Answer:
(A) [tex]9.14\times 10^{-9}sec[/tex]
(B) [tex]6.20\times 10^{-3}A[/tex]
Explanation:
We have given inductance [tex]L=5.41\mu H=5.41\times 10^{-6}H[/tex]
Resistance [tex]R=0.949kohm=0.949\times 10^3ohm[/tex]
Time constant of RL circuit is equal to [tex]\tau =\frac{L}{R}[/tex]
[tex]\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec[/tex]
Battery voltage e = 16 volt
(a) It is given current becomes 79.9% of its final value
Current in RL circuit is given by
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
According to question
[tex]0.799i_0=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]e^{\frac{-t}{\tau }}=0.201[/tex]
[tex]{\frac{-t}{\tau }}=ln0.201[/tex]
[tex]{\frac{-t}{5.7\times 10^{-9} }}=-1.6044[/tex]
[tex]t=9.14\times 10^{-9}sec[/tex]
(b) Current at [tex]t=\tau sec[/tex]
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})[/tex]
[tex]i=6.20\times 10^{-3}A[/tex]
A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum of reflection with normally incident orange light (wavelength 600 nm in air). Assuming the maximum occurs in the first order, determine the thickness of the oil slick. nm
To solve this problem we will apply the concepts related to the principle of destructive and constructive interference. Mathematically this expression can be given as
[tex]2nt = m\lambda[/tex]
Here,
n = Index of refraction
t = Thickness
m = Order of the reflection
[tex]\lambda[/tex] = Wavelength
We have all of this values, therefore replacing,
[tex]2(1.28)t = (1)(600nm)[/tex]
[tex]t = 233nm[/tex]
Therefore the thickness of the oil slick is 233nm
In a mass spectrometer, a singly ionized 24Mg ion has a mass equal to 3.983 10-26 kg and is accelerated through a 3.00-kV potential difference. It then enters a region where it is deflected by a magnetic field of 526 G. Find the radius of curvature of the ion's orbit. Note: There are 10,000 G in 1 T and 1,000 V in 1 kV.
Answer:
The radius of curvature of the ion's orbit is 0.59 meters
Explanation:
Given that,
Mass of the 24 Mg ion, [tex]m=3.983\times 10^{-26}\ kg[/tex]
Potential difference, V = 3 kV
Magnetic field, B = 526 G
Charge on single ionized ion, [tex]q=1.6\times 10^{-19}\ C[/tex]
The radius of the the path traveled by the charge is circular. Its radius is given by :
[tex]r=\dfrac{mv}{Bq}[/tex]
v is speed of particle.
v can be calculated using conservation of energy as :
[tex]\dfrac{1}{2}mv^2=qV\\\\v=\sqrt{\dfrac{2qV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^3}{3.983 \times 10^{-26}}} \\\\v=1.26\times 10^5\ m/s[/tex]
Radius,
[tex]r=\dfrac{3.983 \times 10^{-26}\times 1.26\times 10^5}{0.0526\times 1.6\times 10^{-19}}\\\\r=0.59\ m[/tex]
So, the radius of curvature of the ion's orbit is 0.59 meters.
A toroidal inductor has a circular cross-section of radius a a . The toroid has N turns and radius R. The toroid is narrow ( a≪R ), so the magnetic field inside the toroid can be considered to be uniform in magnitude. What is the self-inductance L of the toroid?
Answer:
L=N/I*magnetic flux
=N/I*BA=N*mu(0)NIA/2pi*R=mu(0)N^2/2pi*R)*Pi*a^2
=mu(0)N^2a^2/2R
Explanation:
The self-inductance L of the toroid is mu(0)N^2a^2/2R.
Calculation of the self-inductance L:Since
The toroidal inductor has a circular cross-section of radius a. The toroid has N turns and radius R.
The toroid is narrow ( a≪R )
So,
L=N/I*magnetic flux
=N/I*BA=N*mu(0)NIA/2pi*R=mu(0)N^2/2pi*R)*Pi*a^2
=mu(0)N^2a^2/2R
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A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released from rest 2.0 mm from a very wide plane of positive charge, after which it accelerates toward the plane and collides with a speed of 4.5 m/s.
What is the surface charge density of the plane?
Answer:
[tex]6.75\mu C/m^2[/tex]
Explanation:
We are given that
Diameter,d=[tex]1\mu m=1\time 10^{-6} m[/tex]
[tex]1\mu m=10^{-6} m[/tex]
Radius,r=[tex]\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m[/tex]
Density,[tex]\rho=900kg/m^3[/tex]
Total number of electrons,n=39
Charge on electron =[tex]1.6\times 10^{-19} C[/tex]
Total charge=[tex]q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C[/tex]
Distance,s=2mm=[tex]2\times 10^{-3} m[/tex]
Mass =[tex]density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg[/tex]
Initial velocity,u=0
Final speed,v=4.5 m/s
[tex]v^2-u^2=2as[/tex]
[tex](4.5)^2-0=2a(2\times 10^{-3})[/tex]
[tex]20.25=4a\times 10^{-3}[/tex]
[tex]a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2[/tex]
Force,F=ma
[tex]qE=ma[/tex]
[tex]q(\frac{\sigma}{2\epsilon_0})=ma[/tex]
[tex]\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}[/tex]
[tex]\epsilon_0=8.85\times 10^{-12}[/tex]
[tex]\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2[/tex]
An elastic conducting material is stretched into a circular loop of 14.8 cm radius. It is placed with its plane perpendicular to a uniform 0.814 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 88.4 cm/s. What emf is induced in volts in the loop at that instant
Answer: 0.666 V
Explanation:
Given
Radius of the loop, r = 14.8 cm = 0.148 m
Magnetic field present, B = 0.814 T
Rate of shrinking, dr/dt = 88.4 cm/s = 0.884 m/s
emf = dΦ/dt , where Φ = BA
emf = d(BA)/dt, where A = πr²
emf = d(Bπr²)/dt
if B is constant, then
emf = Bπ d(r²)/dt, on differentiating, we have,
emf = Bπ * 2r dr/dt
emf = 2πrB dr/dt, now if we substitute the values, we have
emf = 2 * 3.142 * 0.148 * 0.814 * 0.884
emf = 6.284 * 0.106
emf = 0.666 V
A string of 18 identical Christmas tree lights are connected in series to a 130 V source. The string dissipates 61 W.
What is the equivalent resistance of the light string?
Answer in units of Ω.
What is the resistance of a single light? Answer in units of Ω.
How much power is dissipated in a single light?
Answer in units of W.
One of the bulbs quits burning. The string has a wire that shorts out the bulb filament when it quits burning, dropping the resistance of that bulb to zero. All the rest of the bulbs remain burning.
What is the resistance of the light string now?
Answer in units of Ω.
How much power is dissipated by the string now?
Answer in units of W.
Answer:
(a) 277.05 Ω
(b) 15.39 Ω
(c) 3.76 W
Explanation:
(a)
Applying,
P = V²/R.......................... Equation 1
Where P = Power dissipated by the string. V = Voltage source, R = equivalent resistance of the light string
Make R the subject of the equation
R = V²/P................... Equation 2
Given: V = 130, P = 61 W
Substitute into equation 2
R = 130²/61
R = 277.05 Ω
(b) The resistance of a single light is given as
R' = R/18 (since the light are connected in series and the are identical)
Where R' = Resistance of the single light.
R' = 277.05/18
R' = 15.39 Ω
(c)
Heat dissipated in a single light is given as
P' = I²R'..................... Equation 3
Where P' = heat dissipated in a single light, I = current flowing through each light.
We can calculate for I using
P = VI
make I the subject of the equation
I = P/V
I = 61/130
I = 0.469 A.
Also given: R' = 15.39 Ω
Substitute into equation 3
P' = 0.496²(15.39)
P' = 3.76 W
(a)The equivalent resistance of the light string is 277.05 Ω
(b)The power is dissipated in a single ligh15.39 Ω
(c)The resistance of the light string now3.76 W
Calculation of Power is dissipates
(a) P is = V²/R.......................... Equation 1
Where P is = Power dissipated by the string.
Then V = Voltage source,
After that R is = the equivalent resistance of the light string
Now Make R the subject of the equation
R is = V²/P................... Equation 2
Then Given: V = 130,
P = 61 W
After that Substitute into equation 2
Then R = 130²/61
Therefore, R = 277.05 Ω
(b) When The resistance of a single light is given as
R' is = R/18 (since the light are connected in series and are identical)
Now Where R' is = Resistance of the single light.
R' is = 277.05/18
Therefore, R' is = 15.39 Ω
(c) When Heat dissipated in a single light is given as
P' is = I²R'..................... Equation 3
Where P' is = heat dissipated in a single light,
Then I = current flowing through each light.
Now We can calculate for I using
P is = VI
Now we make I the subject of the equation
After that I = P/V
Then I = 61/130
I is = 0.469 A.
Also given: R' is = 15.39 Ω
Then Substitute into equation 3
P' is = 0.496²(15.39)
Therefore, P' is = 3.76 W
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A 0.62112 slug uniform disc A is pinned to a 0.12422 slug uniform rod CB which is pinned to a 0.03106 slug collar C. C slides on a smooth vertical rod and the disc rolls without slipping. Determine: a) the angular acceleration ofA and the acceleration of C when released from rest and b)The velocity of B and C when the rod is horizontal. Ans:b) vc
Answer:
Explanation: check the attached document for step by step solution.
Although the transmission of light, and electromagnetic radiation generally, is correctly described by wave (physical) optics, there are situations for which ray (geometric) optics gives a sufficiently good approximation. For each of the situations described in the following, determine whether ray optics may be used or wave optics must be used. Part (a) Green laser light of wavelength 530 nm is incident on a 26-cm diameter mirror. Ray None of these. WavePart (b) Red laser light of wavelength 722 nm is incident on a molecule of size 114 nm. Wave Ray None of these
Answer:
a) the geometric optics is adequate (Ray)
b) wave optics should be used for the second case
Explanation:
In general, the approximation of the geometric optics is adequate when the dimensions of the system are much greater than the wavelengths and the wave optics should be used for cases in which the size of the system is from the length of the wave.
Let's apply this to our case
a) in this case the size of the system is d = 26 cm=0.26 m and the wavelength is alm = 530 10⁺⁹ m
in this case
d >>> λ
therefore the geometric optics is adequate (Ray)
b) in this case the system has a size of d = 114 10⁻⁹ m with a wavelength of λ= 722 nm
for this case d of the order of lam
therefore wave optics should be used for the second case
Does the box contain positive charge, negative charge, or no charge? Does the box contain positive charge, negative charge, or no charge? Positive charge, since there is net electric flux passing outward through the surface of the closed box, it contains positive charge. Negative charge, since there is net electric flux passing outward through the surface of the closed box, it contains negative charge. Positive charge, since there is net electric flux passing intward through the surface of the closed box, it contains positive charge. Negative charge, since there is net electric flux passing inward through the surface of the closed box, it contains negative charge. No charge, because the flux into the bos is canceled by the flux out of it.
Due to the net electric flux that is moving outward through the closed box's surface, the box has a negative charge.
Given:-
The electric field is constant over all the faces.
From Gauss law:
[tex]\int \vec E \cdot d\vec A = \dfrac{Q}{e_o}[/tex]
[tex]\vec E =[/tex] An electric field throughout the cube.
[tex]d \vec A =[/tex] Elemental area of the cube surface.
[tex]Q =[/tex] Electric charge around the cube.
[tex]e_o =[/tex] Electric constant.
[tex]\dfrac{Q}{e_o} = (-15-20-15+20+15+10)\\ \\ = -5A[/tex]
According to Gauss's law, the box contains a negative charge.
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The box's charge type (being positive, negative, or neutral) is determined by the net electric flux's direction. Outward flux signifies positive charge, whilst inward flux indicates a negative charge. Boxes with no net electric flux have no charge.
Explanation:The determination of whether a closed box contains a positive charge, negative charge, or no charge is dependent on the direction of the net electric flux. If there is net electric flux passing outward through the surface of the box, this indicates the box contains a positive charge. Conversely, if the net electric flux is passing inward through the surface, the box contains a negative charge. Lastly, if there is no net electric flux (flux in cancelled by flux out), it suggests there is no charge in the box.
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A particle enters a uniform magnetic field initially traveling perpendicular to the field lines and is bent with a circular arc of radius R. If this particle was traveling two times as fast, the radius of its circular arc would be? EXPLAIN
Answer:
Explanation:
In a scenario where a particle of charge overrightarrow{B} enters a magnetic field with a velocity overrightarrow{V}, it experiences a force overrightarrow{F} given by: overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B}).
implies F=BqVSin\theta.
Where theta is the angle between the velocity vector of the particle and the magnetic field vector.
When a particle enters the magnetic field at an angle 90, it moves in a circular path as it experiences a centripetal force, given by F=\frac{mV^2}{R}.
Where R is the radius of the circle, V is its velocity and m is its mass
Thus, magnetic force becomes F=BqVSin90^o=\frac{mV^2{R}\implies R=\frac{mV}{Bq}.
The equation changes as below, when velocity is doubled, let us assume that the radius is given by R_1.
R_1=\frac{2mV}{Bq}=2R.
Therefore, it is obvious that the velocity of a charged particle in a circular arc is directly proportional to the radius of the arc. The radius of the circular arc doubles when the velocity of the charged particle in the circular orbit doubles only if the mass, charge and magnetic field of the particle remains constant
Hence when velocity is doubled radius of the circle also gets doubled.
When the speed of the particle doubles, the radius of the arc also doubles.
The magnetic force on the particle is calculated as follows;
[tex]F = qvB[/tex]
The centripetal force on the particle is calculated as follows;
[tex]F_c = \frac{mv^2}{R}[/tex]
The speed of the particle is calculated as follows;
[tex]\frac{mv^2}{R} = qvB\\\\mv = qBR\\\\v = \frac{qBR}{m} \\\\\frac{v_1}{R_1} = \frac{v_2}{R_2}[/tex]
when the speed of the particle doubles;
[tex]\frac{v_1}{R_1} = \frac{2v_1}{R_2} \\\\R_2v_1 = 2R_1v_1\\\\R_2 = 2R_1[/tex]
Thus, we can conclude that when the speed of the particle doubles, the radius of the arc also doubles.
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An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E. Part A What is the system's potential energy when its kinetic energy is equal to 34E
The correct question is;
An object of mass m attached to a spring of force constant K oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.
What is the system's potential energy when its kinetic energy is equal to ¾E?
Answer:
P.E = ⅛KA²
Explanation:
From conservation of energy, the total energy in the system is given as the sum of potential and kinetic energy.
Thus,
Total Energy; E = K.E.+P.E.
In simple harmonic motion, the total energy is given by;
E = ½KA²
We are told that kinetic energy is ¾E.
Thus, ½KA² = ¾(½KA²) + P.E
P.E = ½KA² - ⅜KA²
P.E = ⅛KA²
g You're a safety engineer reviewing plans for a university's new high-rise dorm. The elevator motors draw 20 A and behave electrically like 2.4-H inductors. You're concerned about dangerous voltages developing across the switch when a motor is turned off, and you recommend that a resistor be wired in parallel with each motor. Part A What should be the resistance in order to limit the emf to 100 V
To solve the problem it is necessary to apply Ohm's law. From this it is established that the voltage is the equivalent to the product between the current and the resistance, therefore we have to,
[tex]V = IR[/tex]
Here,
V = Voltage
I = Current
R = Resistance
Rearranging to find the resistance,
[tex]R = \frac{V}{ I}[/tex]
Replacing,
[tex]R = \frac{100V}{20A}[/tex]
[tex]R = 5\Omega[/tex]
Therefore the resistance should be [tex]5\Omega[/tex]
The resistance should be [tex]\bold { 5\Omega}[/tex] in order to limit the emf to 100 V.
Ohm's law:
It states that the voltage is the equival to the product of the current and the resistance.
[tex]\bold{V =I\times R}[/tex]
Where,
V = Voltage = 100 Volts
I = Current = 20 Ampere
R = Resistance
Put the values and solve it for R
[tex]\bold {R =\dfrac{100}{20}}\\\\\bold {R = 5\Omega}[/tex]
Therefore, the resistance should be [tex]\bold { 5\Omega}[/tex] in order to limit the emf to 100 V.
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Domestic cats have vertical pupils. Imagine a cat is observing two small birds sitting side-by-side on a telephone wire. If the slit width of the cat's pupils is a = 0.550 mm and the average wavelength of the ambient light is λ = 519 nm, what is the angular resolution (in rad) for the two birds?
To solve this problem we will apply the concepts related to angular resolution based on wavelength and the slit width at this case of the cat's pupils. This relationship is given as,
[tex]\theta = \frac{\lambda}{a}[/tex]
Here,
[tex]\lambda[/tex] = Wavelength
a = Slit width
[tex]\theta = \frac{519*10^{-9}m}{0.55*10^{-3}m}[/tex]
[tex]\theta = 9.43*10^{-4} Rad[/tex]
Therefore the angular resolution is [tex]9.43*10^{-4}rad[/tex]
Final answer:
The angular resolution for a cat observing birds can be estimated using the formula θ = 1.22 λ / D, with a pupil width of 0.550 mm and a wavelength of 519 nm, yields an angular resolution of approximately 1.152 × 10⁻⁶ radians.
Explanation:
The question refers to angular resolution in optics, specifically related to the diffraction limit of a cat's eyes observing birds. In physics, the angular resolution for a circular aperture, like the pupil of a cat's eye, can be estimated using the formula θ = 1.22 λ / D, where θ is the angular resolution in radians, λ is the wavelength of the light, and D is the diameter of the aperture (the pupil in this case). Given that the cat's pupil has a slit width of a = 0.550 mm (which we will use in place of the diameter for this rough estimate) and the light has an average wavelength of λ = 519 nm, the angular resolution θ can be calculated as follows:
θ = 1.22 × 519 x 10⁻⁹ m / 0.550 x 10^-3 m
θ = 1.22 × 519 / 550 × 10⁻⁶
θ = 1.22 × 0.944 × 10⁻⁶
θ = 1.152 × 10⁻⁶ radians
Given the specified conditions, this calculation estimates the angular resolution for the cat's eyes. Note, however, that this is a simplification and doesn't take into account the complexities of a vertical slit pupil versus a circular aperture.
1. If only the vertical magnet were present, the field would point _____ at the point P.2. If only the horizontal magnet were present, the field would point _______.3. The vector combination of two magnetic fields from the two magnets points __________.Options areUpward and to the leftDownward and to the rightto the leftto the rightdownward and to the leftupwardupward and to the rightdownward
Answer:
1. downward
2. to the left
3. downward and to the left
Explanation:
This is gotten by using vector law of triangle Addition which states that If 2 vectors acting simultaneously on a body are represented both in magnitude and direction by 2 sides of a triangle taken in an order then the resultant(both magnitude and direction) of these vectors is given by 3rd side of that triangle taken in opposite order.
You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel with the clock to the distant planet CornTeen and measure the period of the clock to be 4s. By what factor is the gravitational acceleration constant g different on planet CornTeen compared to g on Earth
Answer:
[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]
Explanation:
The period of the simple pendulum is:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex]
Where:
[tex]l[/tex] - Cord length, in m.
[tex]g[/tex] - Gravity constant, in [tex]\frac{m}{s^{2}}[/tex].
Given that the same pendulum is test on each planet, the following relation is formed:
[tex]T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}[/tex]
The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:
[tex]\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}[/tex]
[tex]\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}[/tex]
[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]
A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length of the loaded spring
Answer:
The total length of the spring would be 0.65 m
Explanation:
The Concept
Hooke's law evaluates the increment of spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;
F = k x ..................1
where F is the force applied to the spring
k is the spring constant
x is the spring stretch or extension
Step by Step Calculations
We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;
x = F/k
but F = m x g
so, x = (m x g)/k
given that, the mass of the person m =150 kg
g is the acceleration due to gravity = 9.81 m/[tex]s^{2}[/tex]
k is the spring constant = 10000 N/m
then x = (9.81 m/[tex]s^{2}[/tex] x 150 kg)/10000 N/m
x = 0.14715 m
the extension experienced by the spring after the compression is 0.14715 m
The total length of the spring would be;
L = 0.14715 m + 0.5 m = 0.64715
L ≈ 0.65 m
Therefore the total length of the spring would be 0.65 m
An ice skater is spinning on frictionless ice with her arms extended outward. She then pulls her arms in toward her body, reducing her moment of inertia. Her angular momentum is conserved, so as she reduces her moment of inertia, her angular velocity increases and she spins faster. Compared to her initial rotational kinetic energy, her final rotational kinetic energy is _________
Answer:
larger, because her angular speed is larger.
Explanation:
The rotational kinetic energy is proportional to the square of the angular velocity while it is linearly proportional to the moment of inertia. So the increase of angular speed will have a larger effect of the kinetic energy than the decrease of the moment of inertia.
Answer:
Rotational kinetic energy will Increase
Explanation:
Rotational kinetic energy KE is
KE = 1/2 x I x w^2
Where I is moment of inertia,
w is angular velocity.
It can be seen that increasing angular velocity increases rotational kinetic energy.
Why might these "Mental Maps" be inaccurate or differ between different people?
Answer:
Catastrophic events of weather related outcomes.
Explanation:
The mental map of a person from a certain place may change due to long periods of time outside, since this gives the mind time to forget certain important details, they may also change according to people's experience and perception, places, regions and environments since these places are also changing and the perception we had is no longer the same. A mental map is a first person perspective of an area that an individual possesses. This type of subconscious map shows a person how a place looks and how to interact with it.
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Tia needs to produce a solenoid that has an inductance of 3.01 μ H 3.01 μH . She constructs the solenoid by uniformly winding 1.13 m 1.13 m of thin wire around a tube. How long, in centimeters, should the tube be?
Answer: 13 cm
Explanation:
Given
Inductance of the solenoid, L = 3.01•10⁻⁶ H
Width of the wire, x = 1.13 m
Length of the tube, z = ?
Now, we know that
L = μ₀N²A/z, where
N = x/2πr, making r subject of formula,
r = x/2πN
Also,
A = πr², on substituting for A, we have
A = πx²/4π²N²
Now finally, we substitute in the initial equation and solve
L = μ₀N²A/z
L = μ₀N²πx²/4π²N²z
L = μ₀x²/4z, making z subject of formula, we have
z = μ₀x²/4L
z = 4π*10⁻⁷ * 1.13² /4 * 3.01*10⁻⁶
z = (4π*10⁻⁷ * 1.2769) / (4 * 3.01*10⁻⁶)
z = 1.6*10^-6 / 1.2*10^-5
z = 0.13 m
Therefore, the length of the tube should be 13 cm
The length of the tube should be 469 centimeters.
To find the required length of the tube for Tia's solenoid, we can use the formula for the inductance of a solenoid:
[tex]\[ L = \frac{{\mu_0 \cdot N^2 \cdot A}}{l} \][/tex]
Where:
L = inductance of the solenoid (3.01 μH)
[tex]\( \mu_0 \)[/tex] = permeability of free space (4π × 10^-7 T*m/A)
N = number of turns of wire
A = cross-sectional area of the solenoid
[tex]\( l \)[/tex] = length of the solenoid
We're given [tex]\( L \)[/tex] and [tex]\( N \cdot l \)[/tex] (the total length of wire wound around the tube). We need to find [tex]\( l \)[/tex], the length of the tube.
First, let's rearrange the formula to solve for [tex]\( l \)[/tex]:
[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot A}}{{L}} \][/tex]
We know the cross-sectional area ( A ) can be represented as [tex]\( A = \pi r^2 \)[/tex], where ( r ) is the radius of the tube. Since the wire is uniformly wound, we can calculate ( A ) using the length of the wire and the number of turns.
[tex]\[ A = \frac{{\text{{Length of wire}}}}{{\text{{Number of turns}}}} \][/tex]
Given that the wire length is 1.13 m and there are [tex]\( N \)[/tex] turns, [tex]\( A = \frac{{1.13}}{{N}} \)[/tex].
Substituting this into the equation for [tex]\( l \)[/tex], we get:
[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot \left(\frac{{1.13}}{{N}}\right)}}{{L}} \][/tex]
[tex]\[ l = \frac{{\mu_0 \cdot 1.13}}{{L}} \][/tex]
Now, plug in the known values:
[tex]\[ l = \frac{{4\pi \times 10^{-7} \cdot 1.13}}{{3.01 \times 10^{-6}}} \][/tex]
[tex]\[ l = \frac{{4\pi \times 1.13}}{{3.01}} \][/tex]
[tex]\[ l = \frac{{4 \times 3.1416 \times 1.13}}{{3.01}} \][/tex]
[tex]\[ l = \frac{{14.1376}}{{3.01}} \][/tex]
[tex]\[ l = 4.69 \, \text{m} \][/tex]
Finally, convert this length to centimeters:
[tex]\[ l = 469 \, \text{cm} \][/tex]
So, the length of the tube should be 469 centimeters.
Suppose you have Avogadro's number of mini marshmallows and use them to cover the state of South Dakota which has a land area of 7.588 × 10 4 mi 2 . Each mini marshmallow has a diameter of 0.635 cm and a height of 2.54 cm. Assuming the marshmallows are packed together so there is no space between them, to what height above the surface, in kilometers, will the mini marshmallows extend?
Answer: 2.44*10^3 km
Explanation:
NOTE: We would be solving this question in "cm" instead of the usual "m"
1 mi² = 2.59 km²
2.59 km² = 2.59*10^10 cm²
Given, area of South Dakota
A = 7.588*10^4 mi²
A = 7.588*10^4 * 2.59*10^10
A = 1.97*10^15 cm² is the area of South Dakota
S = πd²/4
S = 3.142 * 0.635² / 4
S = 3.142 * 0.4/4
S = 3.142 * 0.1
S = 0.3142 cm² is the area of 1 marshmallow
1.97*10^15 / 0.3142 = 6.27*10^15, thus is the number of marshmallows in one single layer to cover the state area
6.02*10^23 / 6.27*10^15 = 9.6*10^7, this is the number of layers
9.6*10^7 * 2.54 = 2.44*10^8, this is the height in cm
Height in km is 2.44*10^3 km
A two-stage rocket is traveling at 1210m/s with respect to the earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of 40m/s relative to the second stage after the explosion. The first stage is three times as massive as the second stage.
What is the speed of the second stage after the separation?
Answer:
R
Explanation:
Given that,
Two stage rocket traveling at
V = 1210 m/s with respect to earth
First stage
When fuel Is run out , explosive bolts releases and push rocket backward at speed is
V1 = 40m/s relative to second stage
Therefore
V1 = 40 - V2
The first stage is 3 times as massive as the second stage
I.e Mass of first stage is 3 times the second stage
Let Mass of second stage be
M2 = M
Then, M1 = 3M
Velocity of second stage V2?
Applying conservation of linear momentum
Momentum before explosion = momentum after explosion
Momentum p=mv
Then,
(M1+M2)V = —M1•V1 + M2•V2
(3M+M)•1210 = —3M•(40-V2) +M•V2
4M × 1210 = —120M + 3M•V2 + MV2
4840M = —120M + 4M•V2
4840M + 120M = 4M•V2
4960M = 4M•V2
Then, V2 = 4960M / 4M
V2 = 1240 m/s
Two in-phase loudspeakers, which emit sound in all directions, are sitting side by side. One of them is moved sideways by 3.0 m, then forward by 7.0 m. Afterward, constructive interference is observed 14, 12, and 34 the distance between the speakers along the line that joins them, and at no other positions along this line.What is the maximum possible wavelength of the sound waves?Express your answer with the appropriate units.
The maximum possible wavelength for constructive interference from two shifted loudspeakers at given interference maxima positions is calculated to be 0.5738 m.
Explanation:The question is about the phenomenon of constructive interference of sound waves emitted from two loudspeakers that are shifted apart. The problem requires identifying the maximum possible wavelength of the sound waves produced by the speakers.
Constructive interference occurs when two waves meet and their crests (high points) and troughs (low points) align. This occurs effectively when the path difference is either zero or a multiple of the wavelength. When the path difference is specifically an integral multiple of the wavelength, the interference is constructive and has a maximum value.
Given that the loudspeakers are moved to a distance of 3.0 m sideways and 7.0 m forward, the distance separating the two is calculated using Pythagoras's theorem as √((7)^2+(3)^2) = √58 m. The maxima of the constructive interference are at positions 14, 12, and 34 the distance between the speakers, thus the maximum possible wavelength is 2 * √58 / 34 = 0.5738 m.
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The rho− meson has a charge of −e, a spin quantum number of 1, and a mass 1 507 times that of the electron. The possible values for its spin magnetic quantum number are −1, 0, and 1. Imagine that the electrons in atoms are replaced by rho− mesons. Select all of the following which are possible sets of quantum numbers (n, ℓ, mℓ, s, ms) for rho− mesons in the 3d subshell.
A. (2, 2, 1, 1, 0)
B. (3, 2, -1, 1, 1)
C. (3, 2. -1, 1, 1/2)
D. (3, 2, 0, 1, 1)
E. (3, 2, 0, 1, -1)
F. (3, 2, -1, 1, 0)
Answer:
Look up attached file
Explanation:
A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to reach maximum. At time t = 0 a switch is thrown that disconnects the voltage source, but leaves the resistor and the inductor connected in their own circuit.
(a) How much time, in milliseconds, is needed for the current in the circuit to drop to 12% of its value at t = 0?
(b) How much energy, in millijoules, does the circuit dissipate during that time?
To find the time it takes for the current to drop to 12% of its value and the energy dissipated by the circuit, we can use the equations for the decay of current in an RL circuit and the energy in an inductor respectively.
Explanation:To determine the time it takes for the current in the circuit to drop to 12% of its value at t = 0, we need to use the equation for the decay of current in an RL circuit. The equation is given by I(t) = I(0) * exp(-t/τ), where I(t) is the current at time t, I(0) is the initial current, t is the time, and τ is the time constant. In this case, we can use I(t) = I(0) * exp(-t/τ) = 0.12 * I(0). To find the time, we rearrange the equation as t = -τ * ln(0.12).
The energy dissipated by the circuit can be calculated using the equation for the energy in an inductor, which is given by E = 1/2 * L * I(0)^2, where E is the energy, L is the inductance, and I(0) is the initial current. In this case, we can substitute the values given to find the energy dissipated.
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The time needed for the current in the circuit to drop to 12% of its value at t = 0 is approximately 20.63 ms. The energy dissipated in the circuit during that time is approximately 7.22 mJ.
Explanation:Given, initial current I0 can be obtained from Ohm's law, I0 = V / R = 108 / 1100 = 0.09818 A. The circuit value diminishes exponentially over time and can be expressed as I = I0 * e-Rt/2L. Therefore, to find out when the current drops to 12% of its initial value we set I / I0 = 0.12 = e-Rt/2L, and solve for t. After calculation, we find t ≈ 20.63 ms.
As for part (b), the energy dissipated in an RL circuit over time is given by W = 1/2 * L * I2, where I is the current at time t, which is given by the relation: I = I0 * e-Rt/2L. Performing the integration over the time period t = 0 to 20.63 ms, we find that the energy dissipated is approximately 7.22 mJ.
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You will find in Chapter 39 that the electrons cannot move indefinite orbits within atoms, like the planets in our solar system.To see why, let us try to "observe" such an orbiting electron byusing a light microscope to measure the electron's presumed orbitalposition with a precision of, say, 8.2 pm(a typical atom has a radius of about 100 pm). The wavelength ofthe light used in the microscope must then be about 8.2 pm.
(a) What would be the photon energy of thislight?
1 keV
(b) How much energy would such a photon impart to an electron in ahead-on collision?
2 keV
(c) What do these results tell you about the possibility of"viewing" an atomic electron at two or more points along itspresumed orbital path? (Hint: The outer electrons of atoms arebound to the atom by energies of only a few electron-volts.)
The energy of a photon is the product of the planks constant and frequency. The photon energy of this light is 151 keV.
The Energy of a photon:It is the product of the planks constant and frequency.
[tex]E = \dfrac {hc}\lambda[/tex]
Where,
[tex]h[/tex] -Plank's constant = [tex]\bold {6.63\times 10^{-34}\rm \ J/s}[/tex]
[tex]c[/tex]- speed of light = [tex]\bold { 3\times 10^8 \rm \ m/s }[/tex]
[tex]\lambda[/tex][tex]\labda[/tex]- wavelength = 8.2 pm = [tex]\bold {8.2 \times 10^{-12 }\rm \ m}[/tex]
Put the values in the equation,
[tex]E = \dfrac { 6.63\times 10^{-34}\rm \ J/s\times 3\times 10^8 \rm \ m/s }{ 8.2 \times 10^{-12 }\rm \ m}\\E = 151 \rm \ keV[/tex]
Therefore, the photon energy of this light is 151 keV.
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The projectile partially fills the end of the 0.3 m pipe. Calculate the force required to hold the projectile in position when the mean velocity in the pipe is 6 m
Final answer:
To hold the projectile in position, the force required is 20 Newtons.
Explanation:
To calculate the force required to hold the projectile in position, we need to consider the acceleration of the projectile in the pipe.
From the given information, we can calculate the acceleration using the formula:
Acceleration = Change in velocity / Time taken
Since the mean velocity is given as 6 m/s, we can use this as the final velocity and assume the initial velocity is 0 m/s.
Substituting the values in the formula, we have:
Acceleration = (6 m/s - 0 m/s) / 0.3 m = 20 m/s^2
Now we can calculate the force using Newton's second law:
Force = Mass * Acceleration
Assuming the mass of the projectile is 1 kg, we have:
Force = 1 kg * 20 m/s^2 = 20 N
g If 6.35 moles of a monatomic ideal gas at a temperature of 320 K are expanded isothermally from a volume of 1.45 L to a volume of 3.95 L, calculate: a) the change in the internal energy of the gas. b) the work done by the gas. c) the heat flow into or out of the gas.
Answer:
(a) change in the internal energy of the gas is zero
(b) the work done by the gas is 16.93 kJ
(c) the heat flow is 16.93 kJ, which is into the gas
Explanation:
Given;
number of moles of gas, n = 6.35 moles
temperature of the gas, T = 320 K
initial volume of the gas, V₁ = 1.45 L
final volume of the gas, V₂ = 3.95 L
Part (a)
For isothermal expansion, temperature is constant and internal energy will also be constant.
Therefore, change in the internal energy of the gas is zero since the gas expanded isothermally (constant temperature).
ΔU = Q - W
where;
ΔU is change in internal energy
Q is heat transferred to the system
W is the work done by the system
Thus, Q = W
ΔU = 0
Part (b)
the work done by the gas
[tex]W = nRTln{[\frac{V_2}{V_1}][/tex]
where;
R is gas constant = 8.314 J/mol.K
[tex]W = (6.35)(8.314)(320)ln{[\frac{3.95}{1.45}]}\\\\W =16930.4\ J\\\\W = 16.93\ kJ[/tex]
Part (c)
the heat flow into or out of the gas
Q = ΔU + W
Q = 0 + 16.93 kJ
Q = 16.93 kJ
Since the heat flow is positive, then it is heat flow into the gas.
The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the wire represent the direction of current). For the following wires, which all also carry current , indicate the magnitude (in mT) and direction of the magnetic field at the center (red point) of each configuration.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The magnetic field is [tex]B_{net} = \frac{1}{4} * mT[/tex]
And the direction is [tex]-\r k[/tex]
Explanation:
From the question we are told that
The magnetic field at the center is [tex]B = 1mT[/tex]
Generally magnetic field is mathematically represented as
[tex]B = \frac{\mu_o I}{2R}[/tex]
We are told that it is equal to 1mT
So
[tex]B = \frac{\mu_o I}{2R} = 1mT[/tex]
From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)
[tex]\frac{\mu_o I}{2R} = 1mT[/tex]
This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)
[tex]B_1 = \frac{1}{2} \frac{\mu_o I}{2R}[/tex]
and for the larger semi-circular loop is
[tex]B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]
Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction
So the net magnetic field would be
[tex]B_{net} = B_1 - B_2[/tex]
[tex]= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]
[tex]=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}[/tex]
[tex]=\frac{\mu_o I}{8R}[/tex]
[tex]= \frac{1}{4} \frac{\mu_o I}{2R}[/tex]
Recall [tex]\frac{\mu_o I}{2R} = 1mT[/tex]
So
[tex]B_{net} = \frac{1}{4} * mT[/tex]
Using the Right-hand rule we see that the direction is into the page which is [tex]-k[/tex]
About how far does the S wave travel through Earth in 13 minutes?
2,000 km
4,000 km
6,000 km
8,000 km
what is the electric potential at point A in the electric field created by a point charge of 5.5 • 10^-12 C? estimate k as 9.00 • 10^9
The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v
Explanation:
Given data,
charge = 5.5 x 10¹² C
k =9.00 x 10⁹
The electric potential V of a point charge can found by,
V= kQ / r
Assuming, r=5.00×10⁻² m
V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m
V= 49.5 x 10⁻³/ 5.00×10⁻²
Electric potential V= 0.099 x 10⁻¹v
The electric potential at a point in the electric field created by a point charge is calculated by the formula V = kQ/r. The provided constants are [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex] and [tex]k = 9.00 \times 10^9[/tex] The distance r from the point charge is necessary to calculate the electric potential.
Explanation:The electric potential at point A in the electric field created by a point charge can be found using the formula V = kQ/r, where V is the electric potential, k is Coulomb's constant, Q is the value of the point charge, and r is the distance from the point charge.
Given, the charge [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex], and Coulomb's constant [tex]k = 9.00 \times 10^9[/tex]
The distance r is not given in the question. Therefore, we can't calculate the electric potential at point A. However, if distance r from the point charge were provided (in meters), we could substitute k, Q, and r into the formula and solve for V.
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