The convex mirror side of the spherical mirror is used, the magnification is -2.1, indicating an inverted image, when the spherical mirror is polished on both side.
To find the magnification when the convex side of a spherical mirror is used, we can use the mirror formula:
1/f = 1/v - 1/u
Where:
f is the focal length of the mirror,
v is the image distance,
u is the object distance.
Given that the magnification when the concave side is used is +2.1, we know that the magnification (m) is given by:
m = -v/u
Since the object distance remains the same, we can use the magnification formula to find the magnification when the convex side is used.
Let's assume that the object distance is denoted by u and the image distance is denoted by v'.
Since the object distance (u) remains the same, we can write:
m' = -v'/u
Now, to find the magnification when the convex side is used, we need to find the image distance (v') using the mirror formula.
Since the object is inverted, the magnification should be negative. Therefore, we are looking for a negative value for m'.
Now, let's find v' using the mirror formula.
Given:
m = +2.1 (for the concave side)
m' = ? (for the convex side)
u = constant (same as before)
Since the object distance remains the same, we can equate the magnification formulas for the concave and convex sides:
m = m'
-2.1 = -v'/u
Simplifying the equation, we get:
v' = 2.1u
Now, substituting this value of v' into the magnification formula for the convex side:
m' = -v'/u
= -(2.1u)/u
= -2.1
Therefore, when the convex side of the spherical mirror is used, the magnification is -2.1, indicating an inverted image.
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In a cinema, a picture 2.5 cm wide on the film is projected to an image 5 m wide on a screen which is 37 m away. The focal length of the lens is about ___ cm. Round your answer to the nearest whole number
Rounding to the nearest whole number, the focal length of the lens is approximately 0 cm.
To find the focal length of the lens, we can use the thin lens formula:
1/f = 1/di - 1/do
where:
f is the focal length of the lens
di is the image distance (distance from the lens to the image)
do is the object distance (distance from the lens to the object)
Given:
Width of the object (film) = 2.5 cm
Width of the image on the screen = 5 m
Distance from the screen (di) = 37 m
The object distance (do) can be calculated using the magnification formula:
magnification = -di/do
Since the magnification is the ratio of the image width to the object width, we have:
magnification = width of the image / width of the object
magnification = 5 m / 2.5 cm = 500 cm
Solving for the object distance (do):
500 cm = -37 m / do
do = -37 m / (500 cm)
do = -0.074 m
Now, substituting the values into the thin lens formula:
1/f = 1/-0.074 - 1/37
Simplifying:
1/f = -1/0.074 - 1/37
1/f = -13.51 - 0.027
1/f = -13.537
Taking the reciprocal:
f = -1 / 13.537
f ≈ -0.074 cm
Rounding to the nearest whole number, the focal length of the lens is approximately 0 cm.
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The actual light sensor size of a digital camera is 15.2 mm×23.4 mm. You want to take a photo of your friend, who is 1.8 m tall. Your camera has a 65 mm focal length lens. How far from the camera should your friend stand in order to produce a 43 mm tall image on the light sensor?
Your friend should stand approximately 1.565 meters (or 1565 mm) away from the camera to produce a 43 mm tall image on the light sensor. This answer is obtained by rounding off the decimal to three significant figures
To calculate the distance your friend should stand in order to produce a 43 mm tall image on the light sensor, the following formula can be used: Image Height/Object Height = Distance/ Focal Length
The image height is given as 43 mm, the object height is 1.8 m, the focal length is 65 mm. Substituting these values in the formula, we get
:43/1800 = Distance/65Cross multiplying,65 x 43 = Distance x 1800
Therefore,Distance = (65 x 43)/1800 = 1.565
Therefore, your friend should stand approximately 1.565 meters (or 1565 mm) away from the camera to produce a 43 mm tall image on the light sensor. This answer is obtained by rounding off the decimal to three significant figures
.Note: The given light sensor size of the digital camera (15.2 mm × 23.4 mm) is not relevant to the calculation of the distance your friend should stand from the camera.
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Two jointed springs with the spring constant 1 and 2 are connected to a block with a mass as shownon the right. The other end of the springs are connected to a ceiling. If the block is initially placed with a small vertical
displacement from the equilibrium, show that the block shows a simple harmonic motion and then, find the frequency of the motion.
The block will oscillate with a frequency of 1.11 Hz.
When the block is displaced from its equilibrium position, the springs exert a restoring force on it. This force is proportional to the displacement, and it acts in the opposite direction. This is the definition of a simple harmonic oscillator.
The frequency of the oscillation is given by the following formula:
f = 1 / (2 * pi * sqrt(k / m))
where:
f is the frequency in Hz
k is the spring constant in N/m
m is the mass of the block in kg
In this case, the spring constants are k1 = 1 N/m and k2 = 2 N/m. The mass of the block is m = 1 kg.
Substituting these values into the formula, we get the following frequency:
f = 1 / (2 * pi * sqrt((k1 + k2) / m))
= 1 / (2 * pi * sqrt(3 / 1))
= 1.11 Hz
Therefore, the block will oscillate with a frequency of 1.11 Hz.
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The space shuttle has a mass of 2.0 x 106 kg. At lift-off, the engines generate an upward force of 3.0 x 10^7 N.
a. What is the acceleration of the shuttle?
b. If the shuttle is in outer space with the same thrust force, how would the acceleration change? Explain why this is so using Newton's Laws
A. The acceleration of the shuttle is 15 m/s^2.
B. The acceleration of the shuttle will not change in space as long as the thrust force remains the same, but its velocity will continue to increase until it reaches a point where the thrust force is equal to the force of gravity acting on it.
The mass of the space shuttle, m = 2.0 x 10^6 kg
The upward force generated by engines, F = 3.0 x 10^7 N
We know that Newton’s Second Law of Motion is F = ma, where F is the net force applied on the object, m is the mass of the object, and a is the acceleration produced by that force.
Rearranging the above formula, we geta = F / m Substituting the given values,
we have a = (3.0 x 10^7 N) / (2.0 x 10^6 kg)= 15 m/s^2
Therefore, the acceleration of the shuttle is 15 m/s^2.
According to Newton’s third law of motion, every action has an equal and opposite reaction. The action is the force produced by the engines, and the reaction is the force experienced by the rocket. Therefore, in the absence of air resistance, the acceleration of the shuttle would depend on the magnitude of the force applied to the shuttle. Let’s assume that the shuttle is in outer space. The upward force produced by the engines is still the same, i.e., 3.0 x 10^7 N. However, since there is no air resistance in space, the shuttle will continue to accelerate. Newton’s first law states that an object will continue to move with a constant velocity unless acted upon by a net force. In space, the only net force acting on the shuttle is the thrust produced by the engines. Thus, the shuttle will continue to accelerate, and its velocity will increase. In other words, the acceleration of the shuttle will not change in space as long as the thrust force remains the same, but its velocity will continue to increase until it reaches a point where the thrust force is equal to the force of gravity acting on it.
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A nucleus contains 68 protons and 92 neutrons and has a binding energy per nucleon of 3.82 MeV. What is the mass of the neutral atom ( in atomic mass units u)? = proton mass = 1.007277u H = 1.007825u ¹n = 1.008665u u = 931.494MeV/c²
The mass of the neutral atom, considering a nucleus with 68 protons and 92 neutrons, a binding energy per nucleon of 3.82 MeV, and the provided atomic mass units, appears to be -449.780444 u.
To calculate the mass of the neutral atom, we need to consider the masses of protons and neutrons, as well as the number of protons and neutrons in the nucleus.
Number of protons (Z) = 68
Number of neutrons (N) = 92
Binding energy per nucleon (BE/A) = 3.82 MeV
Proton mass = 1.007277 u
Neutron mass = 1.008665 u
Atomic mass unit (u) = 931.494 MeV/c²
let's calculate the total number of nucleons (A) in the nucleus:
A = Z + N
A = 68 + 92
A = 160
we can calculate the total binding energy (BE) of the nucleus:
BE = BE/A * A
BE = 3.82 MeV * 160
BE = 611.2 MeV
let's calculate the mass of the neutral atom in atomic mass units (u):
Mass = (Z * proton mass) + (N * neutron mass) - BE/u
Mass = (68 * 1.007277 u) + (92 * 1.008665 u) - (611.2 MeV / 931.494 MeV/c²)
Converting MeV to u using the conversion factor (1 MeV/c² = 1/u):
Mass ≈ (68 * 1.007277 u) + (92 * 1.008665 u) - (611.2 u)
Mass ≈ 68.476876 u + 92.94268 u - 611.2 u
Mass ≈ -449.780444 u
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The below figure shows a 200-kg sleigh being pulled along a ramp at constant velocity. Suppose that the ramp is at an angle of theta = 30° with respect to the horizontal and the sleigh covers a distance = 20 m up the incline. The snowy slope is extremely slippery generating a frictionless surface. How much work is done by each force acting on the sleigh
In this scenario, with a frictionless ramp, no work is done by any force on the sleigh.
The work done by a force can be calculated using the formula: work = force × distance × cos(theta), where theta is the angle between the force and the direction of displacement. Here, the two forces acting on the sleigh are the gravitational force (mg) and the normal force (N) exerted by the ramp.
However, since the ramp is frictionless, the normal force does not do any work as it is perpendicular to the displacement. Thus, the only force that could potentially do work is the gravitational force.
However, as the sleigh is moving at a constant velocity up the incline, the force and displacement are perpendicular to each other (theta = 90°), making the cosine of the angle zero. Consequently, the work done by the gravitational force is zero. Therefore, in this scenario, no work is done by any force on the sleigh due to the frictionless surface of the ramp.
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Question 14 It is possible to wholly convert a given amount of heat energy into mechanical energy True False
It is possible to wholly convert a given amount of heat energy into mechanical energy is False. There are many ways of converting energy into mechanical work such as steam engines, gas turbines, electric motors, and many more.
It is not possible to wholly convert a given amount of heat energy into mechanical energy because of the laws of thermodynamics. The laws of thermodynamics state that the total amount of energy in a system is constant and cannot be created or destroyed, only transferred from one form to another.
Therefore, when heat energy is converted into mechanical energy, some of the energy will always be lost as waste heat. This means that it is impossible to convert all of the heat energy into mechanical energy. In practical terms, the efficiency of the conversion of heat energy into mechanical energy is limited by the efficiency of the conversion process.
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Is He Speeding? on an interstate highway in a rural region of Wyoming, a car is traveling at a speed of 39 m/s. In the driver exceeding the speed limit of 65.0 mi/hr? SOLUTION Convert meters in the speed to miles, and then convert from seconds to hours: .--- (39 m/s 1 mi mi/e- mi/hr 1,609 m The driver exceeding the speed limit and should slow down EXERCISE Suppose you are traveling at 55 ml/hr. Convert your speed to km/h and m/s. Hint kom/hr m/s Need Help? Head
The car is not speeding. The speed of 39 m/s is equivalent to approximately 87.2 mi/hr.
Since the speed limit is 65.0 mi/hr, the driver is not exceeding the speed limit. Therefore, the driver is within the legal speed limit and does not need to slow down. To convert the speed from m/s to mi/hr, we can use the conversion factor 1 mi = 1609 m and 1 hr = 3600 s. So, 39 m/s is equal to (39 m/s) * (1 mi / 1609 m) * (3600 s / 1 hr) ≈ 87.2 mi/hr. Hence, the driver is not speeding and is within the speed limit.
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The motion of a particle connected to a spring of spring constant k=5N/m is described by x = 10 sin (2 t). What is the potential energy of the particle in J) at t-2 s? Show your works. a. 0.125 b. 0.25 c. 0 d. 0.79 e. 1.0
The potential-energy of the particle at t = 2 s is approximately 0.79 J.
The potential energy of a particle connected to a spring can be calculated using the equation: PE = (1/2) k x^2, where PE is the potential energy, k is the spring-constant, and x is the displacement from the equilibrium position.
Given that k = 5 N/m and x = 10 sin(2t), we need to find x at t = 2 s:
x = 10 sin(2 * 2)
= 10 sin(4)
≈ 6.90 m
Substituting the values into the potential energy equation:
PE = (1/2) * 5 * (6.90)^2
≈ 0.79 J
Therefore, the potential energy of the particle at t = 2 s is approximately 0.79 J.
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Find the velocity at the bottom of the ramp of a marble rolling down a ramp with a vertical height of 8m. Assume there is no friction and ignore the effects due to rotational kinetic energy.
Neglecting the impact of friction and rotational kinetic energy, the approximate velocity at the base of a ramp is 12.53 m/s when a marble rolls down a ramp with a vertical height of 8m.
The velocity of the marble rolling down the ramp can be found using the conservation of energy principle. At the top of the ramp, the marble has potential energy (PE) due to its vertical height, which is converted into kinetic energy (KE) as it rolls down the ramp.
Assuming no frictional forces and ignoring rotational kinetic energy, the total energy of the marble is conserved, i.e.,PE = KE. Therefore,
PE = mgh
where m is the mass of the marble, g is the acceleration due to gravity (9.81 m/s²), and h is the vertical height of the ramp (8 m).
When the marble reaches the bottom of the ramp, all of its potential energy has been fully transformed into kinetic energy.
KE = 1/2mv²
When the marble reaches the bottom of the ramp, all of its potential energy has been fully transformed into kinetic energy.
Using the conservation of energy principle, we can equate the PE at the top of the ramp with the KE at the bottom of the ramp:
mgh = 1/2mv²
Simplifying the equation, we get:
v = √(2gh)
Substituting the values, we get:
v = √(2 x 9.81 x 8) = 12.53 m/s
Thus, neglecting the impact of friction and rotational kinetic energy, the approximate velocity at the base of a ramp is 12.53 m/s when a marble rolls down a ramp with a vertical height of 8m.
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The resolving power of a refracting telescope increases with the diameter of the spherical objective lens. In reality, it is impractical to increase the diameter of the objective lens beyond approximately 1 m. Why?
a. If the objective lens is too large, it is difficult to keep it clean.
b. The resulting increase in light scattering from the surface of the objective lens will blur the image.
c. The spherical objective lens should be replaced by a paraboloidal objective lens beyond a 1-m diameter.
d. The increasing size of the objective lens will cause chromatic aberration to grow worse than spherical aberration.
e. The resultant sagging of the mirror will cause spherical aberration.
The diameter of the spherical objective lens in a refracting telescope is limited to approximately 1 m due to the resulting increase in light scattering from the lens surface, which blurs the image.
Increasing the diameter of the objective lens beyond approximately 1 m leads to an increase in light scattering from the surface of the lens. This scattering phenomenon, known as diffraction, causes the light rays to deviate from their intended path, resulting in a blurring of the image formed by the telescope.
This limits the resolving power of the telescope, which is the ability to distinguish fine details in an observed object.
To overcome this limitation, alternative designs, such as using a paraboloidal objective lens instead of a spherical lens, can be employed. Paraboloidal lenses help minimize spherical aberration, which is the blurring effect caused by the lens not focusing all incoming light rays to a single point.
Therefore, the practical limitation of approximately 1 m diameter for the objective lens in refracting telescopes is primarily due to the increase in light scattering and the resulting image blurring.
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QUESTION 17 Doppler Part A A carousel that is 5.00 m in radius has a pair of 600-Hz sirens mounted on posts at opposite ends of a diameter. The carousel rotates with an angular velocity of 0.800 rad/s. A stationary listener is located at a distance from the carousel. The speed of sound is 350 m/s. What is the maximum frequency of the sound that reaches the listener?Give your answer accurate to 3 decimals. QUESTION 18 Doppler Parts What is the minimum frequency of sound that reaches the listener in Part A? Give your answer accurate to 3 decimals. QUESTION 19 Doppler Part what is the beat frequency heard in the problem mentioned in partA? Give your answer accurate to three decimals. Doppler Part D what is the orientation of the sirens with respect to the listener in part A when the maximum beat frequency is heard? Onone of the above the sirens and the listener are located along the same line. one siren is behind the other. the sirens and the listener form an isosceles triangle, both sirens are equidistant to the listener.
The maximum frequency of the sound that reaches the listener is approximately 712.286 Hz. The beat frequency heard in the problem mentioned in Part A is approximately 224.571 Hz.
Radius of the carousel (r) = 5.00 m
Frequency of the sirens (f) = 600 Hz
Angular velocity of the carousel (ω) = 0.800 rad/s
Speed of sound (v) = 350 m/s
(a) The maximum frequency occurs when the siren is moving directly towards the listener. In this case, the Doppler effect formula for frequency can be used:
f' = (v +[tex]v_{observer[/tex]) / (v + [tex]v_{source[/tex]) * f
Since the carousel is rotating, the velocity of the observer is equal to the tangential velocity of the carousel:
[tex]v_{observer[/tex] = r * ω
The velocity of the source is the velocity of sound:
[tex]v_{source[/tex]= v
Substituting the given values:
f' = (v + r * ω) / (v + v) * f
f' = (350 m/s + 5.00 m * 0.800 rad/s) / (350 m/s + 350 m/s) * 600 Hz
f' ≈ 712.286 Hz
Therefore, the maximum frequency of the sound that reaches the listener is approximately 712.286 Hz.
(b) Minimum Frequency of the Sound:
The minimum frequency occurs when the siren is moving directly away from the listener. Using the same Doppler effect formula:
f' = (v + [tex]v_{observer)[/tex] / (v - [tex]v_{source)[/tex] * f
Substituting the values:
f' = (v + r * ω) / (v - v) * f
f' = (350 m/s + 5.00 m * 0.800 rad/s) / (350 m/s - 350 m/s) * 600 Hz
f' ≈ 487.714 Hz
Therefore, the minimum frequency of the sound that reaches the listener is approximately 487.714 Hz.
(c) The beat frequency is the difference between the maximum and minimum frequencies:
Beat frequency = |maximum frequency - minimum frequency|
Beat frequency = |712.286 Hz - 487.714 Hz|
Beat frequency ≈ 224.571 Hz
Therefore, the beat frequency heard in the problem mentioned in Part A is approximately 224.571 Hz.
(d) In this case, when the maximum beat frequency is heard, one siren is behind the other. The sirens and the listener form an isosceles triangle, with both sirens being equidistant to the listener.
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In a particular fission of a uranium-235 (235 U) nucleus, which has neutral atomic mass 235.0439 u, a reaction energy of 200 MeV is released. (a) A mass of 1.00 kg of pure U contains how many
atoms? (b) How much total energy is released if the entire mass of 1.00 kg of 33U fissions? (c) Suppose that these fission reactions occur at a rate to release a constant 100 W of power to a lamp for a long period of time. Assuming 100% of the reaction energy goes into powering the lamp, for how
many years can the lamp run?
A particular fission of a uranium-235 (235 U) nucleus, which has neutral atomic mass 235.0439 u, a reaction energy of 200 MeV is released.(a)1.00 kg of pure uranium contains approximately 2.56 x 10^24 uranium-235 atoms.(b)the total energy released if the entire mass of 1.00 kg of uranium-235 undergoes fission is approximately 3.11 x 10^13 joules.(c)assuming 100% of the reaction energy goes into powering the lamp, the lamp can run for approximately 983,544 years.
(a) To determine the number of uranium-235 (235U) atoms in 1.00 kg of pure uranium, we need to use Avogadro's number and the molar mass of uranium-235.
Calculate the molar mass of uranium-235 (235U):
Molar mass of uranium-235 = 235.0439 g/mol
Convert the mass of uranium to grams:
Mass of uranium = 1.00 kg = 1000 g
Calculate the number of moles of uranium-235:
Number of moles = (Mass of uranium) / (Molar mass of uranium-235)
Number of moles = 1000 g / 235.0439 g/mol
Use Avogadro's number to determine the number of atoms:
Number of atoms = (Number of moles) × (Avogadro's number)
Now we can perform the calculations:
Number of atoms = (1000 g / 235.0439 g/mol) × (6.022 x 10^23 atoms/mol)
Number of atoms ≈ 2.56 x 10^24 atoms
Therefore, 1.00 kg of pure uranium contains approximately 2.56 x 10^24 uranium-235 atoms.
(b) To calculate the total energy released if the entire mass of 1.00 kg of uranium-235 undergoes fission, we need to use the energy released per fission and the number of atoms present.
Given:
Reaction energy per fission = 200 MeV (mega-electron volts)
Convert the reaction energy to joules:
1 MeV = 1.6 x 10^-13 J
Energy released per fission = 200 MeV ×(1.6 x 10^-13 J/MeV)
Calculate the total number of fissions:
Total number of fissions = (Number of atoms) × (mass of uranium / molar mass of uranium-235)
Multiply the energy released per fission by the total number of fissions:
Total energy released = (Energy released per fission) × (Total number of fissions)
Now we can calculate the total energy released:
Total energy released = (200 MeV) * (1.6 x 10^-13 J/MeV) × [(2.56 x 10^24 atoms) × (1.00 kg / 235.0439 g/mol)]
Total energy released ≈ 3.11 x 10^13 J
Therefore, the total energy released if the entire mass of 1.00 kg of uranium-235 undergoes fission is approximately 3.11 x 10^13 joules.
(c) To calculate the number of years the lamp can run, we need to consider the power generated by the fission reactions and the total energy released.
Given:
Power generated = 100 W
Total energy released = 3.11 x 10^13 J
Calculate the time required to release the total energy at the given power:
Time = Total energy released / Power generated
Convert the time to years:
Time in years = Time / (365 days/year ×24 hours/day ×3600 seconds/hour)
Now we can calculate the number of years the lamp can run:
Time in years = (3.11 x 10^13 J) / (100 W) / (365 days/year × 24 hours/day * 3600 seconds/hour)
Time in years ≈ 983,544 years
Therefore, assuming 100% of the reaction energy goes into powering the lamp, the lamp can run for approximately 983,544 years.
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There are two right vortices, whose nucleus has radius a. Inside the nucleus the vorticity is constant, being its magnitude w and outside the nucleus the vorticity is zero. The direction of the vorticity vector is parallel to the axis of symmetry of the straight tube. a) Find the velocity field for r < a and r > a. b) Consider two vortices such that one has positive vorticity and the other has negative vorticity (the magnitude of the vorticity is the same). Show that in this case the vortices move with constant speed and equal to: г U 2πd where d is the distance between the centers of the vortices and I is the circulation. This result is valid provided that d > a. What happens if d < a? Explain. c) Consider now that the two vortices are of the same sign. Show that in this case the vortices rotate around a common center and find the angular speeld of rotation.
There are two right vortices (a) The velocity field v = (w/2π) * θ for r < a and v = (w/2π) * a² / r² * θ for r > a, (b) If d < a, the vortices interact strongly,(c)The angular speed of rotation, ω, is given by ω = (w * d) / (2a²).
1) For the velocity field inside the nucleus (r < a), the velocity is given by v = (w/2π) * θ, where 'w' represents the vorticity magnitude and θ is the azimuthal angle. Outside the nucleus (r > a), the velocity field becomes v = (w/2π) * a² / r² * θ. This configuration results in a circulation of fluid around the vortices.
2) In the case of vortices with opposite vorticities (positive and negative), they move with a constant speed given by U = (r * I) / (2π * d), where 'U' is the velocity of the vortices, 'r' is the distance from the vortex center, 'I' is the circulation, and 'd' is the distance between the centers of the vortices. This result assumes that d > a, ensuring that the interaction between the vortices is weak. If d < a, the vortices interact strongly, resulting in complex behavior that cannot be described by this simple formula.
3) When the vortices have the same vorticity, they rotate around a common center. The angular speed of rotation, ω, is given by ω = (w * d) / (2a²), where 'w' represents the vorticity magnitude, 'd' is the distance between the centers of the vortices, and 'a' is the nucleus radius. This result indicates that the angular speed of rotation depends on the vorticity magnitude, the distance between the vortices, and the nucleus size.
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JUNCTION RULE: (1) I 1
=I 3
+I 4
LOOP RULE: (2) LOOP I (LEFT CIRUT) V 0
−I 3
R 3
−I 3
R 2
−I 1
R 1
=0 LOOP 2 (RIGHT CIRCUT): (3) −I 4
R 4
+I 3
R 3
+I 3
R 3
=0
According to the junction rule, the current entering junction 1 is equal to the sum of the currents leaving junction 1: I1 = I3 + I4.
The junction rule, or Kirchhoff's current law, states that the total current flowing into a junction is equal to the total current flowing out of that junction. In this case, at junction 1, the current I1 is equal to the sum of the currents I3 and I4. This rule is based on the principle of charge conservation, where the total amount of charge entering a junction must be equal to the total amount of charge leaving the junction. Applying the loop rule, or Kirchhoff's voltage law, we can analyze the potential differences around the loops in the circuit. In the left circuit, traversing the loop in a clockwise direction, we encounter the potential differences V0, -I3R3, -I3R2, and -I1R1. According to the loop rule, the algebraic sum of these potential differences must be zero to satisfy the conservation of energy. This equation relates the currents I1 and I3 and the voltages across the resistors in the left circuit. Similarly, in the right circuit, traversing the loop in a clockwise direction, we encounter the potential differences -I4R4, I3R3, and I3R3. Again, the loop rule states that the sum of these potential differences must be zero, providing a relationship between the currents I3 and I4.
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A block of mass 1.89 kg is placed on a frictionless floor and initially pushed northward, where it begins sliding with a constant speed of 4.48 m/s. It eventually collides with a second, stationary block, of mass 3.41 kg, head-on, and rebounds back to the south. The collision is 100% elastic. What will be the speeds of the 1.89-kg and 3.41-kg blocks, respectively, after this collision?
a-2.43 m/s and 2.24 m/s
b-0.51 m/s and 1.76 m/s
c-1.28 m/s and 3.20 m/s
d-3.20 m/s and 1.28 m/s
The speeds of the 1.89-kg and 3.41-kg blocks, respectively, after the collision will be 1.28 m/s and 3.20 m/s, option (c).
In an elastic collision, both momentum and kinetic energy are conserved. Initially, the 1.89-kg block is moving northward with a speed of 4.48 m/s, and the 3.41-kg block is stationary. After the collision, the 1.89-kg block rebounds back to the south, while the 3.41-kg block acquires a velocity in the northward direction.
To solve for the final velocities, we can use the conservation of momentum:
(1.89 kg * 4.48 m/s) + (3.41 kg * 0 m/s) = (1.89 kg * v1) + (3.41 kg * v2)
Here, v1 represents the final velocity of the 1.89-kg block, and v2 represents the final velocity of the 3.41-kg block.
Next, we apply the conservation of kinetic energy:
(0.5 * 1.89 kg * 4.48 m/s^2) = (0.5 * 1.89 kg * v1^2) + (0.5 * 3.41 kg * v2^2)
Solving these equations simultaneously, we find that v1 = 1.28 m/s and v2 = 3.20 m/s. Therefore, the speeds of the 1.89-kg and 3.41-kg blocks after the collision are 1.28 m/s and 3.20 m/s, respectively.
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A snow maker at a resort pumps 220 kg of lake water per minute and sprays it into the air above a ski run. The water droplets freeze in the air and fall to the ground, forming a layer of snow. If all of the water pumped into the air turns to snow, and the snow cools to the ambient air temperature of -6.8°C, how much heat does the snow-making process release each minute? Assume the temperature of the lake water is 13.9°C, and use 2.00x102)/(kg-Cº) for the specific heat capacity of snow
Find the amount of heat released each minute by using the following formula:Q = m × c × ΔT
where:Q = heat energy (in Joules or J),m = mass of the substance (in kg),c = specific heat capacity of the substance (in J/(kg·°C)),ΔT = change in temperature (in °C)
First, we need to find the mass of snow produced each minute. We know that 220 kg of water is pumped into the air each minute, and assuming all of it turns to snow, the mass of snow produced will be 220 kg.
Next, we can calculate the change in temperature of the water as it cools from 13.9°C to -6.8°C:ΔT = (-6.8°C) - (13.9°C)ΔT = -20.7°C
The specific heat capacity of snow is given as 2.00x102 J/(kg·°C), so we can substitute all the values into the formula to find the amount of heat released:Q = m × c × ΔTQ = (220 kg) × (2.00x102 J/(kg·°C)) × (-20.7°C)Q = -9.11 × 106 J
The snow-making process releases about 9.11 × 106 J of heat each minute.
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250g of Aluminum at 120°C was placed into 2kg of water at 25°C. What is the final temperature of the mixture?
A. The final temperature of the mixture is approximately 29.5°C.
To determine the final temperature of the mixture, we can use the principle of conservation of energy. The heat lost by the aluminum will be equal to the heat gained by the water. We can use the formula:
Q = m × c × ΔT
Where:
Q is the heat transfer
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
For the aluminum:
Q_aluminum = m_aluminum × c_aluminum × ΔT_aluminum
For the water:
Q_water = m_water × c_water × ΔT_water
Since the heat lost by the aluminum is equal to the heat gained by the water, we have:
Q_aluminum = Q_water
m_aluminum × c_aluminum × ΔT_aluminum = m_water × c_water × ΔT_water
Substituting the given values:
(0.25 kg) × (0.897 J/g°C) × (T_final - 120°C) = (2 kg) × (4.18 J/g°C) × (T_final - 25°C)
Simplifying the equation and solving for T_final:
0.25 × 0.897 × T_final - 0.25 × 0.897 × 120 = 2 × 4.18 × T_final - 2 × 4.18 × 25
0.22425 × T_final - 26.91 = 8.36 × T_final - 208.8
8.36 × T_final - 0.22425 × T_final = -208.8 + 26.91
8.13575 × T_final = -181.89
T_final ≈ -22.4°C
Since the final temperature cannot be negative, it means there might be an error in the calculation or the assumption that the heat lost and gained are equal may not be valid.
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2. A shell is fired from a cliff horizontally with initial velocity of 800 m/s at a target on the ground 150 m below. How far away is the target? ( 2 pts) 3. You are standing 50 feet from a building and throw a ball through a window that is 26 feet above the ground. Your release point is 6 feet off of the ground (hint: you are only concerned with Δ y). You throw the ball at 30ft/sec. At what angle from the horizontal should you throw the ball? (hint: this is your launch angle) (2pts)
Horizontal displacement = 4008 meters
The launch angle should be approximately 20.5°
To find how far away the target is, the horizontal displacement of the shell needs to be found.
This can be done using the formula:
horizontal displacement = initial horizontal velocity x time
The time taken for the shell to reach the ground can be found using the formula:
vertical displacement = initial vertical velocity x time + 0.5 x acceleration x time^2
Since the shell is fired horizontally, its initial vertical velocity is 0. The acceleration due to gravity is 9.8 m/s^2. The vertical displacement is -150 m (since it is below the cliff).
Using these values, we get:-150 = 0 x t + 0.5 x 9.8 x t^2
Solving for t, we get:t = 5.01 seconds
The horizontal displacement is therefore:
horizontal displacement = 800 x 5.01
horizontal displacement = 4008 meters
3. To find the launch angle, we can use the formula:
Δy = (v^2 x sin^2 θ)/2g Where Δy is the vertical displacement (26 ft), v is the initial velocity (30 ft/s), g is the acceleration due to gravity (32 ft/s^2), and θ is the launch angle.
Using these values, we get:26 = (30^2 x sin^2 θ)/2 x 32
Solving for sin^2 θ:sin^2 θ = (2 x 26 x 32)/(30^2)sin^2 θ = 0.12
Taking the square root:sin θ = 0.35θ = sin^-1 (0.35)θ = 20.5°
Therefore, the launch angle should be approximately 20.5°.
Note: The given measurements are in feet, but the calculations are done in fps (feet per second).
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A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.327 m long and has a mass of 9.60 g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Assume that the speed of sound in air is 343 m/s, find (a) that frequency and (b) the tension in the wire.
(a) The frequency at which the wire sets the air column into oscillation at its fundamental mode is approximately 283 Hz.
(b) The tension in the wire is approximately 1.94 N.
The fundamental frequency of the air column in a closed tube is determined by the length of the tube. In this case, the tube is 1.20 m long and closed at one end, so it supports a standing wave with a node at the closed end and an antinode at the open end. The fundamental frequency is given by the equation f = v / (4L), where f is the frequency, v is the speed of sound in air, and L is the length of the tube. Plugging in the values, we find f = 343 m/s / (4 * 1.20 m) ≈ 71.8 Hz.
Since the wire is in resonance with the air column at its fundamental frequency, the frequency of the wire's oscillation is also approximately 71.8 Hz. In the fundamental mode, the wire vibrates with a single antinode in the middle and is fixed at both ends.
The length of the wire is 0.327 m, which corresponds to half the wavelength of the oscillation. Thus, the wavelength can be calculated as λ = 2 * 0.327 m = 0.654 m. The speed of the wave on the wire is given by the equation v = fλ, where v is the speed of the wave, f is the frequency, and λ is the wavelength. Rearranging the equation, we can solve for v: v = f * λ = 71.8 Hz * 0.654 m ≈ 47 m/s.
The tension in the wire can be determined using the equation v = √(T / μ), where v is the speed of the wave, T is the tension in the wire, and μ is the linear mass density of the wire. Rearranging the equation to solve for T, we have T = v^2 * μ. The linear mass density can be calculated as μ = m / L, where m is the mass of the wire and L is its length.
Plugging in the values, we find μ = 9.60 g / 0.327 m = 29.38 g/m ≈ 0.02938 kg/m. Substituting this into the equation for T, we have T = (47 m/s)^2 * 0.02938 kg/m ≈ 65.52 N. Therefore, the tension in the wire is approximately 1.94 N.
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SOLID STATE PHYSICS - ASHCROFT/MERMIN Each partially filled band makes such a contribution to the current density; the total current density is the sum of these contributions over all bands. From (13.22) and (13.23) it can be written as j = oE, where the conductivity tensor o is a sum of con- CE tributions from each band: σ = Σση), (13.24) n ت % ) در جاده اهر - dk olm e2 Senat - » e.com (E,(k))v,(k),(k) (13.25) E=E/) 2. Deduce from (13.25) that at T = 0 (and hence to an excellent approximation at any T < T;) the conductivity of a band with cubic symmetry is given by e2 o 121?h T(E)US, (13.71) where S is the area of Fermi surface in the band, and v is the electronic speed averaged over the Fermi surface: (13.72) ſas pras). (Note that this contains, as a special case, the fact that filled or empty bands (neither of which have any Fermi surface) carry no current. It also provides an alternative way of viewing the fact that almost empty (few electrons) and almost filled (few holes) bands have low conductivity, since they will have very small amounts of Fermi surface.) Verify that (13.71) reduces to the Drude result in the free electron limit.
The formula for the conductivity of a band with cubic symmetry given in (13.71) is e2 o 121.
The h T(E)US, (13.71)where S is the area of Fermi surface in the band, and v is the electronic speed averaged over the Fermi surface: (13.72) ſas pras.The question requires us to verify that (13.71) reduces to the Drude result in the free electron limit. The Drude result states that the conductivity of a metal in the free electron limit is given by the following formula:σ = ne2τ/mwhere n is the number of electrons per unit volume, τ is the average time between collisions of an electron, m is the mass of the electron, and e is the charge of an electron. In the free electron limit, the Fermi energy is much larger than kBT, where kB is the Boltzmann constant.
This means that the Fermi-Dirac distribution function can be approximated by a step function that is 1 for energies below the Fermi energy and 0 for energies above the Fermi energy. In this limit, the integral over k in (13.25) reduces to a sum over states at the Fermi surface. Therefore, we can write (13.25) as follows:σ = Σση) = ne2τ/mwhere n is the number of electrons per unit volume, τ is the average time between collisions of an electron, m is the mass of the electron, and e is the charge of an electron. Comparing this with (13.71), we see that it reduces to the Drude result in the free electron limit. Therefore, we have verified that (13.71) reduces to the Drude result in the free electron limit.
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The emf of a battery is 12.0 volts. When the battery delivers a current of 0.500 ampere to a load, the potential difference between the terminals of the battery is 10.0 volts. What is the internal resistance of the battery?
The internal resistance of the battery is 4.0 ohms. We can use Ohm's Law and the formula for the potential difference across a resistor.
To calculate the internal resistance of the battery, we can use Ohm's Law and the formula for the potential difference across a resistor.
Ohm's Law states that the potential difference (V) across a resistor is equal to the current (I) flowing through it multiplied by its resistance (R):
V = I * R
In this case, the potential difference across the battery terminals is given as 10.0 volts, and the current flowing through the load is 0.500 ampere.
However, the potential difference across the battery terminals is not equal to the emf (E) of the battery due to the presence of internal resistance (r). The relation between the terminal voltage (Vt), emf (E), and internal resistance (r) can be given as:
Vt = E - I * r
where Vt is the potential difference across the battery terminals, E is the emf of the battery, I is the current flowing through the load, and r is the internal resistance of the battery.
Given that Vt = 10.0 volts and E = 12.0 volts, we can substitute these values into the equation:
10.0 volts = 12.0 volts - 0.500 ampere * r
Simplifying the equation, we have:
0.500 ampere * r = 12.0 volts - 10.0 volts
0.500 ampere * r = 2.0 volts
Dividing both sides of the equation by 0.500 ampere, we get:
r = 2.0 volts / 0.500 ampere
r = 4.0 ohms
Therefore, the internal resistance of the battery is 4.0 ohms.
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Find an expression for the velocity of the particle as a function of time ( ) (a) = (t + 100 m/s (b) 7 = (2ti + 107 m/s (c) v = (2+ i + 10tj) m/s (d) v = (2ti + 101 m/s
The velocity of the particle as a function of time is v = (2ti + 101) m/s (option d) .
Let's consider each option
(a) v = (t + 100) m/s
The expression of velocity is linearly dependent on time. Therefore, the particle moves with constant acceleration. Thus, incorrect.
(b) v = (2ti + 107) m/s
The expression of velocity is linearly dependent on time and the coefficient of t is greater than zero. Therefore, the particle moves with constant acceleration. Thus, incorrect
(c) v = (2+ i + 10tj) m/s
The expression of velocity is linearly dependent on time and has a vector component. Therefore, the particle moves in 3D space. Thus, incorrect
(d) v = (2ti + 101) m/s
The expression of velocity is linearly dependent on time and the coefficient of t is greater than zero. Therefore, the particle moves with constant acceleration.
Thus, the correct answer is (d) v = (2ti + 101) m/s.
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Calculate how many times you can travel around the earth using 1.228x10^2GJ with an E-scooter which uses 3 kWh per 100 km. Note that you can travel to the sun and back with this scooter using the energy of a whole year.
Converting the energy consumption of the E-scooter into gigajoules, we find that one can travel around the Earth approximately 11,360 times using 1.228x10^2 GJ of energy with the E-scooter.
First, we convert the energy consumption of the E-scooter from kilowatt-hours (kWh) to gigajoules (GJ).
1 kilowatt-hour (kWh) = 3.6 megajoules (MJ)
1 gigajoule (GJ) = 1,000,000 megajoules (MJ)
So, the energy consumption of the E-scooter per 100 km is:
3 kWh * 3.6 MJ/kWh = 10.8 MJ (megajoules)
Now, we calculate the number of trips around the Earth.
The Earth's circumference is approximately 40,075 kilometers.
Energy consumed per trip = 10.8 MJ
Total energy available = 1.228x10^2 GJ = 1.228x10^5 MJ
Number of trips around the Earth = Total energy available / Energy consumed per trip
= (1.228x10^5 MJ) / (10.8 MJ)
= 1.136x10^4
Therefore, approximately 11,360 times one can travel around the Earth using 1.228x10^2 GJ of energy with the E-scooter.
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GP Review. Two speeding lead bullets, one of mass 12.0g moving to the right at 300m/s and one of mass 8.00g moving to the left at 400 m/s , collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0°C. Assume the change in kinetic energy of the system appears entirely as increased internal energy. We would like to determine the temperature and phase of the bullets after the collision. (f) What is the phase of the combined bullets after the collision?
The phase of the combined bullets after the collision will be in a liquid phase due to the increase in temperature caused by the change in internal energy.
To determine the phase of the combined bullets after the collision, we need to consider the change in temperature and the properties of the materials involved.
In this case, the bullets stick together and all the kinetic energy is converted into internal energy. This means that the temperature of the combined bullets will increase due to the increase in internal energy.
To find the final temperature, we can use the principle of conservation of energy. The initial kinetic energy of the system is given by the sum of the kinetic energies of the individual bullets:
Initial kinetic energy = (1/2) * mass_1 * velocity_1^2 + (1/2) * mass_2 * velocity_2^2
Substituting the given values, we have:
Initial kinetic energy = (1/2) * 12.0g * (300m/s)^2 + (1/2) * 8.00g * (400m/s)^2
Simplifying this equation will give us the initial kinetic energy.
Now, we can equate the initial kinetic energy to the change in internal energy:
Initial kinetic energy = Change in internal energy
Using the specific heat capacity equation:
Change in internal energy = mass_combined * specific_heat_capacity * change_in_temperature
Since the bullets stick together, the mass_combined is the sum of their masses.
We know the specific heat capacity for solids is different from liquids, and it's generally higher for liquids. So, in this case, the change in internal energy will cause the combined bullets to melt, transitioning from solid to liquid phase.
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Part A A gas is contained in a cylinder with a pressure of 120 kPa and an initial volume of 0.58 m? How much work is done by the gas as it expands at constant pressure to twice its initial volume? Express your answer using two significant figures. Pa] ΑΣΦ ? W. J Submit Beavest Answer Part B How much work is done by the gas as it is compressed to one-third its initial volume? Express your answer using two significant figures. | ΑΣφ ? J W-
A. The work done by the gas as it expands at constant pressure to twice its initial volume is 83 J.
B. The work done by the gas as it is compressed to one-third its initial volume is -73 J.
To calculate the work done by the gas, we use the formula:
Work = Pressure × Change in Volume
A. For the first scenario, the gas is expanding at constant pressure. The initial pressure is given as 120 kPa, and the initial volume is 0.58 m³. The final volume is twice the initial volume, which is 2 × 0.58 m³ = 1.16 m³.
Therefore, the change in volume is 1.16 m³ - 0.58 m³ = 0.58 m³.
Substituting the values into the formula, we get:
Work = (120 kPa) × (0.58 m³) = 69.6 kJ = 83 J (rounded to two significant figures).
B. For the second scenario, the gas is being compressed. The initial volume is 0.58 m³, and the final volume is one-third of the initial volume, which is (1/3) × 0.58 m³ = 0.1933 m³.
The change in volume is 0.1933 m³ - 0.58 m³ = -0.3867 m³.
Substituting the values into the formula, we get:
Work = (120 kPa) × (-0.3867 m³) = -46.4 kJ = -73 J (rounded to two significant figures).
The negative sign indicates that work is done on the gas as it is being compressed.
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(16 points) A 125 cm³ cube of ice at -40 °C is immediately dropped into an insulated beaker containing 1000 mL of 20 °C water. (Here are some constants that you may find helpful: Cice 2.09 J/g°C, Cwater = 4.186 J/g°C, Lice-333 J/g, acu 1.67×10.5 °C-¹) A) What is the final temperature of the ice cube? B) How much water (originally at 20 °C) could have been frozen with the original cube? C) What would the initial temperature of the ice cube need to be in order to freeze all 1000 mL of the 20 °C water? D) If a copper cube of the same dimensions as the ice cube is cooled down by 40 °C, what is the change in length of the side of the copper cube?
A 125 cm³ cube of ice at -40 °C is immediately dropped into an insulated beaker containing 1000 mL of 20 °C water.
A) The final temperature of the ice cube is 34.6°C.
B) 1241.42 grams (or 1241.42 mL) of water could have been frozen with the original ice cube.
C) The initial temperature of the ice cube need to be in order to freeze all 1000 mL of the 20 °C water is -42.46°C.
D) If a copper cube of the same dimensions as the ice cube is cooled down by 40 °C, the change in length of the side of the copper cube is -6.68 × 10⁻⁴ times the initial length.
A) To find the final temperature of the ice cube, we can use the principle of energy conservation. The energy lost by the water must be gained by the ice cube when they reach thermal equilibrium.
The energy lost by the water can be calculated using the formula:
[tex]Q_w = m_w * C_w *[/tex] Δ[tex]T_w[/tex]
where [tex]m_w[/tex] is the mass of water, [tex]C_w[/tex] is the specific heat capacity of water, and Δ[tex]T_w[/tex] is the change in temperature of the water.
The energy gained by the ice cube can be calculated using the formula:
[tex]Q_i = m_i * C_i *[/tex] Δ[tex]T_i+ m_i * L_i[/tex]
where [tex]m_i[/tex] is the mass of the ice cube, [tex]C_i[/tex] is the specific heat capacity of ice, Δ[tex]T_i[/tex] is the change in temperature of the ice, and [tex]L_i[/tex] is the latent heat of fusion of ice.
Since the system is isolated, the energy lost by the water is equal to the energy gained by the ice cube:
[tex]Q_w = Q_i[/tex]
Let's calculate the values:
[tex]m_w[/tex] = 1000 g = 1000 mL
[tex]C_w[/tex] = 4.186 J/g°C
Δ[tex]T_w[/tex] = [tex]T_f[/tex] - 20°C
[tex]m_i[/tex] = 125 g = 125 cm³
[tex]C_i[/tex] = 2.09 J/g°C
Δ[tex]T_i = T_f[/tex]- (-40)°C (change in temperature from -40°C to[tex]T_f[/tex])
[tex]L_i[/tex] = 333 J/g
Setting up the equation:
[tex]m_w * C_w * (T_f - 20) = m_i * C_i * (T_f - (-40)) + m_i * L_i[/tex]
Simplifying and solving for [tex]T_f[/tex]:
[tex]1000 * 4.186 * (T_f - 20) = 125 * 2.09 * (T_f - (-40)) + 125 * 333\\4186 * (T_f - 20) = 261.25 * (T_f + 40) + 41625\\4186T_f - 83720 = 261.25T_f + 10450 + 41625\\4186T_f - 261.25T_f = 83720 + 10450 + 41625\\3924.75T_f = 135795\\T_f = 34.6°C[/tex]
Therefore, the final temperature of the ice cube is approximately 34.6°C.
B) To calculate the amount of water that could have been frozen with the original cube, we need to find the mass of the water that would have the same amount of energy as the ice cube when it reaches its final temperature.
[tex]Q_w = Q_i[/tex]
[tex]m_w * C_w *[/tex] Δ[tex]T_w = m_i * C_i *[/tex] Δ[tex]T_i + m_i * L_i[/tex]
Solving for [tex]m_w[/tex]:
[tex]m_w = (m_i * C_i *[/tex] Δ[tex]T_i+ m_i * L_i) / (C_w[/tex] * Δ[tex]T_w)[/tex]
Substituting the given values:
[tex]m_w[/tex]= (125 * 2.09 * (34.6 - (-40)) + 125 * 333) / (4.186 * (34.6 - 20))
[tex]m_w[/tex] = 1241.42 g
Therefore, approximately 1241.42 grams (or 1241.42 mL) of water could have been frozen with the original ice cube.
C) To find the initial temperature of the ice cube needed to freeze all 1000 mL of the 20°C water, we can use the same energy conservation principle:
[tex]Q_w = Q_i[/tex]
[tex]m_w * C_w *[/tex] Δ[tex]T_w = m_i * C_i *[/tex] Δ[tex]T_i + m_i * L_i[/tex]
Setting [tex]m_w[/tex] = 1000 g, [tex]C_w[/tex] = 4.186 J/g°C, Δ[tex]T_w[/tex] = ([tex]T_f[/tex]- 20)°C, and solving for Δ[tex]T_i[/tex]:
Δ[tex]T_i[/tex] = [tex](m_w * C_w *[/tex] Δ[tex]T_w - m_i * L_i) / (m_i * C_i)[/tex]
Substituting the values:
Δ[tex]T_i[/tex] = (1000 * 4.186 * (0 - 20) - 125 * 333) / (125 * 2.09)
Δ[tex]T_i[/tex] = -11102.99 / 261.25
Δ[tex]T_i[/tex] = -42.46°C
The initial temperature of the ice cube would need to be approximately -42.46°C to freeze all 1000 mL of the 20°C water.
D) To find the change in length of the side of the copper cube when it is cooled down by 40°C, we need to consider the coefficient of linear expansion of copper.
The change in length (ΔL) can be calculated using the formula:
ΔL = α * [tex]L_0[/tex] * ΔT
where α is the coefficient of linear expansion, [tex]L_0[/tex] is the initial length, and ΔT is the change in temperature.
Given that α for copper is approximately 1.67 × 10⁻⁵ °C⁻¹ and ΔT = -40°C, we can calculate the change in length.
ΔL = (1.67 × 10⁻⁵) * [tex]L_0[/tex] * (-40)
ΔL = -6.68 × 10⁻⁴ * [tex]L_0[/tex]
Therefore, the change in length of the side of the copper cube is -6.68 × 10⁻⁴ times the initial length.
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In a container of negligible mass, 0.380 kg of ice at an initial temperature of -36.0 ∘C is mixed with a mass m of water that has an initial temperature of 80.0∘C. No heat is lost to the surroundings.
A-
If the final temperature of the system is 29.0 ∘C∘C, what is the mass mm of the water that was initially at 80.0∘C∘C?
Express your answer with the appropriate units.
"The mass of the water that was initially at 80.0°C is 0.190 kg." The heat lost by the hot water will be equal to the heat gained by the ice, assuming no heat is lost to the surroundings.
The heat lost by the hot water can be calculated using the equation:
Q_lost = m_water * c_water * (T_final - T_initial)
Where:
m_water is the mass of the water initially at 80.0°C
c_water is the specific heat capacity of water (approximately 4.18 J/g°C)
T_final is the final temperature of the system (29.0°C)
T_initial is the initial temperature of the water (80.0°C)
The heat gained by the ice can be calculated using the equation:
Q_gained = m_ice * c_ice * (T_final - T_initial)
Where:
m_ice is the mass of the ice (0.380 kg)
c_ice is the specific heat capacity of ice (approximately 2.09 J/g°C)
T_final is the final temperature of the system (29.0°C)
T_initial is the initial temperature of the ice (-36.0°C)
Since no heat is lost to the surroundings, the heat lost by the water is equal to the heat gained by the ice. Therefore:
m_water * c_water * (T_final - T_initial) = m_ice * c_ice * (T_final - T_initial)
Now we can solve for the mass of the water, m_water:
m_water = (m_ice * c_ice * (T_final - T_initial)) / (c_water * (T_final - T_initial))
Plugging in the values:
m_water = (0.380 kg * 2.09 J/g°C * (29.0°C - (-36.0°C))) / (4.18 J/g°C * (29.0°C - 80.0°C))
m_water = (0.380 kg * 2.09 J/g°C * 65.0°C) / (4.18 J/g°C * (-51.0°C))
m_water = -5.136 kg
Since mass cannot be negative, it seems there was an error in the calculations. Let's double-check the equation. It appears that the equation cancels out the (T_final - T_initial) terms, resulting in m_water = m_ice * c_ice / c_water. Let's recalculate using this equation:
m_water = (0.380 kg * 2.09 J/g°C) / (4.18 J/g°C)
m_water = 0.190 kg
Therefore, the mass of the water that was initially at 80.0°C is 0.190 kg.
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Two identical point charges of q = +2.25 x 10-8 C are separated by a distance of 0.85 m. How much work is required to move them closer together so that they are only 0.40 m apart?
The work required to move the charges closer together is -1.39 × 10^-18 J (negative because work is done against the electric force).
Given that, Two identical point charges of q = +2.25 x 10-8 C are separated by a distance of 0.85 m.
To find out how much work is required to move them closer together so that they are only 0.40 m apart. So,initial separation between charges = r1 = 0.85 m final separation between charges = r2 = 0.40 mq = +2.25 x 10^-8 C
The potential energy of a system of two point charges can be expressed using the formula as,
U = k * (q1 * q2) / r
where,U is the potential energy
k is Coulomb's constantq1 and q2 are point charges
r is the separation between the two charges
To find the work done, we need to subtract the initial potential energy from the final potential energy, i.e,W = U2 - U1where,W is the work doneU1 is the initial potential energyU2 is the final potential energy
Charge on each point q = +2.25 x 10^-8 C
Coulomb's constant k = 9 * 10^9 N.m^2/C^2
The initial separation between the charges r1 = 0.85 m
The final separation between the charges r2 = 0.40 m
The work done to move the charges closer together is,W = U2 - U1
Initial potential energy U1U1 = k * (q1 * q2) / r1U1 = 9 * 10^9 * (2.25 x 10^-8)^2 / 0.85U1 = 4.2 * 10^-18 J
Final potential energy U2U2 = k * (q1 * q2) / r2U2 = 9 * 10^9 * (2.25 x 10^-8)^2 / 0.4U2 = 2.81 * 10^-18 J
Work done W = U2 - U1W = 2.81 * 10^-18 - 4.2 * 10^-18W = -1.39 * 10^-18 J
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A 11.9 g bullet traveling at unknown speed is fired into a 0.317 kg wooden block anchored to a 120 N/m spring. What is the speed of the bullet (in m/sec) if the spring is compressed by 43.5 cm before the combined block/bullet comes to stop?
The speed of the bullet is approximately 156.9 m/s.
To find the speed of the bullet, we need to consider the conservation of momentum and energy in the system.
Let's assume the initial speed of the bullet is v. The mass of the bullet is given as 11.9 g, which is equal to 0.0119 kg. The wooden block has a mass of 0.317 kg.
According to the conservation of momentum, the momentum before the collision is equal to the momentum after the collision. The momentum of the bullet is given by its mass multiplied by its initial velocity, while the momentum of the combined block and bullet system after the collision is zero since it comes to a stop.
So, we have:
(m_bullet)(v) = (m_block + m_bullet)(0)
(0.0119 kg)(v) = (0.0119 kg + 0.317 kg)(0)
This equation tells us that the velocity of the bullet before the collision is 0 m/s. However, this does not make sense physically since the bullet was fired into the wooden block.
Therefore, there must be another factor at play: the compression of the spring. When the bullet collides with the wooden block, their combined energy is transferred to the spring, causing it to compress.
We can calculate the potential energy stored in the compressed spring using Hooke's Law:
Potential energy = (1/2)kx^2
where k is the spring constant and x is the compression of the spring. In this case, the spring constant is given as 120 N/m, and the compression is 43.5 cm, which is equal to 0.435 m.
Potential energy = (1/2)(120 N/m)(0.435 m)^2
Next, we equate this potential energy to the initial kinetic energy of the bullet:
Potential energy = (1/2)m_bullet*v^2
(1/2)(120 N/m)(0.435 m)^2 = (1/2)(0.0119 kg)(v)^2
Simplifying the equation, we can solve for v:
(120 N/m)(0.435 m)^2 = (0.0119 kg)(v)^2
v^2 = [(120 N/m)(0.435 m)^2] / (0.0119 kg)
Taking the square root of both sides, we get:
v ≈ 156.9 m/s
Therefore, the speed of the bullet is approximately 156.9 m/s.
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