Answer:
Explanation:
Given that,
A capacitor of capacitance
C = 500pF
Charge on capacitor is
Q = 10μC
Capacitor is then connected to an inductor of inductance 10H
L = 10H
Since we want to calculate the maximum energy stored by the inductor, then, we will assume all the energy from the capacitor is transfer to the inductor
So energy stored in capacitor can be determined by using
U = ½CV²
Then, Q = CV
Therefore V = Q/C
U = ½ C • (Q/C)² = ½ C × Q²/C²
U = ½Q² / C
Then,
U = ½ × (10 × 10^-6)² / (500 × 10^-9)
U = 1 × 10^-4 J
U = 0.1 mJ
So the energy stored in this capacitor is transfers to the inductor.
So, energy stored in the inductor is 0.1mJ
B. Current through the inductor
Energy in the inductor is given as
U = ½Li²
1 × 10^-4 = ½ × 10 × i²
1 × 10^-4 = 5× i²
i² = 1 × 10^-4 / 5
i² = 2 × 10^-5
I = √(2×10^-5)
I = 4.47 × 10^-3 Amps
Then,
I = 4.47 mA
Understand Key Concepts
Which is NOT a fluid?
A. helium
B. ice
C. milk
D.Water
Answer:
The answer is A, friend.
How do you think air resistance affects measured values of g? If you used a ping pong ball, for example, how would this affect the fall-time? Would you expect the ping pong ball and the steel ball to land at the same time if they were both dropped from a height of ϭ.ϱm? What would happen if you dropped both balls from a height of ϮϬm?
Answer:
A) Air resistance acts in a direction opposite the the fall of an object reducing it by doing work against the weight of the object due to gravity.
B) using a ping pong ball, the time of fall will be greatly reduced since it has little weight (its mass x acceleration due to gravity) against the air resistance. The net downward force of the weight and the air resistance will be small.
C) No, I wouldn't expect them to fall at the same time. The steel ball will have more weight compared to the ping pong ball and hence it will have a larger net force downwards.
D) If they are both released from a 6 m height, the steel ball will fall to the ground first since it has a larger net force downwards.
The shaft is supported by a smooth thrust bearing at B and a journal bearing at C. Determine the resultant internal loadings acting on the cross section at E. Suppose that P1 = 360 lb , P2 = 830 lb . Part a, b, and c please.
Final answer:
To determine the resultant internal loadings at the cross section at E, we calculate the reaction forces at points B and C. The reaction force at B is 360 lb directed towards the left due to the thrust bearing, while the reaction force at C is 830 lb directed towards the right due to the journal bearing. Combining these forces, the resultant force at E is 1190 lb directed towards the right.
Explanation:
To determine the resultant internal loadings acting on the cross-section at E, we need to consider the forces acting on the shaft at points B and C. Part a requires determining the reaction force at B due to the thrust bearing, part b requires determining the reaction force at C due to the journal bearing, and part c requires combining the forces to find the resultant internal loadings at E.
For part a, since the shaft is supported by a smooth thrust bearing at B, the reaction force at B will be equal in magnitude and opposite in direction to the applied force P1. Therefore, the reaction force at B is 360 lb directed towards the left.
For part b, since the shaft is supported by a journal bearing at C, the reaction force at C will be equal in magnitude to the applied force P2. Therefore, the reaction force at C is 830 lb directed towards the right.
For part c, to find the resultant force at E, we need to combine the forces at B and C. Since the forces are along the same line of action, we can simply add the magnitudes of the forces. The resultant force at E is equal to the sum of the reaction forces at B and C, which is (360 lb) + (830 lb) = 1190 lb directed towards the right.
a 150 kg person stands on a compression spring with spring constant 10000 n/m and nominal length of 0.50 what is the total length
Answer:
The total length is 0.65m.
Explanation:
The total length [tex]x_{tot}[/tex] of the spring is equal to its length [tex]x_0[/tex] right now (0.50 m ) plus the amount [tex]x[/tex] by which it is compressed due to weight of the man:
[tex]x_{tot} = x_0+x[/tex]
The spring compression is given by Hooke's law:
[tex]F = -kx[/tex]
which in our case gives
[tex]-Mg = -kx[/tex]
solving for [tex]x[/tex] we get:
[tex]x= \dfrac{Mg}{k }[/tex]
putting in [tex]M = 150kg, g= 10m/s^2[/tex] and [tex]k = 10,000N/m[/tex] we get:
[tex]x= \dfrac{(150kg)(10m/s^2)}{10,000N/m^2}[/tex]
[tex]x = 0.15m[/tex]
Hence, the total length of the spring is
[tex]x_{tot} = 0.50m+0.15m[/tex]
[tex]\boxed{x_{tot} = 0.65m.}[/tex]
(6 points) A cylinder has a radius of 12 mm and a length of 2.5 m. It is made of steel with a thermal conductivity of LaTeX: 53.4 \frac{\mathrm{W}}{\mathrm{m K}}53.4 W m K. One end of the cylinder is held at a temperature of LaTeX: 361^{\circ}\text{C}361 ∘ C, the other at LaTeX: 105^{\circ}\text{C}105 ∘ C. In steady state how much power will flow through the cylinder in the form of heat?
Answer:
2.46 W
Explanation:
Thermal conductivity k = 53.4 W/m-K
Radius r = 12 mm = 12x10^-3 m
Lenght = 2.5 m
T1 = 361 °C
T2 = 105 ∘C
Area A = ¶r^2 = 3.142 x (12x10^-3)^2
= 0.00045 m^2
Power P = -AkdT/dx
Where dT = 361 - 105 = 256
dx = lenght of heat travel = 2.5 m
P = -0.00045 X 53.4 X (256/2.5)
= -0.024 X 102.4 = -2.46 W
The negative sign indicates that heat is lost from the metal.
To introduce you to the concept of escape velocity for a rocket. The escape velocity is defined to be the minimum speed with which an object of mass mmm must move to escape from the gravitational attraction of a much larger body, such as a planet of total mass MMM. The escape velocity is a function of the distance of the object from the center of the planet RRR, but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question "How fast does my rocket have to go to escape from the surface of the planet?"
Answer:
11.206 km/s
Explanation: to leave the surface of the earth, you velocity Ve must be;
Escape velocity Ve = (2gRe)^0.5
Where g = acceleration due to gravity 9.81 m/s^2
Re = earth's radius 6400 km = 6.4x10^6
Ve = (2 x 9.81 x 6.4 x 10^6)^0.5
Ve = (125568000)^0.5
Ve = 11205.7 m/s
= 11.206 km/s
Review the multiple-concept example as an aid in solving this problem. In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so that the ball in her hand moves on a circle. In one instance, the radius of the circle is 0.685 m. At one point on this circle, the ball has an angular acceleration of 63.8 rad/s2 and an angular speed of 15.0 rad/s. (a) Find the magnitude of the total acceleration (centripetal plus tangential) of the ball.
Answer:
Total acceleration will be [tex]160.20rad/sec^2[/tex]
Explanation:
We have given radius of the circle r = 0.685 m
Angular acceleration [tex]\alpha =63.8rad/sec^2[/tex]
Angular speed [tex]\omega =15rad/sec[/tex]
Centripetal acceleration will be
[tex]a_c=\omega ^2r=15^2\times 0.685=154.125rad/sec^2[/tex]
Tangential acceleration will be
[tex]a_t=r\alpha =0.685\times 63.8=43.7rad/sec^2[/tex]
(a) Total acceleration will be equal to
[tex]a=\sqrt{a_t^2+a_c^2}[/tex]
[tex]a=\sqrt{43.7^2+154.125^2}=160.20rad/sec^2[/tex]
So total acceleration will be [tex]160.20rad/sec^2[/tex]
Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, how much kinetic energy (J) has the electron gained
Answer:
[tex]{\rm K} = 2.4\times 10^{-19}~J[/tex]
Explanation:
The electric field inside a parallel plate capacitor is
[tex]E = \frac{Q}{2\epsilon_0 A}[/tex]
where A is the area of one of the plates, and Q is the charge on the capacitor.
The electric force on the electron is
[tex]F = qE = \frac{qQ}{2\epsilon_0 A}[/tex]
where q is the charge of the electron.
By definition the capacitance of the capacitor is given by
[tex]C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60[/tex]
Plugging this identity into the force equation above gives
[tex]F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q[/tex]
The work done by this force is equal to change in kinetic energy.
W = Fx = (30q)(0.05) = 1.5q = K
The charge of the electron is [tex]1.6 \times 10^{-19}[/tex]
Therefore, the kinetic energy is [tex]2.4\times 10^{-19}[/tex]
This problem has been solved! See the answer A proton with a speed of 3.5x10^6 m/s is shot into a region between two plates that are separated by distance of 0.23 m. A magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field so the proton just misses colliding with the opposite plate?
Answer:
The magnitude of the magnetic field 'B' is 0.16 Tesla.
Explanation:
The magnitude of the magnetic field 'B' can be determined by;
B = [tex]\frac{mV}{qR}[/tex]
where: m is the mass of proton, V is its speed , q is the charge of proton and R is the distance between the plates.
Given that: speed 'V' of the proton = 3.5 × [tex]10^{6} ms^{-2}[/tex], distance 'R' between the plates = 0.23m, the charge 'q' on proton = 1.9 × [tex]10^{-19}[/tex] C and mass of proton = 1.67 × [tex]10^{-27}[/tex]Kg.
Thus,
B = (1.67 × [tex]10^{-27}[/tex] × 3.5 ×[tex]10^{6}[/tex]) ÷ (1.6 × [tex]10^{-19}[/tex] × 0.23)
= [tex]\frac{5.845 * 10^{-21} }{3.68 * 10^{-20} }[/tex]
= 0.15883
B = 0.16 Tesla
The magnitude of the magnetic field 'B' is 0.16 Tesla.
Suppose that the inverse market demand for silicone replacement tips for Sony earbud headphones is p = pN - 0.1Q, where p is the price per pair of replacement tips, pN is the price of a new pair of headphones, and Q is the number of tips per week. Suppose that the inverse supply function of the replacement tips is p = 2 + 0.012 Q. a. The effect of a change in the price of a new pair of headphones on the equilibrium price of replacement tips ( dp/dpN) at the equilibrium is given by nothing (Round your answer to three places.)
The effect of a change in the price of new Sony headphones on the equilibrium price of silicone replacement tips can't be accurately calculated from the provided equations without numerical values. An increase in the price of new headphones would likely lead to a surge in demand for replacement parts, thereby increasing their equilibrium price.
Explanation:The student is asking about the impact of a change in the price of the new Sony headphones on the equilibrium price of replacement tips. The inverse demand function is given by p = pN - 0.1Q, and the inverse supply function is given by p = 2 + 0.012Q. The equilibrium condition is that the quantity demanded equals the quantity supplied, so we set the demand equation equal to the supply equation and solve for Q and p. Now to find the effect of a change in the price of the new headphones on the equilibrium price of the replacement tips, we need to differentiate the equilibrium price with respect to the price of the new headphones. This essentially involves implicit differentiation of the equilibrium condition.
However, due to the nature of the equations, getting a definitive value for dp/dpN would be difficult without numerical values to plug in. As such, the phrase 'dp/dpN is given by nothing' may have been a typographical error or misunderstanding. An increase in the price of new headphones would likely increase demand for replacement parts, thus raising their equilibrium price. But we cannot compute dp/dpN definitively from the provided equations.
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A current of 6 A runs through a coffee machine connected to a 120 V circuit. What is the resistance of the coffee machine?
Answer:
20 ohm
Explanation:
V = I x R
R = V/ I
= 120/6
R = 20 ohm
Considering the Ohm's law, the resistance of the coffee machine is 20 Ω.
Definition of currentThe current (I) is a measure of the speed at which the charge passes a given reference point in a specified direction. Its unit of measure is amps (A).
Definition of voltageThe driving force (electrical pressure) behind the flow of a current is known as voltage and is measured in volts (V). That is, voltage is a measure of the work required to move a charge from one point to another.
Definition of resistanceResistance (R) is the difficulty that a circuit opposes to the flow of a current and it is measured in ohms (Ω).
Ohm's lawOhm's law establishes the relationship between current, voltage, and resistance in an electrical circuit.
This law establishes that the intensity of the current that passes through a circuit is directly proportional to the voltage of the same and inversely proportional to the resistance that it presents:
I= V÷R
Where:
I is the current measured in amps (A).V the voltage measured in volts (V).R the resistance that is measured in ohms (Ω).Resistance of the coffee machineIn this case, you know that:
The voltage is 120 V.The current through the coffee machine is 6 A.Replacing in the Ohm's Law:
6 A= 120 V÷R
Solving:
6 A× R= 120 V
R= 120 V÷6 A
R= 20 Ω
Finally, the resistance of the coffee machine is 20 Ω.
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Which of the following careers would require a degree in physics?
A. Art historian
B. War correspondent
C. Nuclear engineer
D. X-ray technician
Answer:
Nuclear engineers require a degree in physics in order to have the job.
Explanation:
A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, the light takes 8.72 ns to travel the length of the tube, but when the tube is filled with a transparent jelly, the light takes 1.82 ns longer to travel its length. What is the refractive index of this jelly?
Answer:
Explanation:
velocity of light in a medium of refractive index V = V₀ / μ
V₀ is velocity of light in air and μ is refractive index of light.
time to travel in tube with air = length of tube / velocity of light
8.72 ns = L / V₀ L is length of tube .
time to travel in tube with jelly = length of tube / velocity of light
8.72+ 1.82 = L / V L is length of tube .
10.54 ns = L / V
dividing the equations
10.54 / 8.72 = V₀ / V
10.54 / 8.72 = μ
1.21 = μ
refractive index of jelly = 1.21 .
21. (a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500-μC charge and flies due west at a speed of 660. m/s over Earth’s south magnetic pole, where the 8.00 × 10−5 − T magnetic field points straight down into the ground. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.
Answer:
Explanation:
The picture attached shows the full explanation and i hope it helps. Thank you
Final answer:
The magnetic force on a supersonic jet with a 0.500-μC static charge, flying over the Earth's south magnetic pole at 660 m/s, is 7.50 × 10-7 N, directed to the north. This effect is considered negligible given the small magnitude of the force.
Explanation:
When an aircraft with a static charge flies through the Earth's magnetic field, it experiences a magnetic force due to the interaction between its charge, its velocity, and the magnetic field. This is a concept from electromagnetism, specifically the Lorentz force. The magnitude of the magnetic force (F) can be calculated using the equation F = qvBsin(θ), where q is the charge on the aircraft, v is the speed of the aircraft, B is the magnetic field strength, and θ is the angle between the velocity of the charge and the direction of the magnetic field.
For a supersonic jet with a 0.500-μC (or 0.500x10-6 C) charge flying due west over the Earth's south magnetic pole, where the magnetic field (B) is 8.00 × 10-5 T and points straight down (θ = 90 degrees), the magnitude of the magnetic force is:
F = (0.500x10-6 C)(660 m/s)(8.00 × 10-5 T)sin(90 degrees) = 7.50 × 10-7 N, perpendicular to both the magnetic field lines and the velocity of the jet (which means it points either north or south). Since the jet is flying due west and the magnetic field is down, the right-hand rule indicates the force will push the jet to the north.
Regarding part (b), since the magnetic force magnitude, 7.50 × 10-7 N, is quite small compared to the typical forces experienced by a supersonic jet, such as thrust, drag, and lift, it can be considered a negligible effect. It's unlikely to have any significant impact on the flight of the aircraft.
A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 2.00 keV and a current of 5.05 mA produced by the generator. (a) What is the speed of the protons (in m/s)? b) How many protons are produced each second?
Answer:
v = 1.95*10^5 m/s
3.13 x 10^16 proton
Explanation:
Identify the unknown:
The speed of the protons
List the Knowns:
Energy of protons KE = 2 keV = 2*10^3 eV
Current produced by the generator: I= 5.05 mA = 5 x 10^-3 A
1 eV = 1.6 x 10-19 Joule
Mass of proton: m = 1.67 x 10^-27 kg
Charge of proton: q_p = 1.6 x 10^-19 C
Set Up the Problem:
Kinetic energy is given by:
KE= 1/2mv^2
v = √2KE/m
Solve the Problem:
v = √2 x 2*10^3 x 1.6 x 10^-19/ 1.67 x 10^-27
v = 1.95*10^5 m/s
b. Identify the unknown:
Number of protons produced each second
Set Up the Problem:
Current is given by:
I =ΔQ/Δt
So, the total charge in one second:
ΔQ =I*Δt = 5 x 10^-3 x 1 = 5 x 10^-3 C
Number of protons in this charge:
n = ΔQ/q_p
Solve the Problem:
n = 5 x 10^-3/1.6 x 10^-19
=3.13 x 10^16 proton
An open container holds ice of mass 0.570 kg at a temperature of -17.2 ∘C . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 820 J/minute . The specific heat of ice to is 2100 J/kg⋅K and the heat of fusion for ice is 334×103J/kg.
The time required to raise the temperature of the ice from -17.2 °C to 0 °C is approximately 6.22 minutes.
Explanation:To calculate the time required for the temperature of the ice to rise from -17.2 °C to 0 °C, we can use the equation Q=mcΔT, where Q is the heat supplied, m is the mass of the ice, c is the specific heat of ice, and ΔT is the change in temperature.
The heat supplied (Q) during this process is used to raise the temperature of the ice without causing a phase change. Given that Q=820J/min and c=2100J/kg⋅K, we rearrange the equation to solve for time (t): t=Q/mcΔT
Substituting the given values (m=0.570kg, ΔT=0°C−(−17.2°C)=17.2°C), we find t≈820J/min/ (0.570kg)(2100J/kg⋅K)(17.2°C) ≈6.22min.
This calculation assumes that all the heat supplied is used to raise the temperature of the ice without considering the heat required for a phase change. To account for the heat of fusion during the phase change from ice to water at 0 °C, an additional calculation would be needed. However, the question specifies raising the temperature to 0 °C, so the heat of fusion is not considered in this particular scenario.
Understanding the principles of heat transfer, specific heat, and phase changes is essential in solving thermodynamics problems. These principles are widely applicable in various scientific and engineering fields, providing insights into the behavior of materials under different conditions.
A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes it difficult for the physicist to read books and journals. Find the focal length and power of a lens that will correct this presbyopia for a reading distance of 25 cm, when worn 2 cm in front of the eye
The issue of farsightedness or presbyopia can be corrected by a converging lens with a focal length of 34 cm and power of 2.94 Diopters.
Explanation:The provided problem can be solved using the lens formula which is 1/f = 1/v - 1/u, where f = focal length, v = image distance, and u = object distance. The image distance (di) for clear vision should match the lens-to-retina distance, which is constant and taken as 2 cm for the given question. The lens that the physicist is recommended to use should create an image with a reading distance of 25 cm and therefore this will be ‘v’. However, the lens is worn 2 cm in front of the eye and hence the actual image distance becomes 27 cm (or 0.27 m). The object is however at a distance of 97 cm (0.97 m) as indicated by the problem. When we plug these values into the lens formula, we get f = 1 / {(1/0.27) - (1/0.97)} which is approximately equal to 0.34 m or 34 cm. The optical power of lens (P) is reciprocal of the focal length in meters. Hence P = 1/f = 1/0.34 = 2.94 D.
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To correct the physicist's presbyopia for clear vision at 25 cm, a lens with a focal length of approximately -30.35 cm (or -0.3035 m) and a power of around -3.29 diopters is needed. The lens formula helped in determining these values.
Correcting Farsightedness
A presbyopic eye has difficulty seeing objects clearly at close range. In this case, the physicist's eye can only see clearly beyond 97 cm. To correct this and allow clear vision at 25 cm, we need to determine the focal length and power of the corrective lens when worn 2 cm in front of the eye.
Using the lens formula:
1/f = 1/v - 1/u
where:
f is the focal length of the lensv is the image distance (97 cm minus 2 cm = 95 cm)u is the object distance (25 cm minus 2 cm = 23 cm)Substituting the values, we get:
1/f = 1/95 - 1/23
1/f = (23 - 95) / (95 * 23)
1/f = -72 / 2185
f = -2185 / 72 ≈ -30.35 cm
Therefore, the focal length of the lens is approximately -30.35 cm (or -0.3035 m).
The power of the lens (P) is given by the formula:
P = 1/f (in meters)
Substituting the focal length:
P = 1 / -0.3035 ≈ -3.29 diopters (D)
Hence, the power of the corrective lens needed is approximately -3.29 D.
To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^v→=(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field B⃗ =(1.63T)i^+(0.980T)j^B→=(1.63T)i^+(0.980T)j^?
Answer:
The magnitude of the acceleration is [tex]a = 0.33 m/s^2[/tex]
The direction is [tex]- \r k[/tex] i.e the negative direction of the z-axis
Explanation:
From the question we are that
The mass of the particle [tex]m = 1.8*10^{-3} kg[/tex]
The charge on the particle is [tex]q = 1.22*10^{-8}C[/tex]
The velocity is [tex]\= v = (3.0*10^4 m/s ) j[/tex]
The the magnetic field is [tex]\= B = (1.63T)\r i + (0.980T) \r j[/tex]
The charge experienced a force which is mathematically represented as
[tex]F = q (\= v * \= B)[/tex]
Substituting value
[tex]F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)[/tex]
[tex]= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))[/tex]
[tex]= (-5.966*10^4 N) \r k[/tex]
Note :
[tex]i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\[/tex]
Now force is also mathematically represented as
[tex]F = ma[/tex]
Making a the subject
[tex]a = \frac{F}{m}[/tex]
Substituting values
[tex]a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}[/tex]
[tex]= (-0.33m/s^2)\r k[/tex]
[tex]= 0.33m/s^2 * (- \r k)[/tex]
A bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes. What is the smallest insect a bat can detect? Assume that bats emit a chirp at a frequency of 47.6 kHz, and that the speed of sound in air is 413 m/s. Answer in units of mm.
Given that,
Frequency emitted by the bat, f = 47.6 kHz
The speed off sound in air, v = 413 m/s
We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :
[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{413}{47.6\times 10^3}\\\\\lambda=0.00867\ m[/tex]
or
[tex]\lambda=8.67\ mm[/tex]
So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.
Which of the following correctly describes the result of a voltage difference?
A. Charges more randomly
B. Charges not to move
C. Charges flow from low voltage areas to high voltage areas.
D. Charges flow from high voltage areas to low voltage areas.
Answer:
D from high to low areas
Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of area density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun?
The critical area density for a mirror in space where radiation pressure cancels out gravitational attraction can be computed using the principles of radiation pressure and gravitational force. By equating the momentum transfer from reflected sunlight to the gravitational pull of the Sun, one can find the area density at which these forces balance.
The question posed is about finding the critical area density for a mirror located in space at a distance R from the Sun, where the radiation pressure from sunlight would balance the gravitational attraction exerted by the Sun. The solution to this problem involves using the principle of momentum transfer from sunlight and equating it to the gravitational force to derive the density at which the forces are balanced.
The radiation pressure P exerted by sunlight can be calculated using the formula P = 2I/c, where I is the intensity of sunlight and c is the speed of light. Given that sunlight above Earth's atmosphere has an intensity of 1.30 kW/m², the radiation pressure P would be twice this value divided by the speed of light, due to the reflection phenomenon (momentum is doubled as the direction is reversed).
On the other hand, the gravitational force acting on an object is given by the formula F = GMm/R², where G is the gravitational constant, M is the mass of the Sun, m is the mass of the object (or spacecraft with the mirror), and R is the distance from the Sun. The mass m can be represented in terms of the area density ρA, where ρ is the area density and A is the area of the mirror.
Setting the radiation pressure equal to the gravitational force and solving for ρ will yield the critical value of area density at which the two forces cancel out.
The critical value of area density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun is:
[tex]\[ \sigma = \frac{P_{sun}}{4\pi G M_{sun} R^2 c} \][/tex]
To find the critical value of area density (σ) for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun, we need to set the radiation pressure equal to the gravitational force per unit area.
Let's denote the radiation pressure as [tex]\( P_{rad} \)[/tex] and the gravitational force per unit area as [tex]\( P_{grav} \)[/tex]
The radiation pressure [tex]\( P_{rad} \)[/tex] can be calculated using the formula:
[tex]\[ P_{rad} = \frac{I}{c} \][/tex]
where [tex]\( I \)[/tex] is the intensity of the sunlight at the mirror's position and [tex]\( c \)[/tex] is the speed of light.
The intensity [tex]\( I \)[/tex] at a distance [tex]\( R \)[/tex] from the sun can be found using the inverse square law:
[tex]\[ I = \frac{P_{sun}}{4\pi R^2} \][/tex]
where [tex]\( P_{sun} \)[/tex] is the total power output of the sun.
The gravitational force per unit area [tex]\( P_{grav} \)[/tex] is given by:
[tex]\[ P_{grav} = \frac{F_{grav}}{A} = \frac{G M_{sun} m_{mirror} / R^2}{A} \][/tex]
where [tex]\( G \)[/tex] is the gravitational constant, [tex]\( M_{sun} \)[/tex] is the mass of the sun, [tex]\( m_{mirror} \)[/tex] is the mass of the mirror, and [tex]\( A \)[/tex] is the area of the mirror.
Since [tex]\( m_{mirror} = \sigma A \), where \( \sigma \)[/tex] is the area density of the mirror, we can write:
[tex]\[ P_{grav} = \frac{G M_{sun} \sigma A / R^2}{A} = \frac{G M_{sun} \sigma}{R^2} \][/tex]
Now, we set [tex]\( P_{rad} = P_{grav} \)[/tex] to find the critical value of [tex]\( \sigma \)[/tex]:
[tex]\[ \frac{I}{c} = \frac{G M_{sun} \sigma}{R^2} \][/tex]
Substituting [tex]\( I \)[/tex] from the intensity equation, we get:
[tex]\[ \frac{P_{sun}}{4\pi R^2 c} = \frac{G M_{sun} \sigma}{R^2} \][/tex]
Solving for [tex]\( \sigma \)[/tex], we find:
[tex]\[ \sigma = \frac{P_{sun}}{4\pi R^2 c} \cdot \frac{1}{G M_{sun}} \][/tex]
[tex]\[ \sigma = \frac{P_{sun}}{4\pi G M_{sun} R^2 c} \][/tex]
Complete question:- Suppose that the mirror described in Part
An inversion in a permutation of the integers 1 to n is a pair of numbers (not necessarily adjacent) such that the larger number is listed first. For example, in the permutation 4, 2, 3, 1, the inverted pairs are (4, 2), (4, 3), (4, 1) (2, 1) and (3, 1). By listing out all 24 permutations and counting the number of inversions in each (if you are lazy you can write a program to do this and attach the code as a separate file), calculate the expected number of inversions in a random permutation of 1, 2, 3 and 4. Then, using this result, posit a guess for the general result, in terms of n for permutations of 1, 2, 3, …, n. Try to prove this guess via a route that uses less calculation, but looks at an arbitrary pair of indexes into the permutation, say i and j with i < j and counts how many permutations for which this pair is "in order" and that this pair is in inverted.
Answer:
The total permutation = nP2
The inversions = nC2
Explanation:
Please look at the solution in the attached Word file
There are two categories of ultraviolet light. Ultraviolet A ( UVA ) has a wavelength ranging from 320 nm to 400 nm . It is not so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 nm and 320 nm , is much more dangerous, because it causes skin cancer.
Find the frequency ranges of UVA.
Enter your answers separated with commas.
Answer:
[tex]7.5x10^{14}Hz[/tex], [tex]9.37x10^{14}Hz[/tex].
Explanation:
The ultraviolet light belongs to the electromagnetic spectrum.
The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.
Any radiation from the electromagnetic spectrum has a speed of [tex]3x10^{8}m/s[/tex] in vacuum.
Therefore, in order to know the frequency the following equation can be used:
[tex]c = \nu \cdot \lambda[/tex] (1)
[tex]\nu = \frac{c}{\lambda}[/tex] (2)
Where [tex]\nu[/tex] is frequency, c is the speed of light and [tex]\lambda[/tex] is the wavelength.
Notice that it is necessary to express the wavelength in units of meters before it can be used in equation 2.
[tex]\lambda = 320nm . \frac{1m}{1x10^{9}nm}[/tex] ⇒ [tex]3.2x10^{-7}m[/tex]
[tex]\nu = \frac{3x10^{8}m/s}{3.2x10^{-7}m}[/tex]
[tex]\nu = 9.37x10^{14}s^{-1}[/tex]
But [tex]1Hz = s^{-1}[/tex]
[tex]\nu = 9.37x10^{14}Hz[/tex]
[tex]\lambda = 400nm . \frac{1m}{1x10^{9}nm}[/tex] ⇒ [tex]4x10^{-7}m[/tex]
[tex]\nu = \frac{3x10^{8}m/s}{4x10^{-7}m}[/tex]
[tex]\nu = 7.5x10^{14}Hz[/tex]
Hence, the frequency range of UVA is [tex]7.5x10^{14}Hz[/tex] to [tex]9.37x10^{14}Hz[/tex].
A capacitor consists of a set of two parallel plates of area A separated by a distance d . This capacitor is connected to a battery that maintains a constant potential difference V across the plates. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be
Answer:
if we double the distance the energy stored will be doubled also
Explanation:
The energy stored in a capacitor is given as
Energy stored =1/2(cv²)
Or
= 1/2(Qv)
Where c = capacitance
Q= charge
But the electric field is expressed as
E= v/d
where v= voltage
d= distance
v=Ed
Substituting into any equation above say
Energy stored =1/2(Qv)
Substituting v=Ev
Energy stored =1/2(QEd)
From the equation above it shows that if we double the distance The energy stored will be doubled also
To understand the formula representing a traveling electromagnetic wave.
Light, radiant heat (infrared radiation), X rays, and radio waves are all examples of traveling electromagnetic waves. Electromagnetic waves comprise combinations of electric and magnetic fields that are mutually compatible in the sense that the changes in one generate the other.
The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given by
E? =E0sin(kx??t)j^,
B? =B0sin(kx??t)k^.
This wave is linearly polarized in the y direction.
1.a. In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0 and B0are the __________ of the electric and magnetic fields.
Choose the best answer to fill in the blank.
1. maxima
2. amplitudes
3. wavelengths
4. velocities
b. The variable ? is called the __________ of the wave.
Choose the best answer to fill in the blank.
1. velocity
2. angular frequency
3. wavelength
c. The variable k is called the __________ of the wave.
Choose the best answer to fill in the blank.
1. wavenumber
2. wavelength
3. velocity
4. frequency
d. What is the mathematical expression for the electric field at the point x=0,y=0,z at time t?
1. E=E0sin(??t)j^
2. E =E0sin(??t)k^
3. E =0
4. E =E0sin(kz??t)i^
5. E =E0sin(kz??t)j^
e. For a given wave, what are the physical variables to which the wave responds?
1. x only
2. t only
3. k only
4. ? only
5. x and t
6. x and k
7. ? and t
8. k and ?
Answer:
please read the answer below
Explanation:
We have that both electric field and magnetic field are given by:
[tex]\vec{E}=E_osin(kx-\omega t)\hat{j}\\\\\vec{B}=B_osin(kx-\omega t)\hat{k}[/tex]
I complete with bold words the answers:
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
a. In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0 and B0are the magnitudes of the electric and magnetic fields.
2. amplitudes
b. The variable w is called the angular frequency of the wave.
2. angular frequency
c. The variable k is called the wavenumber of the wave.
1. wavenumber
c.
1. E =E0sin(wt)k^
d.
6. x and t
hope this helps!!
What is the best way to determine levelness?
A.See if the Acceleration versus Time for both directions are both zero.
B.Plot the Acceleration versus Time for one direction and acceleration should be zero.
C.Plot Position versus Time and compare the linear fit from each direction of travel; slopes should be equal.
D.Plot Position versus Time and compare the linear fit from each direction of travel; slopes should be equal but opposite.
E.Plot Velocity versus Time for both directions; slopes should be equal.
F.See if slope of Velocity versus Time for both directions are equal but opposite.
Answer:
F. See if the slope of Velocity versus Time for both directions are equal but opposite.
Explanation: Velocity is the rate of change of position. The slope of a distance or position-time graph is velocity. Acceleration is the rate of change of velocity. The slope of a velocity-time graph is acceleration.
A velocity-time graph shows changes in the velocity of a moving object over time. The slope of a velocity-time graph represents an acceleration of the moving object.
So, the value of the slope at a particular time represents the acceleration of the object at that instant.
A quantity of gas with an initial volume of 5 cubic feet and a pressure of 1700 pounds per square foot expands to a volume of 9 cubic feet. Find the work done by the gas for the given volume and pressure. Round your answer to two decimal places. Assume the temperature of the gas in this process remain constant.
Answer:
Work done by the gas for the given volume and pressure [tex]= 4996.18[/tex] pounds foot
Explanation:
Given
Pressure applied [tex]= 1700[/tex] pounds per square foot
Initial Volume [tex]= 5[/tex] cubic feet
Final Volume [tex]= 9[/tex] cubic feet
As we know pressure is inversely proportional to V
[tex]P = \frac{k}{V}[/tex]
where k is the proportionality constant
V is the volume and
P is the pressure
Work done
[tex]\int\limits^{V_2}_{V_1} {P} \, dV[/tex]
[tex]\int\limits^{V_2}_{V_1} {\frac{k}{V} } \, dV\\= \int\limits^{V_2}_{V_1} {\frac{1700 * 5}{V} } \, dV\\= 8500* \int\limits^{V_2}_{V_1} {\frac{1}{V} } \, dV[/tex]
Integrating the above equation, we get-
[tex]8500 ln \frac{V_2}{V_1} \\8500 * 2.303 * \frac{9}{5} \\= 4996.18[/tex]
Work done by the gas for the given volume and pressure [tex]= 4996.18[/tex] pounds foot
You are working in a laser laboratory. Your current project involves suspending spherical glass beads in the Earth's gravitational field using a vertically directed laser beam. Today's experiment involves a black bead of radius r and density rho. Assuming that the radius of the laser beam is the same as that of the bead and that the beam is centered on the bead, determine the minimum laser power required to suspend this bead in equilibrium. (Use any variable or symbol stated above along with the following as necessary: g and c.)
Final answer:
The minimum laser power required to suspend the bead in equilibrium is (3c/8πr³ρg).
Explanation:
In order to suspend the bead in equilibrium, the radiation pressure from the laser beam must be equal to the gravitational force on the bead.
The radiation pressure is given by P = 2I/c, where P is the pressure, I is the laser intensity, and c is the speed of light. The gravitational force is given by F = (4/3)πr³ρg, where F is the force, r is the radius of the bead, ρ is the density, and g is the acceleration due to gravity.
Equating the radiation pressure and gravitational force, we have 2I/c = (4/3)πr³ρg. Rearranging this equation, we get I = (3c/8πr³ρg).
Therefore, the minimum laser power required to suspend the bead in equilibrium is (3c/8πr³ρg).
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.630 m. The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.
How much is the wavelength?
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.630 m. The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.
How much is the wavelength?
How fast are the waves traveling ?
What is the amplitude A of wave?
Given Information:
time = t = 2.60 s
wavelength = λ = 5.50 m
distance = d = 0.630 m
Required Information:
a) wavelength = λ = ?
b) speed = v = ?
c) Amplitude = A = ?
Answer:
a) wavelength = 5.50 m
b) speed = 1.056 m/s
c) Amplitude = 0.315 m
Step-by-step explanation:
a)
It is given that wave crests are spaced a horizontal distance of 5.50 m apart that is basically the wavelength so,
λ = 5.50 m
b)
We know that the speed of the wave is given by
v = λf
where λ is the wavelength and f is the frequency of the wave given by
f = 1/T
Where T is the period of the wave.
Since the it is given that boat takes 2.60 s to travel from its highest point to its lowest that is basically half of the period so one full period is
T = 2*2.60
T = 5.2 s
So the frequency is,
f = 1/5.2
f = 0.192 Hz
Therefore, the speed is
v = λf
v = 5.50*0.192
v = 1.056 m/s
c)
The amplitude of the wave is given by
A = d/2
where d is the distance from the highest point to the lowest, therefore, the amplitude is half of it.
A = 0.630/2
A = 0.315 m
Suppose that you are visiting Champaign, IL and on July 10 you wake up early and note the rising azimuth of the Sun. How would the direction of the rising Sun change if you measured
Answer:
It would raise further south
Explanation:
To those of us who live on earth, the most important astronomical object by far is the sun. It provides light and warmth. Its motions through our sky cause day and night, the passage of the seasons, and earth's varied climates.
On any given day, the sun moves through our sky in the same way as a star. It rises somewhere along the eastern horizon and sets somewhere in the west. If you live at a mid-northern latitude (most of North America, Europe, Asia, and northern Africa), you always see the noon sun somewhere in the southern sky.
The sun's path through the rest of the sky is similarly farther north in June and farther south in December. In summary:
In late March and late September (at the "equinoxes"), the sun's path follows the celestial equator. It then rises directly east and sets directly west. The exact dates of the equinoxes vary from year to year but are always near March 20 and September 22.
After the March equinox, the sun's path gradually drifts northward. By the June solstice (usually June 21), the sun rises considerably north of due east and sets considerably north of due west. For mid-northern observers, the noon sun is still toward the south, but much higher in the sky than at the equinoxes.
After the June solstice, the sun's path gradually drifts southward. By the September equinox, its path is again along the celestial equator. The southward drift then continues until the December solstice (usually December 21), when the sun rises considerably south of due east and sets considerably south of due west.
Answer:
It would rise further south
Explanation: