic motion (SHM) is given by:
Frequency (f) = (1 / 2π) * √(k / m)
In this case, the displacement of the web caused by fly landing is given as 0.0430 mm (or 0.0430 * 10^-3 m). The displacement represents the amplitude of the oscillation.
The equilibrium position of the web is when it is initially horizontal. This means that the displacement is also the amplitude of oscillation.
To find the frequency, we need to know the spring constant (k) and the mass (m) of the web. Without that information, it is not possible to calculate the frequency accurately.
Therefore, the frequency of oscillation when the fly lands on the web cannot be determined without additional details about the spring constant and mass of the web.
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4 pts What is the required radius of a cyclotron designed to accelerate protons to energies of 34.0 MeV using a magnetic field of 5.5 T? Note: 1eV = 1.60 × 10-¹9 J. Neglect relativity, even though at this energy it would make a small difference. m ( + 0.002 m)
The required radius of the cyclotron to accelerate protons to energies of 34.0 MeV using a magnetic field of 5.5 T can be determined using the equation for the cyclotron's radius. Required radius is approximately 2.89 × 10⁻² meters.
The equation is given by:
r = (mv) / (qB)
Where:
r is the radius of the cyclotron,
m is the mass of the proton,
v is the velocity of the proton,
q is the charge of the proton, and
B is the magnetic field strength.
To find the radius, we need to calculate the velocity of the proton first. The energy of the proton can be converted to joules using the conversion factor, and then we can use the equation:
E = (1/2)mv²
Rearranging the equation to solve for v:
v = √(2E/m)
Plugging in the values:
v = √(2 × 34.0 MeV × 1.60 × 10⁻¹⁹ J / (1.67 × 10⁻²⁷ kg)
Calculating the velocity, we can substitute it into the formula for the radius to find the required radius of the cyclotron.
To calculate the required radius of the cyclotron, we'll follow the given steps:
1. Convert the energy of the proton to joules:
E = 34.0 MeV × (1.60 × 10⁻¹⁹ J/1 MeV)
E = 5.44 × 10⁻¹² J
2. Calculate the velocity of the proton:
v = √(2E/m)
v = √(2 × 5.44 × 10⁻¹² J / (1.67 × 10⁻²⁷ kg))
v ≈ 3.74 × 10⁷m/s
3. Substitute the values into the formula for the radius of the cyclotron:
r = (mv) / (qB)
r = ((1.67 × 10⁻²⁷ kg) × (3.74 × 10⁷ m/s)) / ((1.60 × 10⁻¹⁹C) × (5.5 T))
r ≈ 2.89 × 10⁻² m
Therefore, the required radius of the cyclotron to accelerate protons to energies of 34.0 MeV using a magnetic field of 5.5 T is approximately 2.89 × 10⁻² meters.
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Set 1: Gravitation and Planetary Motion NOTE. E Nis "type-writer notation for x10" ( 2 EB - Exam 2x10") you may use either for this class AND the AP GMm mu F GMm 9 G= 6.67 11 Nm /kg F = mg 9 GMm = mg GM 12 т GM V = 1 GM 9 GM V = - 21 T F 9 = mac T 1. A whale shark has a mass of 2.0 E4 kg and the blue whale has a mass of 1.5 E5 kg a. If the two whales are 1.5 m apart, what is the gravitational force between them? b. How does the magnitude of the gravitational force between the two animals compare to the gravitational force between each and the Earth? c. Explain why objects on Earth do not seem to be attracted 2. An asteroid with a mass of 1.5 E21 kg orbits at a distance 4E8 m from a planet with a mass of 6 E24 kg a. Determine the gravitational force on the asteroid. b. Determine the gravitational force on the planet. C Determine the orbital speed of the asteroid. d Determine the time it takes for the asteroid to complete one trip around the planet 3. A 2 2 14 kg comet moves with a velocity of 25 E4 m/s through Space. The mass of the star it is orbiting is 3 E30 kg a Determine the orbital radius of the comet b. Determine the angular momentum of the comet. (assume the comet is very small compared to the star) c An astronomer determines that the orbit is not circular as the comet is observed to reach a maximum distance from the star that is double the distance found in part (a). Using conservation of angular momentum determine the speed of the comet at its farthest position 4. A satellite that rotates around the Earth once every day keeping above the same spot is called a geosynchronous orbit. If the orbit is 3.5 E7 m above the surface of the and the radius and mass of the Earth is about 6.4 E6 m and 6.0 E24 kg respectively. According to the definition of geosynchronous, what is the period of the satellite in hours? seconds? a. Determine the speed of the satellite while in orbit b. Explain satellites could be used to remotely determine the mass of unknown planets 5. Two stars are orbiting each other in a binary star system. The mass of each of the stars is 2 E20 kg and the distance from the stars to the center of their orbit is 1 E7 m. a. Determine the gravitational force between the stars.. b. Determine the orbital speed of each star
In this set of questions, we are exploring the concepts of gravitation and planetary motion. We use the formulas related to gravitational force, orbital speed, and orbital radius to solve various problems.
Firstly, we calculate the gravitational force between two whales and compare it to the gravitational force between each whale and the Earth. Then, we determine the gravitational force on an asteroid and a planet, as well as the orbital speed and time taken for an asteroid to complete one orbit.
Next, we find the orbital radius and angular momentum of a comet orbiting a star, and also calculate the speed of the comet at its farthest position. Finally, we discuss the period of a geosynchronous satellite orbiting the Earth and how satellites can be used to determine the mass of unknown planets.
a. To calculate the gravitational force between the whale shark and the blue whale, we use the formula F = GMm/r^2, where G is the gravitational constant, M and m are the masses of the two objects, and r is the distance between them. Plugging in the values, we find the gravitational force between them.
b. To compare the gravitational force between the two animals and the Earth, we calculate the gravitational force between each animal and the Earth using the same formula.
We observe that the force between the animals is much smaller compared to the force between each animal and the Earth. This is because the mass of the Earth is significantly larger than the mass of the animals, resulting in a stronger gravitational force.
c. Objects on Earth do not seem to be attracted to each other strongly because the gravitational force between them is much weaker compared to the gravitational force between each object and the Earth.
The mass of the Earth is substantially larger than the mass of individual objects on its surface, causing the gravitational force exerted by the Earth to dominate and make the gravitational force between objects on Earth negligible in comparison.
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In conservation of energy experiment, the relation between the hanging mass (m) and the increase in the length (x) is given by: mg =kx Where (k) is the spring constant and (g) is the acceleration due to gravity (g = 9.81 m/s2). The graph below shows (m vs. x) for three different springs. Which spring has the lowest spring constant (ks)? Spring 2 Spring 3 Spring 1 All the springs have the same Tidliging mass (m) and the increase in the length (x) is given mg =kx Spring (slope - 2km) Spring 2 (slope 1.7km) Spring 3 slope 2.5km) g) is the acceleration due to gravity (g = 9.81 m/s2). The graph below shows (m vs.
Spring 2 has the lowest spring constant among the three springs in the experiment.
In the given conservation of energy experiment, the relation between the hanging mass (m) and the increase in length (x) is given by mg = kx, where k is the spring constant and g is the acceleration due to gravity (9.81 m/s²).
The graph provided shows the relationship between m and x for three different springs. To determine which spring has the lowest spring constant, we need to compare the slopes of the graph lines. The spring with the lowest slope, which represents the smallest value of k, has the lowest spring constant.
The slope of the graph represents the spring constant (k) in the relation mg = kx. A steeper slope indicates a higher spring constant, while a flatter slope indicates a lower spring constant. Looking at the graph lines for the three springs, we can compare their slopes to determine which one has the lowest spring constant.
If the slope of Spring 1 is 2k, the slope of Spring 2 is 1.7k, and the slope of Spring 3 is 2.5k, we can conclude that Spring 2 has the lowest spring constant (ks). This is because its slope is the smallest among the three, indicating a smaller value for k.
Therefore, Spring 2 has the lowest spring constant among the three springs in the experiment.
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A 400-kg box is lifted vertically upward with constant velocity by means of two cables pulling at 50.0° up from the horizontal direction. What is the tension in each cable?
The tension in each cable used to lift the 400-kg box vertically upward, we can use the equilibrium condition and resolve the forces in the vertical and horizontal directions.
Let's denote the tension in each cable as T₁ and T₂.In the vertical direction, the net force is zero since the box is lifted with constant velocity. The vertical forces can be represented as:
T₁sinθ - T₂sinθ - mg = 0, where θ is the angle of the cables with the horizontal and mg is the weight of the box. In the horizontal direction, the net force is also zero:
T₁cosθ + T₂cosθ = 0
Given that the weight of the box is mg = (400 kg)(9.8 m/s²) = 3920 N and θ = 50.0°, we can solve the system of equations to find the tension in each cable:
T₁sin50.0° - T₂sin50.0° - 3920 N = 0
T₁cos50.0° + T₂cos50.0° = 0
From the second equation, we can rewrite it as:
T₂ = -T₁cot50.0°
Substituting this value into the first equation, we have:
T₁sin50.0° - (-T₁cot50.0°)sin50.0° - 3920 N = 0
Simplifying and solving for T₁:
T₁ = 3920 N / (sin50.0° - cot50.0°sin50.0°)
Using trigonometric identities and solving the expression, we find:
T₁ ≈ 2826.46 N
Finally, since T₂ = -T₁cot50.0°, we can calculate T₂:
T₂ ≈ -2826.46 N * cot50.0°
Therefore, the tension in each cable is approximately T₁ ≈ 2826.46 N and T₂ ≈ -2202.11 N.
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Q|C As in Example 28.2, consider a power supply with fixed emf E and internal resistance r causing current in a load resistance R. In this problem, R is fixed and r is a variable. The efficiency is defined as the energy delivered to the load divided by the energy delivered by the emf.(a) When the internal resistance is adjusted for maximum power transfer, what is the efficiency?
When the internal resistance is adjusted for maximum power transfer, the efficiency of the power supply is 50%.
The efficiency of a power supply is defined as the energy delivered to the load divided by the energy delivered by the emf. In this problem, we are given a power supply with fixed emf E and internal resistance r, causing current in a load resistance R. We are asked to find the efficiency when the internal resistance is adjusted for maximum power transfer.
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Mass on Incline Points:2 A spring, of negligible mass and which obeys Hooke's Law, supports a mass M on an incline which has negligible friction. The figure below shows the system with mass M in its equilibrium position. The spring is attached to a fixed support at P. The spring in its relaxed state is also illustrated. 80 70 60 WWWWWWWWWUnstreched spring Mamma SA y (in cm) 40 30 20 10 0 10 20 30 40 50 60 70 80 90 100110 6 X (in cm) Mass M has a value of 195 g. Calculate k, the spring constant. Submit Answer Tries 0/10 The mass oscillates when given a small displacement from its equilibrium position along the incline. Calculate the period of oscillation. Sukamil Answer Tries 0/10
The period of oscillation of the mass is 0.86 seconds (approx).
Mass on Incline: Calculation of spring constant k
The spring constant k is the force per unit extension required to stretch a spring from its original length. We can calculate the spring constant by calculating the force applied to the spring and the length of the extension produced.
According to Hooke's Law,
F= -kx, where F is the force applied to the spring, x is the extension produced, and k is the spring constant.
Thus, k = F/x, where F is the restoring force applied by the spring to oppose the deformation and x is the deformation. From the given problem, we have the mass of the object M as 195 g or 0.195 kg.
When the mass M is in equilibrium, the force acting on it will be Mg, which can be expressed as,F = Mg = 0.195 kg × 9.8 m/s2 = 1.911 N.
Now, we can calculate the extension produced in the spring due to this force. At equilibrium, the spring is neither stretched nor compressed. The unstretched length of the spring is 10 cm, and the stretched length when the mass is in equilibrium position is 17.5 cm, as given in the figure above.
Hence, the extension produced in the spring is,
x = 17.5 − 10
= 7.5 cm
= 0.075 m.
Hence, the spring constant k can be calculated ask =
F/x = 1.911/0.075
= 25.48 N/m.
Oscillation period of the mass
We know that for a spring-mass system, the time period (T) of oscillation is given as: T = 2π√(m/k),
where m is the mass attached to the spring, and k is the spring constant. From the given problem,
m = 195 g or 0.195 kg, and k = 25.48 N/m.
Thus, the oscillation period can be calculated as:
T = 2π√(0.195/25.48)
= 0.86 s (approx).
Therefore, the period of oscillation of the mass is 0.86 seconds (approx).
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Why is it use or found in our every lives or certain in the industries?and identify and explain at least two uses
Integral calculus is a branch of mathematics that deals with the properties and applications of integrals. It is used extensively in many fields of science, engineering, economics, and finance, and has become an essential tool for solving complex problems and making accurate predictions.
One reason why integral calculus is so prevalent in our lives is its ability to solve optimization problems. Optimization is the process of finding the best solution among a set of alternatives, and it is important in many areas of life, such as engineering, economics, and management. Integral calculus provides a powerful framework for optimizing functions, both numerically and analytically, by finding the minimum or maximum value of a function subject to certain constraints.
Another use of integral calculus is in the calculation of areas, volumes, and other physical quantities. Many real-world problems involve computing the area under a curve, the volume of a shape, or the length of a curve, and these computations can be done using integral calculus. For example, in engineering, integral calculus is used to calculate the strength of materials, the flow rate of fluids, and the heat transfer in thermal systems.
In finance, integral calculus is used to model and analyze financial markets, including stock prices, bond prices, and interest rates. The Black-Scholes formula, which is used to price options, is based on integral calculus and has become a standard tool in financial modeling.
Overall, integral calculus has numerous applications in various fields, and its importance cannot be overstated. Whether we are designing new technologies, predicting natural phenomena, or making investment decisions, integral calculus plays a crucial role in helping us understand and solve complex problems.
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350 g of ice at -10.00oC are added 2.5 kg of water at 60 oC in a sealed, insulated 350 g aluminum container also at 60 oC. At the same time 50.0 g of steam at 140oC is added to the water and ice. Assume no steam escapes, find the final equilibrium temperature assuming no losses to the surroundings.
The final equilibrium temperature assuming no losses is 16.18 oC.
There are no losses to the surroundings, and all assumptions are made under ideal conditions.
When the ice and water are mixed, some of the ice begins to melt. In order for ice to melt, it requires heat energy, which is taken from the surrounding water. This causes the temperature of the water to decrease. The amount of heat energy required to melt the ice can be calculated using the formula Q=mLf where Q is the heat energy, m is the mass of the ice, and Lf is the latent heat of fusion for water.
The heat energy required to melt the ice is
(0.35 kg)(334 J/g) = 117.1 kJ
This causes the temperature of the water to decrease to 45 oC.
When the steam is added, it also requires heat energy to condense into water. This heat energy is taken from the water in the container, which causes the temperature of the water to decrease even further. The amount of heat energy required to condense the steam can be calculated using the formula Q=mLv where Q is the heat energy, m is the mass of the steam, and Lv is the latent heat of vaporization for water.
The heat energy required to condense the steam is
(0.05 kg)(2257 J/g) = 112.85 kJ
This causes the temperature of the water to decrease to 16.18 oC.
Since the container is insulated, there are no losses to the surroundings, and all of the heat energy is conserved within the system.
Therefore, the final equilibrium temperature of the system is 16.18 oC.
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Review. The cosmic background radiation is blackbody radiation from a source at a temperature of 2.73K. (b) In what part of the electromagnetic spectrum is the peak of the distribution?
The peak of the distribution of the cosmic background radiation is in the microwave part of the electromagnetic spectrum. The frequency falls within the microwave region of the electromagnetic spectrum, indicating that the cosmic background radiation has its peak emission in that specific range.
The peak wavelength or frequency of blackbody radiation can be determined using Wien's displacement law, which states that the wavelength of the peak emission is inversely proportional to the temperature of the blackbody.
The formula for Wien's displacement law is:
λ_peak = b/T
where λ_peak is the peak wavelength, T is the temperature of the blackbody, and b is Wien's displacement constant, which is approximately equal to 2.898 × 10^(-3) m·K.
Substituting the given temperature T = 2.73 K into the formula, we can calculate the peak wavelength:
λ_peak = (2.898 × 10^(-3) m·K) / 2.73 K
≈ 1.06 × 10^(-3) m
To determine the corresponding region of the electromagnetic spectrum, we can use the relationship between wavelength and frequency:
c = λ · ν
where c is the speed of light (approximately 3.00 × 10^8 m/s), λ is the wavelength, and ν is the frequency.
Rearranging the equation, we get:
ν = c / λ
Substituting the calculated peak wavelength into the equation and solving for the frequency, we find:
ν = (3.00 × 10^8 m/s) / (1.06 × 10^(-3) m)
≈ 2.83 × 10^11 Hz
The frequency obtained corresponds to the microwave region of the electromagnetic spectrum.
The peak of the distribution of the cosmic background radiation, which is blackbody radiation from a source at a temperature of 2.73 K, is in the microwave part of the electromagnetic spectrum. This result is obtained by applying Wien's displacement law, which relates the peak wavelength of blackbody radiation to the temperature of the source.
The peak wavelength is determined to be approximately 1.06 × 10^(-3) m, which corresponds to a frequency of approximately 2.83 × 10^11 Hz. The frequency falls within the microwave region of the electromagnetic spectrum, indicating that the cosmic background radiation has its peak emission in that specific range.
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A 1.41 kg bowling trophy is held at arm's length, a distance of 0.645 m from the shoulder joint.(1) What torque does the trophy exert about the shoulder if the arm is horizontal? τ = (?) N⋅m
(2) What torque does the trophy exert about the shoulder if the arm is at an angle of 20.5 ∘ below the horizontal? τ = (?) N⋅m
1. The torque exerted by the trophy about the shoulder joint when the arm is horizontal The torque is the product of the magnitude of the force applied and the perpendicular distance from the line of action of the force to the axis of rotation. We have to first figure out the force acting on the trophy.
The force acting on the trophy is equal to the weight of the trophy.
= mg
= (1.41)(9.81)
= 13.8321 N
= r × FW
= force acting on the
= distance between the shoulder joint and the trophyτ
= rFW
= (0.645)(13.8321
= 8.913 N⋅m2.
= r × F × sinθWhere,θ
= 20.5ºr
= 0.645 mF
= 13.8321 Nτ
= (0.645)(13.8321)sin(20.5º)τ
= 3.60 N⋅mThe torque exerted by the trophy about the shoulder if the arm is horizontal is 8.913 N⋅m.
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(a) What is the width of a single slit that produces its first minimum at 60.0⁰ for 591-nm light? nm (b) Using the slit from part (a), find the wavelength of light that has its first minimum at 64.3º. nm
To determine the width of a single slit that produces its first minimum at a given angle for a specific wavelength of light, we can use the formula for single-slit diffraction. By rearranging the formula and substituting the known values, we can calculate the width of the slit. In part (b), using the same slit from part (a), we can find the wavelength of light that produces its first minimum at a different angle by rearranging the formula and solving for the wavelength.
a. For part (a), we can use the formula for single-slit diffraction:
sin(θ) = m * λ / w
Where:
θ is the angle at which the first minimum occurs
m is the order of the minimum (in this case, m = 1)
λ is the wavelength of light
w is the width of the slit
By rearranging the formula and substituting the known values (θ = 60.0⁰, λ = 591 nm, m = 1), we can solve for the width of the slit (w).
b. For part (b), we can use the same formula and rearrange it to solve for the wavelength of light:
λ = w * sin(θ) / m
Given the width of the slit (w) determined in part (a), the angle at which the first minimum occurs (θ = 64.3º), and the order of the minimum (m = 1), we can substitute these values into the formula to find the wavelength of light (λ).
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8 (20 points) You have been out deer hunting with a bow. Just after dawn you see a large 8 point buck. It is just at the outer range of your bow. You take careful aim, and slowly release your arrow. It's a clean hit. The arrow is 0.80 meters long, weighs 0.034 kg, and has penetrated 0.18 meter. Your arrows speed was 1.32 m/s. a Was it an elastic or inelastic collision? b What was its momentum? c How long was the time of penetration? d What was the impulse? e What was the force.
a. Elastic collision.
b. Momentum is mass x velocity.
Therefore, momentum = 0.034 x 1.32 = 0.04488 kgm/s
c. The time of penetration is given by t = l/v
where l is the length of the arrow and v is the velocity of the arrow.
Therefore, t = 0.8 / 1.32 = 0.6061 s.
d. Impulse is the change in momentum. As there was no initial momentum, impulse = 0.04488 kgm/s.
e. Force is the product of impulse and time.
Therefore, force = 0.04488 / 0.6061 = 0.0741 N.
a. Elastic collision.
b. Momentum = 0.04488 kgm/s.
c. Time of penetration = 0.6061 s.
d. Impulse = 0.04488 kgm/s
.e. Force = 0.0741 N.
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2. [20 points] In each of following (a) through (e), use all of the listed words in any order in one sentence that makes scientific sense. You may use other words, including conjunctions; however, simple lists of definitions will not receive credit. Underline each of those words where they appear. You will be assessed on the sentence's grammatical correctness and scientific accuracy. (a) Popper, theory, falsification, science, prediction, [name of a celebrity] (b) vibration, pitch, music, stapes, power, [name of a singer] (c) harmonic, pendulum, frequency, spring, energy, [name of a neighbor] (d) Kelvin, joule, calorie, absorption, heat, [name of a food] (e) Pouiselle, millimeters, pressure, bar, over, [any metal]
When measuring the absorption of heat, one must consider the conversion between Kelvin, joules, and calories, as it relates to the specific properties of the food.
(a) Popper's theory of falsification is a cornerstone of science, emphasizing the importance of making testable predictions to validate or refute hypotheses, and even [name of a celebrity] could not escape its scrutiny.
(b) The vibration of the stapes bone in the ear contributes to perceiving different pitches in music, and [name of a singer]'s powerful voice can create a mesmerizing auditory experience.
(c) The harmonic motion of a pendulum, governed by its frequency and influenced by the spring's energy, can be observed by [name of a neighbor] in their backyard.
(d) When measuring heat absorption, the conversion between Kelvin, joules, and calories is crucial, and [name of a food] can release a specific amount of energy upon combustion.
(e) The Pouiselle effect describes the flow of fluids through narrow tubes, where millimeters of diameter can greatly affect the pressure drop across a bar made of any metal.
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Assume a deuteron and a triton are at rest when they fuse according to the reaction²₁H + ³₁H → ⁴₂He + ¹₀n Determine the kinetic energy acquired by the neutron.
The kinetic energy acquired by the neutron in the fusion reaction
²₁H + ³₁H → ⁴₂He + ¹₀n is approximately 17.6 MeV (million electron volts).
In a fusion reaction, two nuclei combine to form a new nucleus. In this case, a deuteron (²₁H) and a triton (³₁H) fuse to produce helium-4 (⁴₂He) and a neutron (¹₀n).
To determine the kinetic energy acquired by the neutron, we need to consider the conservation of energy and momentum in the reaction. Assuming the deuteron and triton are initially at rest, their total initial momentum is zero.
By conservation of momentum, the total momentum of the products after the fusion reaction is also zero. Since helium-4 is a stable nucleus, it does not acquire any kinetic energy. Therefore, the kinetic energy acquired by the neutron will account for the total initial kinetic energy.
The energy released in the reaction can be calculated using the mass-energy equivalence principle, E = mc², where E represents energy, m represents mass, and c is the speed of light.
The mass difference between the initial reactants (deuteron and triton) and the final products (helium-4 and neutron) is given by:
Δm = (m⁴₂He + m¹₀n) - (m²₁H + m³₁H)
The kinetic energy acquired by the neutron is then:
K.E. = Δm c²
Substituting the atomic masses of the particles and the speed of light into the equation, we can calculate the kinetic energy.
Using the atomic masses: m²₁H = 1.008665 u, m³₁H = 3.016049 u, m⁴₂He = 4.001506 u, and converting to kilograms (1 u = 1.66 × 10⁻²⁷ kg), the calculation gives:
Δm = (4.001506 u + 1.674929 u) - (2.016331 u + 3.016049 u)
≈ 0.643 u
K.E. = (0.643 u) × (1.66 × 10⁻²⁷ kg/u) × (3.00 × 10⁸ m/s)²
≈ 17.6 MeV
Therefore, the kinetic energy acquired by the neutron in the fusion reaction is approximately 17.6 MeV.
In the fusion reaction ²₁H + ³₁H → ⁴₂He + ¹₀n, the neutron acquires a kinetic energy of approximately 17.6 MeV. This value is obtained by calculating the mass difference between the initial reactants and the final products using the mass-energy equivalence principle, E = mc². The conservation of momentum ensures that the total initial momentum is equal to the total final momentum, allowing us to consider the kinetic energy acquired by the neutron as accounting for the total initial kinetic energy.
Understanding the energy released and the kinetic energy acquired by particles in fusion reactions is essential in fields such as nuclear physics and energy research, as it provides insights into the dynamics and behavior of atomic nuclei during nuclear reactions.
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