Answer:
3.37x10^6J
Explanation:
The external energy is given as KQ1Q2/r
=(9*10^9)*(3*10^-3)*(25*10^-3)/(0.2)
=3.375 *10^6 J
The external energy required can be calculated using the electric potential energy formula. The external energy required to bring the charge of 25µC from infinity to the center of the circle is 337.5 joules.
Explanation:The external energy required to bring a charge of 25µC from infinity to the center of the circle can be calculated using the electric potential energy formula, which is given by:
U = k * (q1 * q2) / r
Where U is the electric potential energy, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.
In this case, the charge at the center of the circle is 25µC, and the charge distributed uniformly along the circumference is 3.0µC. The radius of the circle is given as 20 cm. Plugging these values into the formula, we have:
U = (9 * 10^9 Nm^2/C^2) * (25 * 10^-6 C) * (3 * 10^-6 C) / (0.20 m) = 337.5 J
Therefore, the external energy required to bring the charge of 25µC from infinity to the center of the circle is 337.5 joules.
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Two very small planets orbit a much larger star. Planet A orbits the star with a period TA . Planet B orbits the star at four times the distance of planet A, but planet B is four times as massive as planet A. Assume that the orbits of both planets are approximately circular.
Planet B orbits the star with a period TB equal to
a. TA/4.
b. 16TA.
c. 8TA.
d. 4TA
d. TA
Answer:
Explanation:
Given that,
Period of Planet A around the star
Pa = Ta
Let the semi major axis of planet A from the star be
a(a) = x
Given that, the distance of Planet B from the star (i.e. it semi major axis) is 4 time that of Planet A
a(b) = 4 × a(a) = 4 × x
a(b) = 4x
Also planet B is 4 times massive as planet A
Using Kepler's law
P² ∝a³
P²/a³ = k
Where
P is the period
a is the semi major axis
Then,
Pa² / a(a)³ = Pb² / a(b)³
Pa denotes Period of Planet A and it is Pa = Ta
a(a) denoted semi major axis of planet A and it is a(a) = x
Pb is period of planet B, which is the required question
a(b) is the semimajor axis of planet B and it is a(b) = 4x
So,
Ta² / x³ = Pb² / (4x)³
Ta² / x³ = Pb² / 64x³
Cross multiply
Ta² × 64x³ = Pb² × x³
Divide both sides by x³
Ta² × 64 = Pb²
Then, Pb = √(Ta² × 64)
Pb = 8Ta
Then, the period of Planet B is eight times the period of Planet A.
The correct answer is C
The period of orbit for planet B is 4 times the period of orbit for planet A.
Explanation:To determine the period of orbit for planet B, we can use Kepler's Third Law, which states that the square of the period of an orbit is proportional to the cube of its average distance from the center of attraction. Since planet B is four times the distance of planet A and four times as massive, its period will be 4^3/4^2 times that of planet A. Simplifying, this means that TB = 4 * TA, so the correct answer is d. 4TA.
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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.
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degrees, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 60.0 w/cm^ 2 after it passes through the stack.
If the incident intensity is kept constant:
1) What is the intensity of the light after it has passed through the stack if the second polarizer is removed?
2) What is the intensity of the light after it has passed through the stack if the third polarizer is removed?
Answer:
1
When second polarizer is removed the intensity after it passes through the stack is
[tex]I_f_3 = 27.57 W/cm^2[/tex]
2 When third polarizer is removed the intensity after it passes through the stack is
[tex]I_f_2 = 102.24 W/cm^2[/tex]
Explanation:
From the question we are told that
The angle of the second polarizing to the first is [tex]\theta_2 = 21^o[/tex]
The angle of the third polarizing to the first is [tex]\theta_3 = 61^o[/tex]
The unpolarized light after it pass through the polarizing stack [tex]I_u = 60 W/cm^2[/tex]
Let the initial intensity of the beam of light before polarization be [tex]I_p[/tex]
Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as
[tex]I_1 = \frac{I_p}{2}[/tex]
Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as
[tex]I_2 = I_1 cos^2 \theta_1[/tex]
[tex]= \frac{I_p}{2} cos ^2 \theta_1[/tex]
The intensity of light that will emerge from the third filter is mathematically represented as
[tex]I_3 = I_2 cos^2(\theta_2 - \theta_1 )[/tex]
[tex]I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)][/tex]
making [tex]I_p[/tex] the subject of the formula
[tex]I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}[/tex]
Note that [tex]I_u = I_3[/tex] as [tex]I_3[/tex] is the last emerging intensity of light after it has pass through the polarizing stack
Substituting values
[tex]I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}[/tex]
[tex]I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}[/tex]
[tex]=234.622W/cm^2[/tex]
When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as
[tex]I_f_3 = \frac{I_p}{2} cos ^2 \theta_2[/tex]
[tex]I_f_3[/tex] is the intensity of the light emerging from the stack
substituting values
[tex]I_f_3 = \frac{234.622}{2} * cos^2(61)[/tex]
[tex]I_f_3 = 27.57 W/cm^2[/tex]
When the third polarizer is removed the second polarizer becomes the
the final polarizer and the intensity of light emerging from the stack would be
[tex]I_f_2 = \frac{I_p}{2} cos ^2 \theta_1[/tex]
[tex]I_f_2[/tex] is the intensity of the light emerging from the stack
Substituting values
[tex]I_f_2 = \frac{234.622}{2} cos^2 (21)[/tex]
[tex]I_f_2 = 102.24 W/cm^2[/tex]
The intensity of light after the second polarizer is removed is 22.88 w/cm² and after the third polarizer is removed is 26.73 w/cm².
Explanation:The intensity of light passing through a polarizing filter can be determined by Malus's Law, which states that the intensity of the transmitted light is equal to the incident light multiplied by the square of the cosine of the angle between the filters. If unpolarized light is incident on the stack, the intensity is halved after passing through the first filter.
1) When the second polarizer is removed, the angle difference will be 40 degrees (61 degrees - 21 degrees). Hence, the intensity would be (60/2) * cos²(40 degree) = 22.88 w/cm².
2) When the third polarizer is removed, the angle difference is only 21 degrees, hence the intensity would be (60/2) * cos²(21 degree) = 26.73 w/cm².
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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
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]
If the bore of an engine is increased without any other changes except for the change to proper-size replacement pistons. the displacement will ________ and the compression rate will ________.
Answer:
increase, increase
Explanation:
If the bore of an engine is increased without any other changes except for the change to proper-size replacement pistons. the displacement will increase and the compression rate will increase.
Further Explanation:
If the bore is increased, there are several benefits to this. There is more area for the valves this improves flow through the engine. The compression ratio will increase, thereby increasing thermal efficiency. The displacement also increases as the power output increase. Also, higher RPM will further increase power.
However, a decrease in bore will result in corresponding reduction in displacement and power output.
A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is the period of small oscillations of the hoop? Type your answer below, accurate to two decimal places, and assuming it is in seconds. [Recall that the moment of inertia of a hoop around its center is Icm=MR2.]
Answer:
Explanation:
Given that,
Mass of the thin hoop
M = 2kg
Radius of the hoop
R = 0.6m
Moment of inertial of a hoop is
I = MR²
I = 2 × 0.6²
I = 0.72 kgm²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
I = 0.72 kgm²
M is the mass of the hoop
M = 2kg
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = r = 0.6m
Then, applying the formula
T = 2π √ (I / MgR)
T = 2π √ (0.72 / (2 × 9.8× 0.6)
T = 2π √ ( 0.72 / 11.76)
T = 2π √0.06122
T = 2π × 0.2474
T = 1.5547 seconds
T ≈ 1.55 seconds to 2d•p
Then, the period of oscillation is 1.55seconds
A firm wants to determine the amount of frictional torque in their current line of grindstones, so they can redesign them to be more energy efficient. To do this, they ask you to test the best-selling model, which is basically a diskshaped grindstone of mass 1.1 kg and radius 0.09 m which operates at 73.3 rad/s. When the power is shut off, you time the grindstone and find it takes 42.4 s for it to stop rotating. What is the frictional torque exerted on the grindstone in newton-meters
Explanation:
Mass of the diskshaped grindstone, m = 1.1 kg
Radius of disk, r = 0.09 m
Angular velocity, [tex]\omega=73.3\ rad/s[/tex]
Time, t = 42.4 s
We need to find the frictional torque exerted on the grindstone. Torque in the rotational kinematics is given by :
[tex]\tau=I\alpha[/tex]
I is moment of inertia of disk, [tex]I=\dfrac{mr^2}{2}[/tex]
[tex]\tau=\dfrac{mr^2\alpha }{2}\\\\\tau=\dfrac{1.1\times (0.09)^2\times 73.3 }{2\times 42.4}\\\\\tau=7.7\times 10^{-3}\ N-m[/tex]
So, the frictional torque exerted on the grindstone is [tex]7.7\times 10^{-3}\ N-m[/tex].
An airplane is initially flying horizontally (not gaining or losing altitude), and heading exactly North. Suppose that the earth's magnetic field at this point is also exactly horizontal, and points from South to North (it really does!). The airplane now starts a different motion. It maintains the same speed, but gains altitude at a constant rate, still heading North. The magnitude of the electric potential difference between the wingtips has... [Remember that the airplane is made of metal, and is a conductor] Group of answer choices
Note: The answer choices are :
a) Increased
b) Decreased
c) stayed the same
Answer:
The correct option is Increased
The magnitude of the electric field potential difference between the wingtips increases.
Explanation:
The magnitude of the electric potential difference is the induced emf and is given by the equation:
[tex]emf = l (v \times B)[/tex]
where l = length
v = velocity
B = magnetic field
As the altitude of the airplane increases, the magnetic flux becomes stronger, the speed of the airplane becomes perpendicular to the magnetic field, i.e. [tex]v \times B = vB sin90 = vB\\[/tex] ,
the induced emf = vlB, and thus increases.
The magnitude of the electric field potential difference between the wingtips increases
What direction does a S wave move
Answer:
As they travel through rock, the waves move tiny rock particles back and forth -- pushing them apart and then back together -- in line with the direction the wave is traveling. These waves typically arrive at the surface as an abrupt thud. Secondary waves (also called shear waves, or S waves) are another type of body wave
Explanation:
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Understand Key Concepts
Which is NOT a fluid?
A. helium
B. ice
C. milk
D.Water
Answer:
The answer is A, friend.
How do the electrons in the bulb's filament, far from the battery and its emf, "know" that they should flow?
Answer:this is obviously because free electrons can easily move in the metal due to potential difference, and they are inserted in the wire from the negative terminal of the battery and pulled away ( come back) from the wire into the battery at positive terminal.
Electrons in a lightbulb filament flow due to the electric fields created by a connected battery, encountering resistance that dissipates energy as heat and allows the bulb to emit light.
Explanation:Electrons 'know' to flow in a bulb's filament due to the electric fields created when a battery is connected to a circuit. These fields exert a force on the electrons, causing them to move. The flow of electrons is sustained by the potential difference (emf) provided by the battery, but it's not perpetual because as electrons move through the resistive filament, they encounter resistance analogous to friction, which causes energy to be dissipated as heat. This is the principle by which the bulb emits light. An open circuit, like if a wire is cut, prevents the flow of charge, demonstrating the necessity of a closed path for current to flow.
The resistance to electron movement through the filament is due to interactions with the material's atoms, as described by Newton's second law (a = Ftotal/m). This resistance leads to the conversion of electrical energy into heat, which in the case of an incandescent lightbulb, is necessary for it to glow. Without resistance, electrons accelerated by the emf would continue gaining kinetic energy, which does not occur due to this energy transformation. The steady flow of electrons is the result of these resisting forces balancing out the force exerted by the electric field, leading to a constant flow of current as long as the circuit is complete.
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]
A laser shines on a pair of vertical slits. The horizontal distance L 1 between the laser and the ruler is 12.3 m. The distance L 2 between the laser and the slits is 0.511 m. The distance d is 0.440 mm. A laser is located a distance L subscript 1 from a ruler and a distance L subscript 2 from a barrier containing two, narrow slits. The slit separation is d. An interference intensity pattern is shown at the position of the ruler. The intensity pattern has a central maximum at 0 centimeters. The first dark fringe is at plus or minus 1.6 centimeters. The first bright fringe is at about plus or minus 3 centimeters. The second dark fringe is at about plus or minus 4.8 centimeters. The second bright fringe is at about plus or minus 6.4 centimeters. The third dark fringe is at plus or minus 8 centimeters. The illustration is not to scale. Note that the ruler measures in centimeters. Calculate the wavelength ? of the light.
To calculate the wavelength of the light, use the equation d sin θ = mλ. Plugging in the values, we find that the wavelength is 563 nm.
Explanation:In order to calculate the wavelength of the light, we can use the equation d sin θ = mλ, where d is the slit separation, θ is the angle of the bright fringe, m is the order of the fringe, and λ is the wavelength of the light. In this case, we are given the slit separation (0.0100 mm) and the angle of the third bright line (10.95°). We can rearrange the equation to solve for λ:
λ = d sin θ / m
Plugging in the values, we get:
λ = (0.0100 mm) sin(10.95°) / 3
λ = 5.63 x 10-7 mm or 563 nm
The same pipe is used to carry both air and water. For the same fluid velocity and friction factor for the air and water flows: the pressure drop for the air flow is greater than that for the water flow. the pressure drop for the air flow equals the pressure drop for the water flow. the pressure drop for the water flow is greater than that for the air flow.
Answer:
the pressure drop for the water flow is greater than that for the air flow.
Explanation:
Detailed analysis of the problem is show below.
Space satellites in the inner solar system (such as those in orbit around Earth) are often powered by solar panels. Satellites that travel into the outer solar system, however, are usually powered by the decay of nuclear isotopes – a more expensive and dangerous technology. Why can’t they use solar panels as well?
a. At greater distances, sunlight is more likely to destructively interfere.
b. At greater distances, sunlight is more likely to constructively interfere.
c. At greater distances from the sun, the frequency of sunlight will be too low.
d. At greater distances from the sun, the intensity of sunlight will be too low.
Answer:
D
Explanation:
See attached file
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
Consider two pulses (a wave with only a single peak) traveling towards each other on a string. When the instant that the peaks of these two pulses cross, the resultant disturbance has a maximum displacement of 3.80 A where A is the amplitude of the first pulse. What must be the amplitude of the second pulse be
Answer:
2.8A
Explanation:
To calculate the total amplitude (when both pulses meet), we need to add up the amplitudes of each pulse. Since A is the amplitude of the first pulse, and we can call B the amplitude of the second pulse and C the total amplitude, we have that A+B=C=3.8A, which means that B=3.8A-A=2.8A, which we have already said is the amplitude of the second pulse.
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
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 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!!
A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 115 N/m and negligible mass. At time t = 0 the mass is released from rest at a distance d = 0.35 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by y(t) = A cos(ωt – φ). The positive y-axis points upward. show answer Correct Answer 17% Part (a) Find the angular frequency of oscillation, in radians per second. ω = 10.22 ✔ Correct! show answer Incorrect Answer 17% Part (b) Determine the value of the coefficient A, in meters.
Answer:
a) = 10.22 rad/s
b) = 0.35 m
Explanation:
Given
Mass of the particle, m = 1.1 kg
Force constant of the spring, k = 115 N/m
Distance at which the mass is released, d = 0.35 m
According to the differential equation of s Simple Harmonic Motion,
ω² = k / m, where
ω = angular frequency in rad/s
k = force constant in N/m
m = mass in kg
So,
ω² = 115 / 1.1
ω² = 104.55
ω = √104.55
ω = 10.22 rad/s
If y(0) = -0.35 m and we want our A to be positive, then suffice to say,
The value of coefficient A in meters is 0.35 m
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.52 10-6 W/m2 at a distance of 123 m from the explosion, at what distance from the explosion is the sound intensity half this value
Answer:
The distance when the intensity is halved is 173.95 m
Explanation:
Given;
initial intensity of the sound, I₁ = 1.52 x 10⁻⁶ W/m²
initial distance from the explosion, d₁ = 123 m
final intensity of the sound, I₂ = ¹/₂ (1.52 x 10⁻⁶ W/m²) = 0.76 x 10⁻⁶ W/m²
Intensity of sound is inversely proportional to the square of distance between the source and the receiver.
I ∝ ¹/d²
I₁d₁² = I₂d²
(1.52 x 10⁻⁶)(123)² = (0.76 x 10⁻⁶)d₂²
d₂² = (1.52 x 10⁻⁶ x 123²) / (0.76 x 10⁻⁶)
d₂² = 30258
d₂ = √30258
d₂ = 173.95 m
Therefore, the distance when the intensity is halved is 173.95 m
The sound intensity varies inversely as the square of the distance from the
explosion source.
The distance from the explosion at which the sound intensity is half of 1.52 × 10⁻⁶ W/m², is approximately 173.95 meters
Reasons:
The sound intensity at 123 m = 1.52 × 10⁻⁶ W/m²
Required:
The distance at which the sound intensity is half the given value.
Solution:
Sound intensity is given by the formula;
[tex]I \propto \mathbf{ \dfrac{1}{d^2}}[/tex]
Which gives;
I × d² = Constant
I₁ × d₁² = I₂ × d₂²
Where;
I₁ = The sound intensity at d₁
I₂ = The sound intensity at d₂
[tex]d_2 = \mathbf{ \sqrt{\dfrac{I_1 \times d_1^2}{I_2} }}[/tex]
When the sound intensity is half the given value, we have;
I₂ = 0.5 × I₁
I₂ = 0.5 × 1.52 × 10⁻⁶ = 7.6 × 10⁻⁷
Therefore;
[tex]d_2 = \sqrt{\dfrac{1.52 \times 10^{-6} \times 123^2}{7.6 \times 10^{-7}} } \approx 173.95[/tex]
The distance from the explosion at which the sound intensity is half of the
sound intensity at 123 meters from the explosion, d₂ ≈ 173.95 m.
Learn more here:
https://brainly.com/question/14695863
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.
Preparing an Accounts Payable Schedule Pilsner Inc. purchases raw materials on account for use in production. The direct materials ected purchases April May June Pilsner typically pays 25% on account in the month of billing and 75% the next month Required: 1. How much cash is required for payments on account in May? $374,400 411,200 416,000 2. How much cash is expected for payments on account in June?
Answer:
How much cash is expected for payments on account in June? = $412,400.
Explanation:
Cash Required for expected payments on account in June
Pilsner Inc. pays 25% on account in the month of billing and the remaining 75% in the next month, therefore, for the month of June the total payment will include 75% of Purchases made in May and 25% of purchases made in June.
Purchases in May = $411,200
Purchases in June = $416,000
Total Payment for May = (75% of 411,200)+(25% of 416,000)
= $308,400 + $104,000
= $412,400.
The neurons of giant squids, for example, consist of axons with very large radii, which allows the squid to react very quickly when confronted with a predator. Assuming no change in the resistivities or membrane thickness of the axon, by what factor must the radius of the axon increase such that the speed of the pulse increases by a factor of 10
Answer:
100
Explanation:
[tex]\rho_m[/tex] = Resistivity of axon
r = Radius of axon
t = Thickness of the membrane
[tex]\rho_a[/tex] = Resistivity of the axoplasm
Speed of pulse is given by
[tex]v=\sqrt{\dfrac{\rho_mrt}{2\rho_a}}[/tex]
So, radius is given by
[tex]r=\dfrac{2\rho_a}{\rho_mt}v^2[/tex]
If radius is increased by a factor of 10 new radius will be
[tex]r_2=\dfrac{2\rho_a}{\rho_mt}(10v)^2\\\Rightarrow r_2=100\dfrac{2\rho_a}{\rho_mt}v^2\\\Rightarrow r_2=100r[/tex]
So, The radius will increase by a factor of 100.
One possible explanation for a galaxy's type invokes the angular momentum of the protogalactic cloud from which it formed. Suppose a galaxy forms from a protogalactic cloud with a lot of angular momentum. Assuming its type has not changed as a result of other interactions, we'd expect this galaxy to be ______.
Answer:
A spiral galaxy
Explanation:
Suppose the Earth orbited the Sun in 4 months rather than 1 year, but had exactly the samerotation speed. How much longer would a solar day be than a sidereal day? Mercury’s siderealday is 58.6 days and its orbital period is 88 days. What is the length of Mercury’s solar day?
Answer:
A) 0.1955 days
B) 35.18 days
Explanation: Given that the Earth orbited the Sun in 4 months rather than 1 year, but had exactly the same rotation speed. That is the orbital period will be
365.24/12 = P/4
P = 121.75 days
We have three periods, the orbital period Po = 121.75 days,
the sidereal period Ps = 23.93419 hours and
The mean solar day Pd =24 hours.
Since they had exactly the same rotation speed, the number of solar days in a year is Po/Pd.
The number of sidereal days in a year would be Po/Pd + 1 = 121.75 + 1 = 122.75 days. Since there are more sidereal days in a year this means that the sidereal day is a little bit shorter, 121.75/122.75 = 0.99185 times shorter.
0.99185 × 24 = 23.80448 hours.
24 - 23.80448 = 0.1955 days
Therefore, solar day is 0.99185 days longer than a sidereal day
Given that the Mercury’s side real day is 58.6 days and its orbital period is 88 days. What is the length of Mercury’s solar day?
Using the formula
Ld = PoPs/(Po + Ps)
Ld = Length of Mercury's solar day
Po = orbital period
Ps = Mercury's side real day
Ld = (88 × 58.6)/ (88 + 58.6)
Ld = 5156.8/146.6
Ld = 35.18 days
2. An electrical heater 200mm long and 15mm in diameter is inserted into a drilled hole normal to the surface of a large block of material having a thermal conductivity of 5W/m·K. Estimate the temperature reached by the heater when dissipating 25 W with the surface of the block at a temperature of 35 °C.
Answer:
The final temperature is 50.8degrees celcius
Explanation:
Pls refer to attached handwritten document
Answer: 50.63° C
Explanation:
Given
Length of heater, L = 200 mm = 0.2 m
Diameter of heater, D = 15 mm = 0.015 m
Thermal conductivity, k = 5 W/m.K
Power of the heater, q = 25 W
Temperature of the block, = 35° C
T1 = T2 + (q/kS)
S can be gotten from the relationship
S = 2πL/In(4L/D)
On substituting we have
S = (2 * 3.142 * 0.2) / In (4 * 0.2 / 0.015)
S = 1.2568 / In 53.33
S = 1.2568 / 3.98
S = 0.32 m
Proceeding to substitute into the main equation, we have
T1 = T2 + (q/kS)
T1 = 35 + (25 / 5 * 0.32)
T1 = 35 + (25 / 1.6)
T1 = 35 + 15.625
T1 = 50.63° C
A nearsighted person cannot see objects further away than 10 m clearly. This person wants to be able to see objects 300 m away without trouble. What is the focal length of the lens you would prescribe? (Ignore the distance between the person’s eye and the lens.)
a. 9.7 m
b. – 9.7 m
c. No such lens can be prescribed.
d. 10.3 m
e. – 30 m
f. 30 m
g. – 10.3 m
Answer:
d.-10.3m
Explanation:
Note for short sightedness the focal length is negative
Let do be object distance=10m
And di= image distance=-300m
Using lens formula
F=do*di/do-di= 10*300/10-300=-10.3m
Answer:
d. 10.3 m
Explanation:
For a nearsighted person,
1/f = (1/v)+(1/u).................... Equation 1
f = focal, v = image distance, u = object distance.
f = vu/(v+u)................. Equation 2
Given: v = -10 m, u = 300 m
Substitute into equation 2
f = (-10×300)/(-10+300)
f = -3000/290
f = -10.34 m
therefore a concave lens of focal length 10.34 m is required.
The right option is d. 10.3 m