Answer:
Fractional calculus is a branch of mathematics that deals with the calculus of functions that are not differentiable at all points. This can be useful for modeling physical processes that involve memory or dissipation, such as viscoelasticity, diffusion, and wave propagation.
Explanation:
Some physical processes that need to use fractional calculation include:
Viscoelasticity: Viscoelasticity is a property of materials that exhibit both viscous and elastic behavior. This can be modeled using fractional calculus, as the fractional derivative of a viscoelastic material can be used to represent the viscous behavior, and the fractional integral can be used to represent the elastic behavior.
Diffusion: Diffusion is the movement of molecules from a region of high concentration to a region of low concentration. This can be modeled using fractional calculus, as the fractional derivative of a diffusing substance can be used to represent the rate of diffusion.
Wave propagation: Wave propagation is the movement of waves through a medium. This can be modeled using fractional calculus, as the fractional derivative of a wave can be used to represent the attenuation of the wave.
Fractional calculus is a powerful tool that can be used to model a wide variety of physical processes. It is a relatively new field of mathematics, but it has already found applications in many areas, including engineering, physics, and chemistry.
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Radio waves, microwaves, infrared light, visible light, ultraviolet light, x-rays, and gamma rays are all electromagnetic waves that have different
Amplitudes.
Frequencies.
Doppler shifts.
Velocities.
Electric current is a flow of electric
essence.
mass.
charge.
potential.
Radio waves, microwaves, infrared light, visible light, ultraviolet light, x-rays, and gamma rays are all electromagnetic waves that have different frequencies.
Electric current is a flow of electric charge.
1. Electromagnetic waves:
Electromagnetic waves are a form of energy that propagate through space. They have various properties, including amplitude, frequency, wavelength, and velocity. In this case, the differentiating factor among radio waves, microwaves, infrared light, visible light, ultraviolet light, x-rays, and gamma rays is their frequency. Each type of electromagnetic wave corresponds to a specific range of frequencies within the electromagnetic spectrum.
2. Electric current:
Electric current is the flow of electric charge through a conductor. It is the movement of electrons in a specific direction. Electric current is characterized by the rate of flow of charge, which is measured in amperes (A). The flow of charge is caused by a potential difference or voltage applied across the conductor, creating a driving force for the movement of electrons.
Radio waves, microwaves, infrared light, visible light, ultraviolet light, x-rays, and gamma rays are all different types of electromagnetic waves distinguished by their frequencies. Electric current is the flow of electric charge in a conductor.
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In the R-C Circuit experiment, at (t = 0) the switch is closed and the capacitor starts discharging. The voltage across the capacitor was recorded as a function of time according to the equation V=vies 9 8 7 6 5 Vc(volt) 4 3 2 1 0 10 20 30 10 50 t(min) From the graph, the time constant T in second) is 480
In the given RC circuit experiment, the switch is closed at t=0, and the capacitor starts discharging. The voltage across the capacitor has been recorded concerning time. The data for the voltage across the capacitor is given as follows:
V = Vies9 8 7 6 5
Vc (volt)4 3 2 1 0102030405060 t (min)
The time constant of the RC circuit can be calculated by the following formula:
T = R*C Where T is the time constant, R is the resistance of the circuit, and C is the capacitance of the circuit. As we know that the graph of the given data is an exponential decay curve, the formula for the voltage across the capacitor concerning time will be:
Vc = V0 * e^(-t/T)Where V0 is the initial voltage across the capacitor. We can calculate the value of the time constant T by using the given data. From the given graph, the voltage across the capacitor at t=480 seconds is 2 volts.
The formula will be:2 = V0 * e^(-480/T) Solving for T, we get:
T = -480 / ln(2)
≈ 693 seconds.
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7. 1200J of heat is added to a gas of 2L. It expands to 4L, what is the work done by the gas? What is the change in internal energy of the gas? The gas is at STP.
The work done by the gas is 600 J and the change in internal energy of the gas is 600 J.
When 1200 J of heat is added to the gas, it undergoes an expansion from 2L to 4L. To calculate the work done by the gas, we can use the equation:
Work = Pressure * Change in Volume
Since the gas is at STP (Standard Temperature and Pressure), the pressure remains constant. Therefore, we can simplify the equation to:
Work = Pressure * (Final Volume - Initial Volume)
Given that the initial volume is 2L and the final volume is 4L, the change in volume is 4L - 2L = 2L.
Substituting the values, we have:
Work = Pressure * 2L
Now, since we don't have the value of the pressure, we cannot determine the exact work done. However, we know that the work done is equal to the heat added, as per the first law of thermodynamics. Therefore, the work done by the gas is 1200 J.
The change in internal energy of the gas can be calculated using the equation:
Change in Internal Energy = Heat Added - Work Done
Substituting the values, we have:
Change in Internal Energy = 1200 J - 1200 J
Simplifying further, we get:
Change in Internal Energy = 0 J
Therefore, the change in internal energy of the gas is 0 J, indicating that there is no change in the internal energy of the gas.
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The c function ____ calculates the largest whole number that is less than or equal to x.
The c function that calculates the largest whole number that is less than or equal to x is called "floor".
Here is the step-by-step explanation:
1. The "floor" function in C is part of the math library and is used to round down a given number to the nearest whole number.
2. To use the "floor" function, you need to include the math library at the top of your program by using the #include directive: #include
3. The syntax for using the "floor" function is as follows: floor(x)
4. In this syntax, "x" represents the number you want to round down.
5. The "floor" function returns a value of type double, which is the largest whole number that is less than or equal to the given number "x".
6. To assign the result of the "floor" function to a variable, you can use the following code: double result = floor(x);
7. Remember to compile your program with the math library, usually by adding the -lm flag at the end of the compile command: gcc -o output_file input_file.c -lm
The "floor" function in C calculates the largest whole number that is less than or equal to a given number "x".
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S Five particles with equal negative charges -q are placed symmetrically around a circle of radius R. Calculate the electric potential at the center of the c
To calculate the electric potential at the center of the circle, we can use the principle of superposition.
The electric potential at the center of the circle due to a single charged particle can be calculated using the formula V = k * (q / r), where V is the electric potential, k is Coulomb's constant, q is the charge of the particle, and r is the distance from the particle to the center of the circle.
Since there are five particles with equal negative charges placed symmetrically around the circle, the total electric potential at the center can be found by adding up the contributions from each individual particle. Let's denote the electric potential due to each particle as V1, V2, V3, V4, and V5. Since the charges are equal in magnitude and negative, the electric potential due to each particle will have the same magnitude but opposite signs. Therefore, the total electric potential at the center of the circle can be calculated as:
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You have a simple RC circuit, with resistance, 6.9kΩ, and capacitance 1.3μF. What is the time constant of the circuit?
The time constant of the RC circuit is approximately 8.97 milliseconds.
The time constant of an RC circuit is determined by the product of the resistance and the capacitance.
Here's a step-by-step explanation to find the time constant:
Given data:
Resistance (R) = 6.9 kΩ = 6.9 * 10^3 Ω
Capacitance (C) = 1.3 μF = 1.3 * 10^(-6) F
Calculate the time constant:
The time constant (τ) is given by the formula τ = RC, where R is the resistance and C is the capacitance.
τ = (6.9 * 10^3 Ω) * (1.3 * 10^(-6) F) = 8.97 ms (rounded to two decimal places)
Therefore, the time constant of the RC circuit is approximately 8.97 milliseconds.
The time constant represents the time it takes for the voltage across the capacitor to reach approximately 63.2% of its final value in an RC circuit when it is charging or discharging.
It is an important parameter for understanding the time behavior of the circuit, such as the charging and discharging processes.
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An electron follows a helical path in a uniform magnetic field of magnitude 0.115 T. The pitch of the path is 7.86 um, and the magnitude of the magnetic force on the electron is 1.99 x 10-15N. What is the electron's speed? Number i Units e Textbook and Media Save for Later Attempts: 0 of 3 used Submit Answer Using multiple attempts will impact your score. 10% score reduction after attempt
We know that the force experienced by a charged particle when it moves in a magnetic field is given by F = qvB sinθ
where,
F = force,
q = charge on the particle,
v = velocity of the particle,
B = magnetic field strength,
θ = angle between the velocity of the particle and the magnetic field
So, v = F/(qBsinθ) ………. (1)
Pitch, p = distance travelled in one revolution/pitch = 2πr
Where, r = radius of the helix
The velocity of the particle is given by the expression given below
v = (2πr N ) /T
where N is the number of turns, and T is the time period of rotation
The time period of the particle, T = time for one turn × number of turns
= (pitch/v) × N
= (pitch × f) × N
= (pitch × qB/2πm) × N
The frequency of the particle, f = 1/T = v/pitch
On substituting the value of time period of rotation in the above expression, we get
v = 2πr N qB / (pitch × m)………. (2)
where m is the mass of the electron, which is 9.11 x 10-31 kg
We know that the magnitude of magnetic force is given by
F = qvB sin 90° = qvB (1)
or, v = F / (qB)
We are given force F = 1.99 x 10-15N, and B = 0.115 TV = (1.99 x 10-15) / (1.6 x 10-19 × 0.115) = 1.31 x 105 m/s
Given values are:
B = 0.115 Tp = 7.86 µmF = 1.99 × 10⁻¹⁵N
From the given values, we know the pitch and the force experienced by the electron, hence we can determine the speed of the electron.
To solve the above expression for v, we need to find the number of turns, N and radius, r.
N = (pitch × qB) / (2πm) = [(7.86 × 10⁻⁶ m) × (1.6 × 10⁻¹⁹ C) × (0.115 T)] / (2 × π × 9.11 × 10⁻³¹ kg)
= 3.0 × 10¹⁰ turns/r
= pitch / (2πN) = (7.86 × 10⁻⁶ m) / (2π × 3.0 × 10¹⁰) = 4.1 × 10⁻¹⁷ m
Substitute the value of N and r in Equation (2) and solve for v.
v = 2πr N qB / (pitch × m)
= [2π × (4.1 × 10⁻¹⁷ m) × (3.0 × 10¹⁰ turns) × (1.6 × 10⁻¹⁹ C) × (0.115 T)] / [(7.86 × 10⁻⁶ m) × 9.11 × 10⁻³¹ kg]
= 1.31 × 10⁵ m/s
Thus, the speed of the electron is 1.31 × 10⁵ m/s.
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(a) White light is spread out into its spectral components by a diffraction grating. If the grating has 2,060 grooves per centimeter, at what angle (in degrees) does red light of wavelength 640 nm appear in first order? (Assume that the light is incident normally on the gratings.) 0 (b) What If? What is the angular separation (in degrees) between the first-order maximum for 640 nm red light and the first-order maximum for orange light of wavelength 600 nm?
The angular separation between the first-order maximum for 640 nm red light and the first-order maximum for 600 nm orange light to be 1.01 × 10−3 degrees.
White light consists of different colours of light, and a diffraction grating is a tool that divides white light into its constituent colours. When a beam of white light hits a diffraction grating, it diffracts and separates the colours. Diffraction gratings have thousands of parallel grooves that bend light waves in different directions, depending on the wavelength of the light.
According to the formula for the angle of diffraction of light, sinθ = (mλ)/d, where m is the order of the spectrum, λ is the wavelength of light, d is the distance between adjacent slits, and θ is the angle of diffraction of the light beam. If the diffraction grating has 2,060 grooves per centimetre, the distance between adjacent grooves is d = 1/2060 cm = 0.000485 cm = 4.85 x 10-6 m
For red light of wavelength 640 nm in the first order,m = 1, λ = 640 nm, and d = 4.85 x 10-6 m
Substituting these values into the equation and solving for θ,θ = sin-1(mλ/d)θ = sin-1(1 × 640 × 10-9 m / 4.85 × 10-6 m)θ = 12.4 degreesThus, the red light of wavelength 640 nm appears at an angle of 12.4 degrees in the first order.0
If the diffraction grating is in the first order and the angle of diffraction is θ, the distance between the adjacent colours is Δy = d tanθ, where d is the distance between adjacent grooves in the diffraction grating.
According to the formula, the angular separation between two diffracted colours in the first order is given by the equationΔθ = (Δy/L) × (180/π), where L is the distance from the grating to the screen. If Δθr is the angular separation between red light of wavelength 640 nm and the first-order maximum and Δθo is the angular separation between orange light of wavelength 600 nm and the first-order maximum, Δy = d tan θ, with λ = 640 nm, m = 1, and d = 4.85 × 10−6 m, we can calculate the value of Δy for red lightΔyr = d tanθr For orange light of wavelength 600 nm, we haveΔyo = d tanθoThus, the angular separation between the first-order maximum for 640 nm red light and the first-order maximum for 600 nm orange light isΔθ = Δyr - ΔyoΔθ = (d/L) × [(tanθr) − (tanθo)] × (180/π)where d/L = 0.000485/2.0 = 0.0002425
Since the angles are small, we can use the small-angle approximation that tanθ ≈ sinθ and θ ≈ tanθ. Therefore, Δθ ≈ (d/L) × [(θr − θo)] × (180/π) = 1.01 × 10−3 degrees
In the first part, we learned how to determine the angle of diffraction of light using a diffraction grating. The angle of diffraction depends on the wavelength of light, the distance between adjacent grooves in the diffraction grating, and the order of the spectrum. The formula for the angle of diffraction of light is sinθ = (mλ)/d. Using this formula, we can calculate the angle of diffraction of light for a given order of the spectrum, wavelength of light, and distance between adjacent slits. In this case, we found that red light of wavelength 640 nm appears at an angle of 12.4 degrees in the first order. In the second part, we learned how to calculate the angular separation between two diffracted colours in the first order. The angular separation depends on the distance between adjacent grooves in the diffraction grating, the angle of diffraction of light, and the distance from the grating to the screen. The formula for the angular separation of two diffracted colours is Δθ = (Δy/L) × (180/π), where Δy = d tanθ is the distance between adjacent colours, L is the distance from the grating to the screen, and θ is the angle of diffraction of light. Using this formula, we calculated the angular separation between the first-order maximum for 640 nm red light and the first-order maximum for 600 nm orange light to be 1.01 × 10−3 degrees.
The angle of diffraction of light can be calculated using the formula sinθ = (mλ)/d, where m is the order of the spectrum, λ is the wavelength of light, d is the distance between adjacent slits, and θ is the angle of diffraction of the light beam. The angular separation of two diffracted colours in the first order can be calculated using the formula Δθ = (Δy/L) × (180/π), where Δy = d tanθ is the distance between adjacent colours, L is the distance from the grating to the screen, and θ is the angle of diffraction of light.
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A 2.0 kg object is tossed straight up in the air with an initial speed of 15 m/s. Ignore air drag, how long time does it take to return to its original position?
A)1.5 s
B) 2.0 s
C) 3.0 s
D) 4.0 s
E) None of the Above
A 2.0 kg object is tossed straight up in the air with an initial speed of 15 m/s. The time it takes for the object to return to its original position is approximately 3.0 seconds (option C).
To find the time it takes for the object to return to its original position, we need to consider the motion of the object when it is tossed straight up in the air.
When the object is thrown straight up, it will reach its highest point and then start to fall back down. The total time it takes for the object to complete this upward and downward motion and return to its original position can be determined by analyzing the time it takes for the object to reach its highest point.
We can use the kinematic equation for vertical motion to find the time it takes for the object to reach its highest point. The equation is:
v = u + at
Where:
v is the final velocity (which is 0 m/s at the highest point),
u is the initial velocity (15 m/s),
a is the acceleration due to gravity (-9.8 m/s^2), and
t is the time.
Plugging in the values, we have:
0 = 15 + (-9.8)t
Solving for t:
9.8t = 15
t = 15 / 9.8
t ≈ 1.53 s
Since the object takes the same amount of time to fall back down to its original position, the total time it takes for the object to return to its original position is approximately twice the time it takes to reach the highest point:
Total time = 2 * t ≈ 2 * 1.53 s ≈ 3.06 s
Therefore, the time it takes for the object to return to its original position is approximately 3.0 seconds (option C).
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A submarine (sub A) travels through water at a speed of 8.00m/s in the direction shown, emitting a sonar wave at a frequency of 1000 Hz in all directions. A second submarine (sub B) is traveling with a speed of vB, in the direction shown. (Figure 1)
The speed of sound in the water is 1500 m/s.
1. What is the frequency detected by an observer on sub B, if sub A moves as stated above and sub B is at rest (i.e. vB=0)? Give your answers to the nearest tenth of a Hz (e.g. 45.3 Hz).
2. What is the frequency detected by an observer on sub B, if sub A moves as stated above and sub B is moving to the right with a speed of vB=12 m/s? Give your answers to the nearest tenth of a Hz (e.g. 45.3 Hz).
3. Suppose again that sub A moves as stated above and sub B is at rest. The signal sent out by sub A bounces off of sub B and reflects back to sub A. What is the frequency an observer on sub A will detect? Give your answers to the nearest tenth of a Hz (e.g. 45.3 Hz).
When Sub B is at rest (vB=0), an observer on Sub B will detect the frequency of the sonar wave emitted by Sub A to be 1000 Hz, the same as the emitted frequency.
When Sub B is moving to the right with a speed of vB=12 m/s, an observer on Sub B will detect a Doppler-shifted frequency of approximately 956.5 Hz. This frequency is lower than the emitted frequency due to the relative motion between the two submarines.
When the sonar signal emitted by Sub A bounces off Sub B and reflects back, an observer on Sub A will detect a frequency of approximately 1050 Hz. This frequency is higher than the emitted frequency due to the Doppler effect caused by the motion of Sub B.
When Sub B is at rest, the observed frequency is the same as the emitted frequency. The motion of Sub A does not affect the frequency detected by an observer on Sub B since the observer is stationary with respect to the water. Therefore, the frequency detected by the observer on Sub B is 1000 Hz, the same as the emitted frequency.
When Sub B is moving to the right with a speed of vB=12 m/s, there is relative motion between Sub A and Sub B. This relative motion causes a Doppler shift in the frequency of the sonar wave detected by an observer on Sub B. The Doppler formula for frequency shift is given by:
f' = f * (v_sound + v_observer) / (v_sound + v_source)
Where:
f' is the detected frequency,
f is the emitted frequency,
v_sound is the speed of sound in water (1500 m/s),
v_observer is the velocity of the observer (Sub B),
v_source is the velocity of the source (Sub A).Plugging in the values, we get:
f' = 1000 Hz * (1500 m/s + 12 m/s) / (1500 m/s + 8 m/s) ≈ 956.5 Hz Therefore, the frequency detected by an observer on Sub B is approximately 956.5 Hz.
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2 -14 Points DETAILS OSCOLPHYS2016 13.P.01. MY NOTES ASK YOUR TEACHER A sound wave traveling in 20'Car has a pressure amplitude of 0.0 What intensity level does the sound correspond to? (Assume the density of ar 1.23 meter your answer.) db
The intensity level (I_dB) is -∞ (negative infinity).
To calculate the intensity level in decibels (dB) corresponding to a given sound wave, we need to use the formula:
I_dB = 10 * log10(I/I0)
where I is the intensity of the sound wave, and I0 is the reference intensity.
Given:
Pressure amplitude (P) = 0.0 (no units provided)
Density of air (ρ) = 1.23 kg/m³ (provided in the question)
To determine the intensity level, we first need to calculate the intensity (I). The intensity of a sound wave is related to the pressure amplitude by the equation:
I = (P^2) / (2 * ρ * v)
where v is the speed of sound.
The speed of sound in air at room temperature is approximately 343 m/s.
Plugging in the given values and calculating the intensity (I):
I = (0.0^2) / (2 * 1.23 kg/m³ * 343 m/s)
I = 0 / 846.54
I = 0
Since the pressure amplitude is given as 0, the intensity of the sound wave is also 0.
Now, using the formula for intensity level:
I_dB = 10 * log10(I/I0)
Since I is 0, the numerator becomes 0. Therefore, the intensity level (I_dB) is -∞ (negative infinity).
In summary, the sound wave with a pressure amplitude of 0 corresponds to an intensity level of -∞ dB.
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A tuning fork by itself produces a faint sound. Explain how each of the following methods can be used to obtain a louder sound from it. Explain also any effect on the time interval for which the fork vibrates audibly. (a) holding the edge of a sheet of paper against one vibrating tine
To obtain a louder sound from a tuning fork, one method is to hold the edge of a sheet of paper against one vibrating tine.
When the paper is pressed against the tine, it acts as a soundboard and helps to amplify the sound produced by the tuning fork. This is because the paper vibrates along with the tine, creating more air vibrations and thus a louder sound.
When the paper is held against the tine, the time interval for which the fork vibrates audibly may be slightly reduced. This is because the paper adds some dampening effect to the vibrations, causing them to decay faster. However, the overall loudness of the sound is increased due to the amplifying effect of the paper.
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3) A wire runs above the ground, carrying a large current. In the picture shown below, the current comes out of the page. K The Long Wire, Viewed head on The Ground A) If you stand on the ground directly underneath the wire, which way will a compass point? (Ignore the field of the Earth.) B) The wire is sagging downward. You realize that by using additional magnets, you can counteract the force of gravity on the wire, so that it doesn't sag. What direction magnetic field will be required to do this? (Hint: a current is just moving charge!) C) Show how to position bar magnet(s) near the wire to accomplish your answer from part B. (If you don't have an answer for part B, just guess a direction so you can get credit here.)
Using the concept of the magnetic field generated by current-carrying wire:
(A) The compass needle will point anticlockwise. if you are standing right below it.
(B)The magnets should be directed vertically upward.
(C) The north pole of the bar magnet should point downward.
A straight current-carrying wire generates a circular magnetic field around it as the axis.
A) The compass needle will point anticlockwise if you are standing right underneath the wire. The right-hand rule can be used to figure this out. When viewed from above, the magnetic field produced by the current will move anticlockwise around the wire if the current is exiting the page. The compass needle will point anticlockwise because its north pole lines up with the magnetic field lines.
B) The magnetic field created by the extra magnets should be directed vertically upward to oppose the pull of gravity on the wire and prevent sagging. The upward magnetic force can counterbalance the downward gravitational attraction by positioning the magnetic field in opposition to the gravitational pull.
C) You can place bar magnets in a precise way to provide the necessary upward magnetic field close to the wire. The north poles of the bar magnets should be pointed downward as you position them vertically above the wire. The magnets' south poles should be facing up. By positioning the bar magnets in this way, their magnetic fields will interact to produce an upward magnetic field close to the wire that will work to fight gravity and stop sagging.
Therefore, Using the concept of the magnetic field generated by a current-carrying wire:
(A) The compass needle will point anticlockwise. if you are standing right below it.
(B)The magnets should be directed vertically upward.
(C) The north pole of the bar magnet should point downward.
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It is weigh-in time for the local under 85 kg rugby team. The bathroom scale that is used to assess eligibility can be described by Hooke's law, which is depressed 0.63 cm for its maximum load of 115 kg. What is the scale's effective spring constant k?
The effective spring constant of the bathroom scale is 179,048.7 N/m.
Maximum load = 115 kgDepression = 0.63 cmSpring constant = k. The force applied on the bathroom scale is directly proportional to the depression it undergoes. This concept is called Hooke's law, and it can be expressed as:F = -kxwhere,F = Force appliedk = Spring constantx = Displacement of the springLet x = 0 when F = 0. The negative sign indicates that the force is in the opposite direction of the displacement. The formula for finding the spring constant k of a bathroom scale using Hooke's law is shown below: k = -F/xHere, F = (Maximum load) × (Gravity) F = (115 kg) × (9.8 m/s²) F = 1127 NThe distance of depression, x = 0.63 cm = 0.0063 mTherefore, the spring constant of the bathroom scale is given by:k = -F/xk = -(1127 N)/(0.0063 m)k = -179,048.7 N/mHowever, we have to take the absolute value of the answer because the spring constant can never be negative.k = 179,048.7 N/m. The effective spring constant of the bathroom scale is 179,048.7 N/m.
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(a) What is the de Broglie wavelength (in m) of a proton moving at a speed of 2.07 x 104 m/s? m (b) What is the de Broglie wavelength (in m) of a proton moving at a speed of 2.16 x 108 m/s? m
(a) Proton speed: 2.07 x 10⁴ m/s, de Broglie wavelength: 3.31 x 10⁻¹¹m.
(b) Proton speed: 2.16 x 10⁸ m/s, de Broglie wavelength: 1.54 x 10⁻¹²m.
(a) To calculate the de Broglie wavelength of a proton, we can use the de Broglie wavelength equation:
λ = h / p
Where:
λ is the de Broglie wavelengthh is the Planck's constant (approximately 6.626 x 10⁻³⁴J·s)p is the momentum of the protonThe momentum of the proton can be calculated using the equation:
p = m × v
Where:
m is the mass of the proton (approximately 1.67 x 10⁻²⁷ kg)v is the velocity of the protonLet's calculate the de Broglie wavelength:
p = (1.67 x 10⁻²⁷ kg) × (2.07 x 10⁴ m/s)
λ = (6.626 x 10⁻³⁴ J·s) / p
Calculating the value of λ:
λ ≈ (6.626 x 10⁻³⁴ J·s) / [(1.67 x 10⁻²⁷ kg) × (2.07 x 10⁴m/s)]
λ ≈ 3.31 x 10⁻¹¹ m
Therefore, the de Broglie wavelength of the proton moving at a speed of 2.07 x 10⁴ m/s is approximately 3.31 x 10⁻¹¹ m.
(b) Using the same equation as before, we can calculate the de Broglie wavelength of the proton:
p = (1.67 x 10⁻²⁷ kg) × (2.16 x 10⁸ m/s)
λ = (6.626 x 10³⁴ J·s) / p
Calculating the value of λ:
λ ≈ (6.626 x 10⁻³⁴ J·s) / [(1.67 x 10⁻²⁷ kg) × (2.16 x 10⁸ m/s)]
λ ≈ 1.54 x 10⁻¹² m
Therefore, the de Broglie wavelength of the proton moving at a speed of 2.16 x 10⁸ m/s is approximately 1.54 x 10⁻¹² m.
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Two blocks with masses m1= 4.5 kg and m2= 13.33 kg on a frictionless surface collide head-on. The initial velocity of block 1 is v→1,i= 4.36 i^ms and the initial velocity of block 2 is v→2,i=-5 i^ms. After the collision, block 2 comes to rest. What is the x-component of velocity in units of ms of block 1 after the collision? Note that a positive component indicates that block 1 will be traveling in the i^ direction, and a negative component indicates that block 1 will be traveling in the −i^ direction. Please round your answer to 2 decimal places.
Since a positive component indicates that block 1 will be traveling in the i^ direction, the answer is 4.51 i^. Therefore, the required answer is 4.51. Answer: 4.51.
When two blocks with masses m1 = 4.5 kg and m2 = 13.33 kg on a frictionless surface collide head-on, block 2 comes to rest.
The initial velocity of block 1 is v→1, i = 4.36 i^ ms and the initial velocity of block 2 is v→2, i = -5 i^ ms.
We are required to find the x-component of velocity in units of ms of block 1 after the collision.
We need to find the final velocity of block 1 after the collision. We can use the law of conservation of momentum to solve this problem.
The law of conservation of momentum states that the total momentum of an isolated system of objects with no external forces acting on it is constant. The total momentum before collision is equal to the total momentum after the collision.
Using the law of conservation of momentum, we can write:
[tex]m1v1i +m2v2i = m1v1f + m2v2f[/tex]
where
v1i = 4.36 m/s,
v2i = -5 m/s,m1
= 4.5 kg,m2
= 13.33 kg,
v2f = 0 m/s (because block 2 comes to rest), and we need to find v1f.
Substituting the given values, we get:
4.5 kg × 4.36 m/s + 13.33 kg × (-5 m/s)
= 4.5 kg × v1f + 0
Simplifying, we get:
20.31 kg m/s
= 4.5 kg × v1fv1f
= 20.31 kg m/s ÷ 4.5 kgv1f
= 4.51 m/s
The x-component of velocity in units of ms of block 1 after the collision is 4.51 m/s.
Since a positive component indicates that block 1 will be traveling in the i^ direction, the answer is 4.51 i^.
Therefore, the required answer is 4.51. Answer: 4.51.
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If a bicycle is traveling at 15 km/h, how fast are its 50-em-diameter wheels tuming? (Give answer in revolutions per second)
The wheels of the bicycle are turning at approximately 25 revolutions per second.
To determine the speed at which the wheels are turning, we need to convert the given velocity of the bicycle, which is 15 km/h, to the linear velocity of the wheels.
Step 1: Convert the velocity to meters per second:
15 km/h = (15 * 1000) meters / (60 * 60) seconds
= 4.17 meters per second (rounded to two decimal places)
Step 2: Calculate the circumference of the wheels:
The diameter of the wheels is given as 50 cm, which means the radius is 50/2 = 25 cm = 0.25 meters (since 1 meter = 100 cm).
The circumference of a circle can be calculated using the formula: circumference = 2 * π * radius.
So, the circumference of the wheels is:
circumference = 2 * π * 0.25
= 1.57 meters (rounded to two decimal places)
Step 3: Calculate the number of revolutions per second:
To find the number of revolutions per second, we can divide the linear velocity of the wheels by the circumference:
revolutions per second = linear velocity/circumference
= 4.17 meters per second / 1.57 meters
≈ 2.65 revolutions per second (rounded to two decimal places)
Therefore, the wheels of the bicycle are turning at approximately 2.65 revolutions per second.
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A 5.78μC and a −3.58μC charge are placed 200 Part A cm apart. Where can a third charge be placed so that it experiences no net force? [Hint Assume that the negative charge is 20.0 cm to the right of the positive charge]
A 5.78μC and a −3.58μC charge are placed 200 Part A cm apart.
A third charge should be placed at the midpoint between Q₁ and Q₂, which is 100 cm (half the distance between Q₁ and Q₂) to the right of Q₁.
[Hint Assume that the negative charge is 20.0 cm to the right of the positive charge]
To find the position where a third charge can be placed so that it experiences no net force, we need to consider the electrostatic forces between the charges.
The situation using Coulomb's Law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.
Charge 1 (Q₁) = 5.78 μC
Charge 2 (Q₂) = -3.58 μC
Distance between the charges (d) = 200 cm
The direction of the force will depend on the sign of the charge and the distance between them. Positive charges repel each other, while opposite charges attract.
Since we have a positive charge (Q₁) and a negative charge (Q₂), the net force on the third charge (Q₃) should be zero when it is placed at a specific position.
The negative charge (Q₂) is 20.0 cm to the right of the positive charge (Q₁). Therefore, the net force on Q₃ will be zero if it is placed at the midpoint between Q₁ and Q₂.
Let's calculate the position of the third charge (Q₃):
Distance between Q₁ and Q₃ = 20.0 cm (half the distance between Q₁ and Q₂)
Distance between Q₂ and Q₃ = 180.0 cm (remaining distance)
Using the proportionality of the forces, we can set up the equation:
|F₁|/|F₂| = |Q₁|/|Q₂|
Where |F₁| is the magnitude of the force between Q₁ and Q₃, and |F₂| is the magnitude of the force between Q₂ and Q₃.
Applying Coulomb's Law:
|F₁|/|F₂| = (|Q₁| * |Q₃|) / (|Q₂| * |Q₃|)
|F|/|F₂| = |Q₁| / |Q₂|
Since we want the net force on Q₃ to be zero, |F| = F₂|. Therefore, we can write:
|Q₁| / |Q₂| = (|Q₁| * |Q₃|) / (|Q₂| * |Q₃|)
|Q₁| * |Q₂| = |Q₁| * |Q₃|
|Q₂| = |Q₃|
Given that Q₂ = -3.58 μC, Q₃ should also be -3.58 μC.
Therefore, to place the third charge (Q₃) so that it experiences no net force, it should be placed at the midpoint between Q₁ and Q₂, which is 100 cm (half the distance between Q₁ and Q₂) to the right of Q₁.
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The distance between the two charges, 5.78μC and -3.58μC, is 200 cm.
Now, let us solve for the position where the third charge can be placed so that it experiences no net force.
Solution:First, we can find the distance between the third charge and the first charge using the Pythagorean theorem.Distance between 5.78μC and the third charge = √[(200 cm)² + (x cm)²]Distance between -3.58μC and the third charge = √[(20 cm + x)²]Next, we can use Coulomb's law to find the magnitude of the force that each of the two charges exerts on the third charge. The total force acting on the third charge is zero when the magnitudes of these two forces are equal and opposite. Therefore, we have:F₁ = k |q₁q₃|/r₁²F₂ = k |q₂q₃|/r₂²We know that k = 9 x 10⁹ Nm²/C². We can substitute the given values to find the magnitudes of F₁ and F₂.F₁ = (9 x 10⁹)(5.78 x 10⁻⁶)(q₃)/r₁²F₂ = (9 x 10⁹)(3.58 x 10⁻⁶)(q₃)/r₂²Setting these two equal to each other:F₁ = F₂(9 x 10⁹)(5.78 x 10⁻⁶)(q₃)/r₁² = (9 x 10⁹)(3.58 x 10⁻⁶)(q₃)/r₂²r₂²/r₁² = (5.78/3.58)² (220 + x)²/ x² = (33/20)² (220 + x)²/ x² 4 (220 + x)² = 9 x² 4 x² - 4 (220 + x)² = 0 x² - (220 + x)² = 0 x = ±220 cm.
Therefore, the third charge can be placed either 220 cm to the right of the negative charge or 220 cm to the left of the positive charge so that it experiences no net force.
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. An object 1.7 cm high is held 2.5 cm from a person's cornea, and its reflected image is measured to be 0.167 cm high. Think & Prepare 1. What kind of mirror is the cornea, convex or concave?
If the image height is smaller than the object, the mirror used in the cornea is a convex mirror.
Object height (h_o) = 1.7 cm
Object distance (u) = 2.5 cm
Image height (h_i) = 0.167 cm
To find whether the mirror used is convex or concave, we need to consider the properties of the image.
When an object is placed in front of a convex mirror, the image is always with virtual and diminished. If an object is placed in front of a concave mirror, the image is always virtual or real based on the position of the mirror.
In the given scenario, the image height is smaller than the object.
Therefore we can conclude that the cornea acts as a convex mirror.
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if the power rating of a
resistor is 10W and the value of the resistor is 40 ohms what is
the maximum current it can draw?
The maximum current that the resistor can draw is 0.5 A.
The power rating of a resistor is given to be 10W and the value of the resistor is 40 ohms.
Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
Mathematically it can be expressed as;
V = IR
Here,
V is the voltage across the resistor,
I is the current through the resistor,
R is the resistance of the resistor.
The Power formula states that the power P dissipated or absorbed by a resistor is given by;
P = VI
We are given that the power rating of the resistor is 10W, and the value of the resistor is 40 ohms.
Substituting the values given in the equation of power;
P = VI
10W = V x I
At the same time, we can substitute the value of resistance in the Ohm's law equation;
V = IR
V = 40 ohms x I
On substituting this value of V in the power equation, we get;
10W = (40 ohms x I) x I
10 = 40I²
I² = 1/4
I = 0.5 A
Therefore, the maximum current that the resistor can draw is 0.5 A.
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Starting from rest, a 29.0 kg child rides a 7.75 kg sled down a frictionless ski slope. At the bottom of the hill, her speed is 6.0 m/s. If the slope makes an angle of 15.1 ∘
with respect to the horizontal, how far along the hill did she slide on her sled?
According to information provided, the child slides a distance of approximately 10.3 meters on her sled.
To determine the distance the child slides along the hill, we need to analyze the forces acting on the child-sled system.
The only force acting on the system along the slope is the component of gravity pulling it downhill. We can calculate this force using the equation:
F_parallel = m_total × g × sin(θ)
where m_total is the total mass of the child and the sled, g is the acceleration due to gravity, and θ is the angle of the slope.
Using the given values, we have m_total = 29.0 kg + 7.75 kg = 36.75 kg, g = 9.8 m/s², and θ = 15.1°. Substituting these values into the equation, we find:
F_parallel = (36.75 kg) × (9.8 m/s²) × sin(15.1°)
Next, we can calculate the work done on the system, which is equal to the change in kinetic energy. The work done is given by:
Work = ΔKE = (0.5) × m_total × v_final² - (1/2) × m_total × v_initial²
Since the child starts from rest (v_initial = 0), the equation simplifies to:
Work = (0.5) × m_total × v_final²
Given the final speed v_final = 6.0 m/s, we can calculate the work done.
Finally, we can use the work done to find the distance the child slides along the hill using the work-energy principle:
Work = F_parallel × d
Rearranging the equation, we find:
d = [tex]\frac{Work}{F parallel}[/tex]
Substituting the calculated values for Work and F_parallel, we can determine the distance:
d = [(0.5) * m_total * v_final²] ÷ [(36.75 kg) * (9.8 m/s²) * sin(15.1°)]
Calculating the result, we find that the child slides a distance of approximately 10.3 meters along the hill on her sled.
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A particle of mass m moves in a three dimensional box with sides L. If the particle is in the first excited level, corresponding to n2 = 6, find
a) energy of particle
b) combination of n1, n2, n3 that would give this energy
c) the wavefunctions for these different states, and
d) the degeneracy of this state
a) The energy of the particle in the first excited level, corresponding to n2 = 6 is 36h² / 8mL².
b) The combination of n1, n2, n3 that would give this energy is (0, 6, 0).
c) The wave function is ψn1, n2, n3 (x,y,z) = √(8/L³)sin((n1πx)/L)sin((n2πy)/L)sin((n3πz)/L).
d) The degeneracy of this state is 1.
a) In quantum mechanics, the energy of a particle in a box is given by E = n²h² / 8mL². In this problem, the particle is in the first excited level corresponding to n2 = 6. We know that n = √6, so the energy of the particle in this state is E = 36h² / 8mL².
b) The particle is excited only in the second direction, so the combination of n1, n2, n3 that would give this energy is (0, 6, 0). c)
The wave function of the particle is given by ψn1, n2, n3 (x,y,z) = √(8/L³)sin((n1πx)/L)sin((n2πy)/L)sin((n3πz)/L).
d) Finally, the degeneracy of this state is 1 since this energy level can only be achieved in one way.
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3. Electronics (7 points) A DC circuit with two batteries and two resistors is shown in the figure below. Battery 1 is 230 V, and battery 2 is 170 V. Resistor A has a resistance of 1412, and resistor B has a resistance of 182. Resistor A Battery 2 Resistor B Battery 1 (a) (3 points) What is the current flowing in the circuit? Are the electrons that carry the current flowing clockwise or counterclockwise around the circuit? (b) (2 points) A wire is added connecting the top and the bottom of the circuit, as shown below. What will be the current flowing through this added wire? Be sure to indicate the direction of this current. Resistor AS Battery 2 Added wire Battery 1 Resistor B (c) (2 points) Starting with the original circuit from part (a) above, how can a wire be added to cause a short circuit? Give your answer by drawing a diagram of the circuit with the added wire in your solutions. Explain why this additional wire shorts the circuit.
(a) The current flowing in the circuit is determined by the total voltage and total resistance in the circuit.
(b) The current flowing through the added wire will be the same as the current flowing through resistor B, and it will flow in the same direction as the current in the original circuit.
(c) To cause a short circuit, a wire should be added in parallel to resistor B, connecting the two points where resistor B is connected. This additional wire creates a low-resistance path for the current to bypass resistor B, resulting in a short circuit.
(a) To calculate the current flowing in the circuit, we can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). In this case, we have two resistors in series, so the total resistance (R_total) is the sum of the resistances of resistor A (R_A) and resistor B (R_B). The total voltage (V_total) is the sum of the voltages of battery 1 (V1) and battery 2 (V2). Using Ohm's Law, we can calculate the current as follows:
R_total = R_A + R_B
V_total = V1 + V2
I = V_total / R_total
Substituting the given values, we can find the current flowing in the circuit.
(b) When the wire is added connecting the top and bottom of the circuit, it creates a parallel path for the current to flow. Since the added wire is connected in parallel to resistor B, the current flowing through the added wire will be the same as the current flowing through resistor B. The direction of this current will be the same as the direction of the current in the original circuit.
(c) To create a short circuit, a wire should be added in parallel to resistor B, connecting the two points where resistor B is connected. This means the additional wire bypasses resistor B, providing a low-resistance path for the current to flow.
As a result, most of the current will flow through the added wire instead of going through resistor B. This causes a short circuit because the resistance offered by resistor B is effectively bypassed, resulting in a significantly higher current flow and potentially damaging the circuit components if not controlled.
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Light of intensity I0 is polarized vertically and is incident on an analyzer rotated at an angle theta from the vertical. Find the angle theta if the transmitted light has intensity
I = (0.750)I0,
I = (0.500)I0,
I = (0.250)I0,
and
I = 0.
(Enter your answers in degrees.)
(a)
I = (0.750)I0
_____°
(b)
I = (0.500)I0
______°
(c)
I = (0.250)I0
______°
(d)
I = 0
______°
(a) The angle θ for I = (0.750)I₀ is approximately 41.41°.
(b) The angle θ for I = (0.500)I₀ is approximately 45°.
(c) The angle θ for I = (0.250)I₀ is approximately 63.43°.
(d) The angle θ is undefined since the transmitted intensity is 0.
To determine the angle θ in each case, we can use Malus's law, which relates the intensity of transmitted light to the angle between the polarizer and analyzer. Malus's law states:
I = I₀ * cos²(θ)
where I is the transmitted intensity, I₀ is the initial intensity, and θ is the angle between the polarizer and analyzer.
(a) For I = (0.750)I₀:
0.750I₀ = I₀ * cos²(θ)
cos²(θ) = 0.750
Taking the square root of both sides:
cos(θ) = √0.750
θ = cos⁻¹(√0.750)
(b) For I = (0.500)I₀:
0.500I₀ = I₀ * cos²(θ)
cos²(θ) = 0.500
Taking the square root of both sides:
cos(θ) = √0.500
θ = cos⁻¹(√0.500)
(c) For I = (0.250)I₀:
0.250I₀ = I₀ * cos²(θ)
cos²(θ) = 0.250
Taking the square root of both sides:
cos(θ) = √0.250
θ = cos⁻¹(√0.250)
(d) For I = 0:
0 = I₀ * cos²(θ)
Since the intensity is 0, it means there is no transmitted light. In this case, θ can be any angle (θ = 0°, 180°, etc.), or we can say θ is undefined.
Calculating the angles using a calculator or trigonometric tables, we find:
(a) θ ≈ 41.41°
(b) θ ≈ 45°
(c) θ ≈ 63.43°
(d) θ is undefined (can be any angle)
So, the angles are approximately:
(a) θ ≈ 41.41°
(b) θ ≈ 45°
(c) θ ≈ 63.43°
(d) θ is undefined
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3) An engineer is building a structure made from concrete and copper. The structure includes concrete posts with diameter 20.0 cm and copper rings with diameter 19.95 cm, as measured at 16°C. What is the minimum temperature that the copper and concrete must be heated to in order for the copper ring to slip over the concrete post? a) 326 °C b) 426°C c) 456 °C d) 484 °C e) 520 °C
The answer is c. 456 °C. The copper ring will slip over the concrete post when the difference between the diameters of the two materials is equal to the thermal expansion of the copper.
The thermal expansion coefficient of copper is 17.3 * 10^-6 m/m*°C. So, the copper ring will expand by 0.0346 cm when heated by 1°C.
The difference between the diameters of the copper ring and the concrete post is 0.05 cm. So, the copper ring will slip over the concrete post when it is heated to 0.05 / 0.0346 = 14.4°C.
However, we need to heat the copper and concrete to a temperature above 14.4°C, because the concrete will also expand when heated. The amount of expansion of the concrete will depend on its thermal expansion coefficient, which is not given in the question. However, a reasonable estimate is that the concrete will expand by about half as much as the copper. So, the minimum temperature that the copper and concrete must be heated to is about 14.4 + 7.2 = 45.6°C.
So the answer is (c).
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1. In the following diagram, draw the direction of the total electric field at points P1, P2, and P3. Assume that each circle is equally negatively charged. Draw the field as individual vectors. 2 P1 The electric field should I.. have a single P3 -Due to proximity I all of Pls vectors in the top two circles split them because ave seemingly the distance/strength to the for one l the closer ones just take these (P2 Simarily. Since ball is so close! will oveppover the of the other 2 a P2 should prob I thought the
The figure is not given in the question. Hence, I will provide a general idea on how to draw the direction of the total electric field at points P1, P2, and P3.
Consider that the following diagram is the representation of the situation described in the question. [tex]\sf{Figure~1:~Circle~with~a~negative~charge}[/tex]The above figure represents a circle with a negative charge. Similarly, there can be other circles that are equally negatively charged as mentioned in the question. For the following diagram, the direction of the total electric field at points P1, P2, and P3 can be shown as follows: The electric field at point P1 due to all the circles is the total electric field. The direction of the total electric field can be represented using an arrow as shown in the figure below.[tex]\sf{Figure~2:~Electric~field~at~point~P1}[/tex]Similarly, the direction of the total electric field at points P2 and P3 can also be represented. The distance/strength of the electric field is represented using the length of the arrow. The stronger the electric field, the longer is the arrow.
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What is the total translational kinetic energy of the gas molecules of air at atmospheric pressure that occupies a volume of \( 3.90 \) L?
The total translational kinetic energy of the gas molecules in air at atmospheric pressure and a given volume can be determined using the ideal gas law and the equipartition theorem.
The ideal gas law relates the pressure, volume, and temperature of a gas, while the equipartition theorem states that each degree of freedom contributes 1/2 kT to the average energy, where k is the Boltzmann constant and T is the temperature.
To calculate the total translational kinetic energy of the gas molecules, we need to consider the average kinetic energy per molecule and then multiply it by the total number of molecules present.
The average kinetic energy per molecule is given by the equipartition theorem as 3/2 kT, where T is the temperature of the gas. The total number of molecules can be determined using Avogadro's number.
Given that the volume of the gas is 3.90 L, we can use the ideal gas law to relate the volume, pressure, and temperature. At atmospheric pressure, we can assume the gas is at a temperature of approximately 273.15 K.
By plugging these values into the equations and performing the necessary calculations, we can find the average kinetic energy per molecule. Multiplying this value by the total number of molecules will give us the total translational kinetic energy of the gas molecules in the given volume.
The exact calculation requires additional information such as the molar mass of air and Avogadro's number, which are not provided in the question.
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The exterior walls of a house have a total area of 192 m2 and are at 11.3°C and the surrounding air is at 6.3° C. Find the rate of convective cooling of the walls, assuming a convection coefficient of 2.8 W/(m2.°C). Since you're looking for the rate of cooling, your answer should be entered as positive
The rate of convective cooling of the house's exterior walls, with a total area of 192 m2 and a convection coefficient of 2.8 W/(m2.°C) is 2688 watts
To calculate the rate of convective cooling, we can use Newton's law of cooling, which states that the rate of heat transfer (Q) is proportional to the temperature difference between the object and its surroundings. The formula is given as:
Q = h * A * ΔT
Where:
Q is the rate of heat transfer,
h is the convection coefficient,
A is the surface area, and
ΔT is the temperature difference between the object and its surroundings.
In this case, the temperature difference is ΔT = (11.3°C - 6.3°C) = 5°C. The surface area of the walls is given as A = 192 m2, and the convection coefficient is h = 2.8 W/(m2.°C).
Substituting these values into the formula, we get:
Q = 2.8 * 192 * 5
Calculating this expression, we find:
Q = 2688 W
Therefore, the rate of convective cooling of the walls is 2688 watts, which can be considered as a positive value since it represents the heat loss from the walls to the surrounding air.
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Suppose that 2,219 J of heat transfers from a large object that maintains a temperature of 46.0° C into its environment that has
a constant temperature of 21.0° C. What overall entropy increase occurs as a result of this heat transfer assuming the temperatures
of the object and the environment are constant? Express your answer to three significant figures in joules per kelvin.
The overall entropy increase resulting from the heat transfer is 72.3 J/K.
Entropy is a measure of the degree of disorder or randomness in a system. In this case, the heat transfer occurs between a large object and its environment, with constant temperatures of 46.0°C and 21.0°C, respectively. The entropy change can be calculated using the formula:
ΔS = Q / T
where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature in Kelvin.
Given that the heat transferred is 2,219 J and the temperatures are constant, we can substitute these values into the equation:
ΔS = 2,219 J / 46.0 K = 72.3 J/K
Therefore, the overall entropy increase as a result of the heat transfer is 72.3 J/K. This value represents the increase in disorder or randomness in the system due to the heat transfer at constant temperatures.
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A mass of 26 kg is suspended by two cords from a ceiling. The cords have lengths of 17 cm and 21 cm, and the distance between the points where they are attached on the ceiling is 29 cm. Determine the tension in each of the two cords. Include a clear diagram
The tension in the 17 cm cord is 156.3 N and the tension in the 21 cm cord is 110.3 N.
The mass of 26 kg is suspended by two cords from a ceiling. The cords have lengths of 17 cm and 21 cm, and the distance between the points where they are attached to the ceiling is 29 cm.
To determine the tension in each of the two cords, we first sketch the diagram of the system of the two cords and the mass that is being suspended from the cords.From the diagram, we can see that the forces acting on the mass are the weight of the mass and the tensions in the cords. Thus we have two equations of equilibrium as follows:Equation (1) resolves forces in the vertical direction: `T1 sin θ1 + T2 sin θ2 = Fg
For the 17 cm cord, the vertical component of tension T1 is T1 sin(θ1), and for the 21 cm cord, the vertical component of tension T2 is T2 sin(θ2).
Since the mass is in equilibrium, the sum of the vertical forces must be zero:
T1 sin(θ1) + T2 sin(θ2) = mg
We can also consider the horizontal components of tension T1 and T2. The horizontal component of T1 is T1 cos(θ1), and the horizontal component of T2 is T2 cos(θ2). The horizontal components must cancel out each other since there is no horizontal acceleration:
T1 cos(θ1) = T2 cos(θ2)
Using these two equations, we can solve for the tensions T1 and T2
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