Answer:
the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m
Explanation:
Using Rayleigh criterion for the limiting angle of resolution of an eye
[tex]\theta = \frac{1.22\lambda }{D } \\ \\ \theta = \frac{1.22*506 *10^{-9} }{4.90*10^{-3}m}[/tex]
[tex]\theta = 1.2598*10^{-4}[/tex] rad
[tex]\theta = 125.98*10^{-6} \ rad[/tex]
Thus; the separation between the two sources is expressed as:
[tex]\theta = \frac{y}{L} \\ \\ y = L \theta \\ \\ y = (250*10^3 )(125.98*10^{-6} \ rad) \\ \\ y = 31.495 \ m[/tex]
Thus; the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m
Final answer:
The approximate size of the smallest object that astronauts can resolve by eye when orbiting 250 km above the Earth is approximately 628 meters.
Explanation:
The approximate size of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth can be determined using the Rayleigh criterion. According to the Rayleigh criterion, the resolution of the eye is determined by the angle of resolution, which is given by:
angle of resolution = 1.22 * (wavelength / pupil diameter)
Using the given values, we can calculate the angle of resolution:
angle of resolution = 1.22 * (506 nm / 4.90 mm) = 1.2611957657 x 10^-3 radians
To determine the approximate size of the smallest object, we need to find the linear size corresponding to this angular resolution at a distance of 250 km:
linear size = 2 * distance * tan(angle of resolution)
linear size = 2 * 250000 m * tan(1.2611957657 x 10^-3 radians) = 628.1079000855 m
Therefore, the approximate size of the smallest object that astronauts can resolve by eye when they are orbiting 250 km above the Earth is approximately 628 meters.
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.
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
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.
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
If a current flowing through a lightbulb is 0.75 ampere and the voltage difference across the lightbulb is 120 volts, how much resistance does the light bulb have
Answer: [tex]160 \Omega[/tex]
Explanation:
According to Ohm's law:
[tex]V=R.I[/tex]
Where:
[tex]V=120 V[/tex] is the voltage difference across the light bulb
[tex]R[/tex] is the resistance of the light bulb (the value we want to find)
[tex]I=0.75 A[/tex] is the electric current
Isolating [tex]R[/tex]:
[tex]R=\frac{V}{I}[/tex]
[tex]R=\frac{120 V}{0.75 A}[/tex]
Finally:
[tex]R=160 \Omega[/tex] This is the resistance of the light bulb
Final answer:
To calculate the resistance of a lightbulb with a current of 0.75 amperes and a voltage difference of 120 volts, use Ohm's law. The formula is R = V / I, which yields a resistance of 160 ohms for the lightbulb.
Explanation:
Calculating the Resistance of a Lightbulb
The student has asked how to calculate the resistance of a lightbulb when a current of 0.75 amperes flows through it and the voltage difference across it is 120 volts. To find the resistance, we can use Ohm's law, which states that the resistance (R) of a circuit is equal to the voltage (V) across it divided by the current (I) flowing through it, which can be written as:
R = V / I
Plugging in the given values, we get:
R = 120 V / 0.75 A
R = 160 ohms
Therefore, the resistance of the lightbulb is 160 ohms.
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.
A strip of copper 130 µm thick and 4.40 mm wide is placed in a uniform magnetic field of magnitude B = 0.79 T, that is perpendicular to the strip. A current i = 26 A is then sent through the strip such that a Hall potential difference V appears across the width. Calculate V. (The number of charge carriers per unit volume for copper is 8.47 × 1028 electrons/m3.)
Answer:
V = 1.1658 × [tex]10^{-5}[/tex] V
Explanation:
given data
strip of copper thick = 130 µm
strip of copper wide = 4.40 mm
uniform magnetic field of magnitude B = 0.79 T
current i = 26 A
number of charge carriers per unit volume = 8.47 × [tex]10^{28}[/tex] electrons/m³
solution
we know that number density is express as
n = \frac{Bi}{Vle} ...............1
B is uniform magnetic field and i is current and V is hall potential difference and l is thickness and e is electron charge 1.6 × [tex]10^{-19}[/tex] C
so V will be as
V = \frac{iB}{nle} .....................2
so put here value and we get V
V = [tex]\frac{26 \times 0.79}{8.47\times 10^{28}\times 130\times10^{-6}\times1.6 \times10^{-19}}[/tex]
V = 1.1658 × [tex]10^{-5}[/tex] V
A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d 5.0 cm. The coil is connected to a battery producing a current of 4.0 A in the wire. (a) What is the magnitude of the magnetic dipole mo- ment of this device? (b) At what axial distance d will the mag- netic field have the magnitude 5.0 mT (approximately one-tenth that of Earth’s magnetic field)?
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
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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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]
he electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the values of their resistances. Unfortunately, stock is low, and all you have are R1 and R2 in parallel and in series - and you can't separate these two resistor combinations. You separately connect each resistor network to a battery with emf 57.0 V and negligible internal resistance and measure the power P supplied by the battery in both cases. For the series combination, P = 48.0 W; for the parallel combination, P = 256 W. You are told that R1>R2. Calculate R1.
Answer:
R₁ = 50.77 Ω
Explanation:
Since, we know that:
Electric Power = P = VI
but from Ohm's Law:
V = IR
(or) I = V/R
Therefore,
P = V²/R
(OR) R = V²/P
where,
V = Battery Voltage
R = Resistance of combination
FOR SERIES COMBINATION:
R = Rs = (57 V)²/48 W
Rs = 67.69 Ω
but, we know that:
Rs = R₁ + R₂
R₁ + R₂ = 67.69 Ω
R₁ = 67.69 Ω - R₂ __________ eqn (1)
FOR PARALLEL COMBINATION:
R = Rp = (57 V)²/256 W
Rp = 12.69 Ω
but, we know that:
Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω
using eqn (1) and value of R₁ + R₂, we get
Rp = 12.69 = R₂(67.69 - R₂)/67.69
859.08 = 67.69 R₂ - R₂²
R₂² - 67.69 R₂ + 859.08 = 0
Solving this quadratic equation we get the answers:
Either, R₂ = 50.76 Ω
Either, R₂ = 16.92 Ω
Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,
R₂ = 16.92 Ω
using this value in eqn (1), we get:
R₁ = 67.69 Ω - 16.92 Ω
R₁ = 50.77 Ω
Final answer:
To calculate R1, first, find the total resistance for series (R1 + R2) and parallel (1/R1 + 1/R2) connections using the power formula P = V^2 / R, with provided power values for each case. Then solve the simultaneous equations.
Explanation:
The problem can be solved by using the formulas for resistances in series and parallel, as well as the formula for the power supplied by the battery.
In series, the resistances add up: Rtotal = R1 + R2. The power provided by the battery to the series circuit is given by P = V2 / Rtotal, where P is 48.0 W and V is 57.0 V. Rearranging to solve for Rtotal, we get Rtotal = V2 / P.
For parallel resistances, we have 1/Rtotal = 1/R1 + 1/R2. The power in the parallel case is P = V2 / Rtotal. With V and P given as 57.0 V and 256 W, respectively, we can solve for Rtotal.
With both Rtotal values calculated, we can set up two equations with two unknowns (R1 and R2), knowing that R1 > R2. Solving these simultaneous equations gives us the value of R1.
Suppose you are in a spaceship traveling at 99% of the speed of light past a long, narrow space station. Your direction of travel is parallel to the length of the station. If you measure lengths of objects on the station and also how time is passing on the station, what results will you get?
A) Lengths will appear shorter and time will appear to pass slower.
B) Lengths will appear shorter and time will appear to pass faster.
C) Lengths will appear shorter and time will appear to pass faster.
Answer:
B) Lengths will appear shorter and time will appear to pass faster.
Explanation:
This is in line with the laws of relativity.
Answer:
A. Lengths will appear shorter and time will appear to pass slower.
Explanation:
From theory of relativity, we know that:
L₀ = length of object, measured in stationary frame of reference
L = length of object measured from a frame moving with respect to object, called ‘relativistic length’
v = relativistic speed between observer and the object
c = speed of light
then,
L = L₀ √(1-v²/c² )
Hence, the length of the object decreases with the increase in its relativistic speed "v"
t₀ = time measured by clock at rest with respect to event.
t = time measured by clock in motion relative to the event
v = relativistic speed between observer and the object
c = speed of light
then,
t = t₀/√(1-v²/c² )
Hence, the time increases with the increase in in relativistic speed and as a result it appears to pass slower.
Hence, the correct option is:
A. Lengths will appear shorter and time will appear to pass slower.
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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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.
Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m be the mass of the moon, and G be the gravitational constant.
Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration the gravitational pull of the moon?
Answer:
The magnitude of the acceleration of earth due to the gravitational pull of earth is a = Gm/r^2
Where r = the center to center distance between the earth and the moon,
m = mass of the moon, and,
G is the gravity constant.
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
[tex]a_e = \frac{Gm}{r^2}[/tex]
Explanation:
We assume that:
M to represent the mass of the earth
m to equally represent the mass of the moon
r should be the distance between the center of the earth to the center of the moon.
Then;
the expression for the gravitational force can be written as:
[tex]F = \frac{GMm}{r^2}[/tex]
Where [tex]a_e[/tex] is the acceleration produced by the earth; then:
[tex]F =M *a_e[/tex]
Then:
[tex]M*a_e = \frac{GMm}{r^2}[/tex]
[tex]a_e = \frac{GMm}{Mr^2}[/tex]
[tex]a_e = \frac{Gm}{r^2}[/tex]
Therefore, the magnitude of the acceleration of the earth due to the gravitational pull of the moon [tex]a_e = \frac{Gm}{r^2}[/tex]
a celestial body moving in an ellipical orbit around a star
Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.
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.
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 convex refracting surface has a radius of 12 cm. Light is incident in air (n = 1) and refracted into a medium with an index of refraction of 2. Light incident parallel to the central axis is focused at a point: Group of answer choices 3 cm from the surface 6 cm from the surface 24 cm from the surface 12 cm from the surface 18 cm from the surface
Light is incident in air (n = 1) and refracted into a medium with an index of refraction of 2. Light incident parallel to the central axis is focused at a point 24cm from the surface
What is refraction?When a light ray passes through from one medium to another medium ,due to the difference in medium index, the light deflect from its original path .
This deflection from its path is called refraction and The phenomenon of refraction can be seen in sound and water waves.
Cornea reflect the light before light reaches to the retina. Change in the speed of light when it travels from one medium to other medium then due to this there is change in the path of light, this phenomenon is known as refraction of light.
Equation for refraction
n2/v- n1/u = n2-n1/R
= u = infinity ( parallel)
2/v= (2-1)/R
V = 2R
V= 24cm
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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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What causes differences in air pressure around the Earth?
Wind is driven due to the differences in air pressure around the Earth.
To find the answer, we need to know about the wind flow.
How does the wind flow?Due to the temperature difference at different places on the earth surface, we get difference in pressures at various regions.For keeping a constant pressure in all regions, the wind flows from high to low pressure region.Thus, we can conclude that wind is driven due to the differences in air pressure around the Earth.
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A block of iron quickly sinks in water, but ships constructed of iron float. A solid cube of iron 1.90 m on each side is made into sheets. From these sheets, to make a hollow cube that will not sink, what should the minimum length of the sides be? (density of iron is 7860 kg/m³)
Answer: 3.78 m
Explanation:
Volume = 1.9 * 1.9 * 1.9
Volume = 6.859 m³
Density of iron assumed to be 7870 kg/m³
Recall,
Density = mass / volume, so
Mass = density * volume
Mass = (7870 * 6.859)
Mass = 53980.33 kg
This mass we calculated, is the mass of water the new cube would have to displace, absolute minimum, so that it can float.
Now density of fresh water is assumed to be (1000 kg/m³).
Thus, the volume of water that will be displaced has to be
Volume = mass / density
Volume = (53980.33 / 1,000)
Volume = 53.98 m³.
Cube root of 53.88 = 3.78 m
Therefore, the minimum side length is 3.78 m
Answer:
3.781488032m or greater
Explanation:
General Rule of thumb is that any thing that is less dense than water will float in it.
So
density of water is 997Kg/m^3 and density of iron on the other hand is 7860kg/m^3.
so to make iron less dense it has to be in larger volume.
mass of the given cube is 53911.74Kg and we have contain it in volume such that density is less than water, mathematically this translates to .
[tex]\frac{53911.74Kg}{V}\leq \frac{997Kg}{m^3}[/tex] left side is density of iron and right side is density of water, solving v in this inequality gives us.
[tex]v \geq 54.0739m^3[/tex] which means each side is [tex]s \geq 3.7814880m[/tex]. so the minum is just greater than that.
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
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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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
Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their mass ratios were 1:4 and their respective speeds as they approached the intersection were both 13 m/s, find the magnitude and direction of the final velocity of the wreck.
Answer with Explanation:
Let mass of one vehicle =m
Mass of other vehicle=m'
[tex]\frac{m}{m'}=\frac{1}{4}[/tex]
[tex]m'=4m[/tex]
Velocity of one vehicle=[tex]v=13 im/s[/tex]
Velocity of other vehicle=[tex]v'=13jm/s[/tex]
In x- direction
By law of conservation of momentum
[tex]mv+0=(m+m')V_x[/tex]
[tex]13m=(m+4m)V_x[/tex]
[tex]13m=5mV_x[/tex]
[tex]V_x=\frac{13}{5}[/tex]
In y- direction
By law of conservation of momentum
[tex]0+m'v'=(m+m')V_y[/tex]
[tex]4m(13)=5mV_y[/tex]
[tex]V_y=\frac{52m}{5m}=\frac{52}{5}[/tex]
Magnitude of velocity of the wreck,V=[tex]\sqrt{V^2_x+V^2_y}=\sqrt{(\frac{13}{5})^2+(\frac{52}{5})^2}=10.72 m/s[/tex]
Direction:[tex]\theta=tan^{-1}(\frac{V_y}{V_x})[/tex]
[tex]\theta=tan^{-1}(\frac{\frac{52}{5}}{\frac{13}{5}})=75.96^{\circ}[/tex]
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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The left end of a long glass rod 7.00 cmcm in diameter has a convex hemispherical surface 3.50 cmcm in radius. The refractive index of the glass is 1.60. The glass rod is immersed in oil (nn = 1.45). An object placed to the left of the rod on the rod's axis is to be imaged 1.15 mm inside the rod.
How far from the left end of the rod must the object be located to form the image?
Answer:
n1/p + n2/q = (n2-n1)/R
1.45/p + 1.6/1.15 = (1.6 - 1.45)/0.035 (m)
1.45/p + 1.39 = 4.28
p = 0.50 m (50 cm)
A particle with a charge of 34.6 $\mu C$ moves with a speed of 68.4 m/s in the positive $x$ direction. The magnetic field in this region of space has a component of 0.466 T in the positive $y$ direction, and a component of 0.876 T in the positive $z$ direction. What is the magnitude of the magnetic force on the particle?
Answer:
The magnitude of the magnetic force on the particle is 67.86 N.
Explanation:
Given that,
Charge, [tex]q=34.6\ \mu C=34.6\times 10^{-6}\ C[/tex]
Speed of particle, v = 68.4 m/s in +x direction
Magnetic field, [tex]B=(0.466 j+0.876 k)\ T[/tex]
We need to find the magnitude of the magnetic force on the particle. The magnetic force is given by :
[tex]F=q(v\times B)\\\\F=34.6\times 10^{-6}(68.4i+0+0\times (0+0.466j+0.876 k))\\\\F=\begin{pmatrix}0&-59.9184&31.8744\end{pmatrix}\\\\F=(-59.9184j+31.8744k)\ N[/tex]
The magnitude of magnetic force on the particle is :
[tex]|F|=\sqrt{(-59.9184)^2+(31.8744)^2} \\\\|F|=67.86\ N[/tex]
So, the magnitude of the magnetic force on the particle is 67.86 N.
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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