Answer:
The thickness is [tex]t = 1.415 *10^{-7 } \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 566 \ nm = 566 *10^{-9} \ m[/tex]
The index of refraction of glass is [tex]n_g = 1.71[/tex]
The index of refraction of the coating is [tex]n= 1.46[/tex]
Generally the condition for destructive interference is
[tex]2 t = (m + \frac{1}{2} ) * \frac{\lambda }{n }[/tex]
Here m is the order of the interference pattern and given from the question that we are considering minimizing reflection m = 0
t = thickness of the coating
substituting values
[tex]2 t = (0 + \frac{1}{2} ) * \frac{ 566 *10^{-9}}{ 1.46 }[/tex]
=> [tex]t = 1.415 *10^{-7 } \ m[/tex]
Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 120 years into the future. According to special relativity, how fast must you travel
Answer:
I must travel with a speed of 2.97 x 10^8 m/s
Explanation:
Sine the spacecraft flies at the same speed in the to and fro distance of the journey, then the time taken will be 6 months plus 6 months
Time that elapses on the spacecraft = 1 year
On earth the people have advanced 120 yrs
According to relativity, the time contraction on the spacecraft is gotten from
[tex]t[/tex] = [tex]t_{0} /\sqrt{1 - \beta ^{2} }[/tex]
where
[tex]t[/tex] is the time that elapses on the spacecraft = 120 years
[tex]t_{0}[/tex] = time here on Earth = 1 year
[tex]\beta[/tex] is the ratio v/c
where
v is the speed of the spacecraft = ?
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
120 = 1/[tex]\sqrt{1 - \beta ^{2} }[/tex]
squaring both sides of the equation, we have
14400 = 1/[tex](1 - \beta ^{2} )[/tex]
14400 - 14400[tex]\beta ^{2}[/tex] = 1
14400 - 1 = 14400[tex]\beta ^{2}[/tex]
14399 = 14400[tex]\beta ^{2}[/tex]
[tex]\beta ^{2}[/tex] = 14399/14400 = 0.99
[tex]\beta = \sqrt{0.99}[/tex] = 0.99
substitute β = v/c
v/c = 0.99
but c = 3 x 10^8 m/s
v = 0.99c = 0.99 x 3 x 10^8 = 2.97 x 10^8 m/s
The intensity level 10 m from a point sound source is 85 dB. What is the intensity level 50 m away from the same source
Answer:
425dBExplanation:
Given the intensity level 10 m from a point sound source is 85 dB, then;
L1 = 10m, I1= 85dB ...1
The intensity level 50 m away from the same source cal be calculated using the equivalent expression;
when L2 = 50m, I2 = ? ... 2
Solving equation 1 nad 2;
10m = 85db
50m = x
Cross multiplying;
50 * 85 = 10 * x
10x = 50*85
10x = 4250
Divide both sides by 10
10x/10 = 4250/10
x = 425 dB
Hence, the intensity level 50 m away from the same source is 425dB
A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. A parallel plate capacitor sets up an electric field E which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by 3 mm and the value of the magnetic field is 0.3 T, what voltage between the plates will allow particles of speed 5 x 105 m/s to pass straight through without deflection? A. 70 V B. 140 V C. 450 V D. 1,400 V E. 2,800 V
Answer:
C. 450v
Explanation:
Using
Voltage= B*distance of separation*velocity
3mm x 0.3T x 5E5m/s
= 450v
Current is the rate at which charge is flowing.
a. True
b. Fals
Answer:
A. True
Explanation:
Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of 0.860 eV from sodium metal. What is the binding energy of electrons to sodium metal?
Answer:
Binding Energy = 2.24 eV
Explanation:
First, we need to find the energy of the photon of light:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)
E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 3.1 eV
Now, from Einstein's Photoelectric Equation:
E = Binding Energy + Kinetic Energy
Binding Energy = E - Kinetic Energy
Binding Energy = 3.1 eV - 0.86 eV
Binding Energy = 2.24 eV
3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000 rpm, completando un total de 12 revoluciones.
d. Determina la aceleración tangencial, centrípeta y la total en m.s-2 cuando el tambor a alcanzado los 60000 rpm
e. Explica lo que ocurre con la magnitud y dirección de los vectores aceleración tangencial, aceleración centrípeta, aceleración total, aceleración angular, velocidad angular cuando la lavadora ha girado desde 3000 rpm hasta 6000 rpm.
Answer:
d) α = 1693.5 rad / s² , a = 392.7 m / s² , a_total = α √(R² +1) ,
e) tan θ = a / α
Explanation:
This is an exercise in linear and angular kinematics.
We initialize reduction of all the magnitudes to the SI system
w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s
w = 6000 rev / mi = 628.32 rad / s
θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad
d) ask for centripetal, tangential and total acceleration.
Let's start by looking for centripetal acceleration, let's use the formula
w² = w₀² + 2 α θ
α = (w²- w₀²) / 2θ
we calculate
α = (628.32²2 - 314.16²) / 2 75.398
α = 1693.5 rad / s²
the quantity is linear and angular are related
the linear or tangential acceleration is
a = α R
where R is the radius of the drum
a = 1693.5 R
Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m
a = 1693.5 0.20
a = 392.7 m / s²
the total acceleration is
a_total = √(a² + α²)
a_total = √ (α² R² + α²)
a_total = α √(R² +1)
e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant
Tangential acceleration is tangency to radius and its value varies proportionally radius
the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry
tan θ = a / α
the angular velocity increases linearly when with centripetal acceleration
A pulley 326 mm in diameter and rotating initially at 4.00 revolutions per second receives a constant angular acceleration of 2.25 radians per second squared by a drive belt. What is the linear velocity of the belt after 5.00 seconds
Answer:
The linear velocity, v = 5.93 m/s
Explanation:
To find the linear velocity after 5 seconds, we find its angular velocity after 5 seconds using
ω' = ω + αt where ω = initial angular speed = 4.00 rev/s = 4.00 × 2π rad/s = 25.13 rad/s, ω' = = final angular speed, α = angular acceleration = 2.25 rad/s² and t = time = 5.00 s
ω' = ω + αt
= 25.13 rad/s + 2.25 rad/s² × 5.00 s
= 25.13 rad/s + 11.25 rad/s
= 36.38 rad/s
The linear velocity v is gotten from v = rω' where r = radius of pulley = 326 mm/2 = 163 mm = 0.163 m
v = rω'
= 0.163 m × 36.38 rad/s
= 5.93 m/s
So, the linear velocity v = 5.93 m/s
Convert 7,348 grams to kilograms
A) Hooke's law is described mathematically using the formula Fsp = -ku. Which statement is correct about the spring force, Fsp?
A.It is a vector quantity
B.It is the force doing the push or pull,
C.It is always a positive force.
D.It is larger than the applied force.
1. Which example best describes a restoring force?
B) the force applied to restore a spring to its original length
2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?
C) The spring exerts a restoring force to the left and returns to its equilibrium position.
3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?
D) 1 m
4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?
D)It is a vector quantity.
5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?
A) It decreases in magnatude.
Hope this Helps!! Sorry its late
e. Your father bought you a pair of shoes. When you wore the shoes, you realized there was a problem. The shoes were too long Why might such a problem arise and how can it be mitigated?
The problem arose due to a difference in length. This was due to father not knowing the exact length of shoe used by the son. And this can be mitigated by the use of shoe fillers.
The length of an object implies how long the object is. And it is one of the fundamental unit of quantities measured in SI unit of meters.
Considering the given question, it can be observed that the father do not know the exact length of shoe that would fit the son appropriately. Thus the realized problem of the pair of shoes too long arose due to difference in length of the pair of shoes and the son's leg. This variation would not have occurred if the exact length of pair of shoes has been bought.
To mitigate this little problem, shoe fillers can be used.
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An interference pattern is produced by light with a wavelength 590 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.580 mm .
Required:
a. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
b. What would be the angular position of the second-order, two-slit, interference maxima in this case?
Answer:
a. 0.058°
b. 0.117°
Explanation:
a. The angular position of the first-order is:
[tex] d*sin(\theta) = m\lambda [/tex]
[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{1* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.058 ^{\circ} [/tex]
Hence, the angular position of the first-order, two-slit, interference maxima is 0.058°.
b. The angular position of the second-order is:
[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{2* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.12 ^{\circ} [/tex]
Therefore, the angular position of the second-order, two-slit, interference maxima is 0.117°.
I hope it helps you!
Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
Required:
What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?
Complete Question
Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
Required:
What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?
Answer:
The velocity is [tex]v = 80.82 \ m/s[/tex]
Explanation:
From the question we are told that
The magnitude of charge one is [tex]q_1 = 3.25 nC = 3.25 *10^{-9} \ C[/tex]
The magnitude of charge two [tex]q_2 = 2.00 \ nC = 2.00 *10^{-9} \ C[/tex]
The distance of separation is [tex]d = 58.0 \ cm = 0.58 \ m[/tex]
Generally the electric potential of the electron at the midway point is mathematically represented as
[tex]V = \frac{ q_1 }{\frac{d}{2} } + \frac{ q_2}{\frac{d}{2} }[/tex]
substituting values
[tex]V = \frac{ 3.25 *10^{-9} }{\frac{ 0.58}{2} } + \frac{ 2 *10^{-9} }{\frac{ 0.58}{2} }[/tex]
[tex]V = 1.8103 *10^{-8} \ V[/tex]
Now when the electron is 10 cm = 0.10 m from charge 1 , it is (0.58 - 0.10 = 0.48 m ) m from charge two
Now the electric potential at that point is mathematically represented as
[tex]V_1 = \frac{q_1}{ 0.10} + \frac{q_2}{ 0.48}[/tex]
substituting values
[tex]V_1 = \frac{3.25 *10^{-9}}{ 0.10} + \frac{2.0*10^{-9}}{ 0.48}[/tex]
[tex]V_1 = 3.67*10^{-8} \ V[/tex]
Now the law of energy conservation ,
The kinetic energy of the electron = potential energy of the electron
i.e [tex]\frac{1}{2} * m * v^2 = [V_1 - V]* q[/tex]
where q is the magnitude of the charge on the electron with value
[tex]q = 1.60 *10^{-19} \ C[/tex]
While m is the mass of the electron with value [tex]m = 9.11*10^{-31} \ kg[/tex]
[tex]\frac{1}{2} * 9.11 *10^{-19} * v^2 = [ (3.67 - 1.8103) *10^{-8}]* 1.60 *10^{-19}[/tex]
[tex]v = \sqrt{6532.4}[/tex]
[tex]v = 80.82 \ m/s[/tex]
Which scientist proposed a mathematical solution for the wave nature of light?
Answer:
Explanation:
Christian Huygens
Light Is a Wave!
Then, in 1678, Dutch physicist Christian Huygens (1629 to 1695) established the wave theory of light and announced the Huygens' principle.
In which type of indicating valve is the valve stem housed in a hollow metal post that contains a movable plate with a small glass window
Answer:
Post indicator valve
Explanation:
Post Indicator Valves are commonly used to control the water flow of sprinkler systems used in public and private buildings, warehouses, and factories for fire suppression. PIVs control water flow from the public system into the building's fire suppression system.
physics approach to study macromoelcues at nanoscales
in detail plx
Answer:
Abstracto
Los ácidos nucleicos y las proteínas comprenden una red de biomacromoléculas que almacenan y transmiten información que sustenta la vida de la célula. El estudio de estos mecanismos es un campo llamado biología molecular. El desarrollo de esta ciencia siempre ha ido acompañado de avances técnicos que permiten romper barreras metodológicas para probar hipótesis novedosas. Entre los métodos disponibles para los biólogos moleculares, destacan cinco: electroforesis, secuenciación, clonación, transferencia y reacción en cadena de la polimerasa. Su impacto llega a la genética, la medicina y la biotecnología. Aquí, se revisan la relevancia histórica, los fundamentos técnicos y las tendencias actuales de estos cinco métodos esenciales. La revisión pretende ser útil tanto para estudiantes como para científicos profesionales que buscan adquirir conocimientos avanzados sobre el valor de estos métodos para investigar los mecanismos moleculares que sostienen la vida.
the treasure map gives the following directions to the buried treasure
Answer:
North
South
East
West
Explanation:
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A typical ten-pound car wheel has a moment of inertia of about 0.35kg *m2. The wheel rotates about the axle at a constant angular speed making 70.0 full revolutions in a time interval of 4.00 seconds. What is the rotational kinetic energy K of the rotating wheel? Express answer in Joules
Answer:
The rotational kinetic energy is [tex]K = 2116.3 \ J[/tex]
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 0.35 \ kg \cdot m^2[/tex]
The number of revolution is N = 70 revolution
The time taken is t = 4.0 s
Generally the angular velocity is mathematically represented as
[tex]w = \frac{2 \pi N }{t }[/tex]
substituting values
[tex]w = \frac{2* 3.142 * 70 }{4 }[/tex]
[tex]w = 109.97 \ rad/s[/tex]
The rotational kinetic energy K i mathematically represented as
[tex]K = \frac{1}{ 2} * I * w^2[/tex]
substituting values
[tex]K = \frac{1}{ 2} * 0.35 * (109.97)^2[/tex]
[tex]K = 2116.3 \ J[/tex]
A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it three wet weather on 3 sepaprate days? Also find it standard deviation
Answer:
The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
Explanation:
We are given that A city of Punjab has 15 percent chance of wet weather on any given day.
So, Probability of wet weather = 0.15
Probability of not being a wet weather = 1-0.15 =0.85
We are supposed to find probability that it will take a week for it three wet weather on 3 separate days
Total number of days in a week = 7
We will use binomial over here
n = 7
p =probability of failure = 0.15
q = probability of success=0.85
r=3
Formula :[tex]P(r=3)=^nC_r p^r q ^{n-r}[/tex]
[tex]P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166[/tex]
Standard deviation =[tex]\sqrt{n \times p \times q}[/tex]
Standard deviation =[tex]\sqrt{7 \times 0.15 \times 0.85}[/tex]
Standard deviation =0.9447
Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
Using a conventional two-slit apparatus with light of wavelength 605 nm, 34 bright fringes per centimeter are observed on a screen 3.1 m away. What is the slit separation
Answer:
d = 6.38 x 10⁻³ m = 6.38 mm
Explanation:
Since, the no. of bright fringes is 34 in a centimeter, therefore, the fringe spacing must be equal to:
Fringe Spacing = Δx = 1 cm/34
Δx = 0.0294 cm = 2.94 x 10⁻⁴ m
But, the formula for fringe spacing in a double slit experiment is:
Δx = λL/d
where,
λ = wavelength of light = 605 nm = 6.05 x 10⁻⁷ m
L = Distance between screen and slits = 3.1 m
d = slit separation = ?
Therefore,
2.94 x 10⁻⁴ m = (6.05 x 10⁻⁷ m)(3.1 m)/d
d = (18.755 x 10⁻⁷ m²)/(2.94 x 10⁻⁴ m)
d = 6.38 x 10⁻³ m = 6.38 mm
A rod bent into the arc of a circle subtends an angle 2θ at the center P of the circle (see below). If the rod is charged uniformly with a total charge Q, what is the electric field at P? (Assume Q is positive. For the magnitude, use the following as necessary: ε0, Q, R, and θ.)
Answer:
Qsinθ/4πε₀R²θ
Explanation:
Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.
Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.
So dq = λRdθ.
Substituting dq into dE', we have
dE' = dqcosθ/4πε₀R²
= λRdθcosθ/4πε₀R²
= λdθcosθ/4πε₀R
E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ
E' = (λ/4πε₀R)[sinθ] from -θ to θ
E' = (λ/4πε₀R)[sinθ]
= (λ/4πε₀R)[sinθ - sin(-θ)]
= (λ/4πε₀R)[sinθ + sinθ]
= 2(λ/4πε₀R)sinθ
= (λ/2πε₀R)sinθ
Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ
Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ
Q = 2λRθ
λ = Q/2Rθ
Substituting λ into E', we have
E' = (Q/2Rθ/2πε₀R)sinθ
E' = (Q/θ4πε₀R²)sinθ
E' = Qsinθ/4πε₀R²θ where θ is in radians
A fisherman in a stream 39 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman
Answer:
30cm
Explanation:
assume that the eyes are substantially above the water so that sin(theta) is approximately theta.
( small angle approximation).
The point at which a ray leaving the fish hits the surface of the water is x to the side of the centreline and the depth of the water is d
x/d = sin( angle of incidence)
if the apparent depth of the water is h then
x/h = sin( angle of refraction)
and applying snells law
1 sin ( theta air) = 1.33 sin( theta water)
1 * x/h = 1.33 * x/d
d/h = 1.33
or h/d = 1/1.33
h/39 = 1.33
h = 39 /1.33 so that is the apparent depth of the stream assuming:-
1. Your eyes are almost directly overhead
and
2. your eyes are a significant distance above the surface of the water.
x/d = 1.33 x/h
h/d =39/1.3
= 30cm
what path would an object have to take to have the distance and the displacement to be equal
Answer:
When an object move in a straight line without moving back.
Explanation:
Distance is covered by an object is the magnitude of length from one position to the another. It is a scalar quantity.
While displacement is the distance covered in a specific direction. Displacement is a vector quantity. It has both magnitude and direction.
If an object move in a straight path without going back, then, the magnitude of distance will be the same with the magnitude of displacement.
Both distance and displacement are measured in the same unit which is metres.
Therefore, an object have to take a straight path without going back to have the distance and the displacement equal.
A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastward. What are the magnitude (in T) and direction of the magnetic field at this instant?
Answer:
The values is [tex]B = 3.2 *10^{-8} \ T[/tex]
The direction is out of the plane
Explanation:
From the question we are told that
The magnitude of the electric field is [tex]E = 9.6 \ V/m[/tex]
The magnitude of the magnetic field is mathematically represented as
[tex]B = \frac{E}{c}[/tex]
where c is the speed of light with value
[tex]B = \frac{ 9.6}{3.0 *10^{8}}[/tex]
[tex]B = 3.2 *10^{-8} \ T[/tex]
Given that the direction off the electromagnetic wave( c ) is northward(y-plane ) and the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page (z-plane )
The electric field 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge?
Answer:
2.1×10¹⁸ C
Explanation:
Using,
E = kq/r²...................... Equation 1
Where E = Electric field, q = charge, r = distance, k = coulombs constant.
make q the subject of the equation
q = Er²/k.................. Equation 2
Given: E = 180000 N/C, r = 2.8 cm = 0.028 m
Constant: k = 9×10⁹ Nm²/C².
Substitute these values into equation 2
q = 180000(9×10⁹)/0.028²
q = 2.1×10¹⁸ C
Hence the object charge is 2.1×10¹⁸ C
A simple arrangement by means of which e.m.f,s. are compared is known
Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
g Assume you are a farsighted person who has a near point distance of 40 (cm). If you use a converging contact lens with focal length of 10 (cm). What is nearest distance you can vision with you contacts now?
Answer:
object distance p = 13.33 cm
Explanation:
For this problem of finding the image of an object we must use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distances to the object and the image, respectively.
In this case they indicate the focal length f = 10 cm, since the person has hyperopia, the image must be formed q = 40 cm, let's find where the object is (p)
1 / p = 1 / f - 1 / q
1 / p = 1/10 - 1/40
1 / p = 0.075
p = 13.33 cm
Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field exceeds 0.100 T. In the absence of any external magnetic field, determine the maximum current a 5.68-mm-diameter niobium wire can carry and remain superconducting.
Answer:
The current is [tex]I = 1420 \ A[/tex]
Explanation:
From the question we are told that
The diameter of the wire is [tex]d = 5.68 \ mm = 0.00568 \ m[/tex]
The magnetic field is [tex]B = 0.100 \ T[/tex]
Generally the radius of the wire is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{ 0.00568}{2}[/tex]
[tex]r = 0.00284 \ m[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * I}{ 2 \pi r }[/tex]
=> [tex]I =\frac{ B * 2 \pi r }{\mu_o}[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]
substituting values
=> [tex]I =\frac{ 0.100 * 2 * 3.142 * 0.00284 }{ 4 \pi * 10^{-7}}[/tex]
=> [tex]I = 1420 \ A[/tex]
Structures on a bird feather act like a diffraction grating having 8500 lines per centimeter. What is the angle of the first-order maximum for 577 nm light shone through a feather?
Answer:
29.5°
Explanation:
To find the distance d
d = 1E10^-2/8500lines
= 1.17x 10-6m
But wavelength in first order maximum is 577nm
and M = 1
So
dsin theta= m. Wavelength
Theta= sin^-1 (m wavelength/d)
= Sin^-1 ( 1* 577 x10^-8m)/1.17*10^-6
= 493*10^-3= sin^-1 0.493
Theta = 29.5°
Specific heat is a measurement of the amount of heat energy input required for one gram of a substance to increase its temperature by one degree Celsius. Solid lithium has a specific heat of 3.5 J/g·°C. This means that one gram of lithium requires 3.5 J of heat to increase 1°C. Plot the temperature of 1g of lithium after 3.5, 7, and 10.5 J of thermal energy are added.
Answer:
ΔT = 1ºC , 2ºCand 3ºC
Explanation:
In this exercise they indicate the specific heat of lithium
let's calculate the temperature increase as a function of the heat introduced
Q = m [tex]c_{e}[/tex] ΔT
ΔT = Q / m c_{e}
calculate
for Q = 3.5 J
ΔT = 3.5 / (1 3.5)
ΔT = 1ºC
For Q = 7.0 J
ΔT = 7 / (1 3.5)
ΔT = 2ºC
for Q = 10.5 J
ΔD = 10.5 / (1 3.5)
ΔT = 3ºC
we see that this is a straight line, see attached
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.2 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.8 cm .
Required:
What is the magnitude of the charge (in nC) on each bead?
Answer:
The magnitude of the charge is 54.9 nC.
Explanation:
The charge on each bead can be found using Coulomb's law:
[tex] F_{e} = \frac{k*q_{1}q_{2}}{r^{2}} [/tex]
Where:
q₁ and q₂ are the charges, q₁ = q₂
r: is the distance of spring stretching = 4.8x10⁻² m
[tex]F_{e}[/tex]: is the electrostatic force
[tex] F_{e} = \frac{k*q^{2}}{r^{2}} \rightarrow q = \sqrt{\frac{F_{e}}{k}}*r [/tex]
Now, we need to find [tex]F_{e}[/tex]. To do that we have that Fe is equal to the spring force ([tex]F_{k}[/tex]):
[tex] F_{e} = F_{k} = -kx [/tex]
Where:
k is the spring constant
x is the distance of the spring = 4.8 - 4.0 = 0.8 cm
The spring constant can be found by equaling the sping force and the weight force:
[tex] F_{k} = -W [/tex]
[tex] -k*x = -m*g [/tex]
where x is 5.2 - 4.0 = 1.2 cm, m = 1.8 g and g = 9.81 m/s²
[tex] k = \frac{mg}{x} = \frac{1.8 \cdot 10^{-3} kg*9.81 m/s^{2}}{1.2 \cdot 10^{-2} m} = 1.47 N/m [/tex]
Now, we can find the electrostatic force:
[tex] F_{e} = F_{k} = -kx = -1.47 N/m*0.8 \cdot 10^{-2} m = -0.0118 N [/tex]
And with the magnitude of the electrostatic force we can find the charge:
[tex]q = \sqrt{\frac{F_{e}}{k}}*r = \sqrt{\frac{0.0118 N}{9 \cdot 10^{9} Nm^{2}/C^{2}}}*4.8 \cdot 10^{-2} m = 54.9 \cdot 10^{-9} C = 54.9 nC[/tex]
Therefore, the magnitude of the charge is 54.9 nC.
I hope it helps you!
The magnitude of the charge (in nC) on each bead is equal to 55.21 nC.
Given the following data:
Original length = 4.0 cm to m = 0.04 mMass = 1.8 grams to kg = 0.0018New length = 5.2 cm to m = 0.052.Final length = 4.8 cm to m = 0.048 m.Extension, e = [tex]0.052 - 0.048[/tex] = 0.012 m
Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]Coulomb's constant = [tex]8.99 \times 10^9\; Nm^2/C^2[/tex]To calculate the magnitude of the charge (in nC) on each bead, we would apply Coulomb's law:
First of all, we would determine the spring constant of this lightweight spring by using this formula:
[tex]W = mg = Ke \\\\K=\frac{mg}{e} \\\\K=\frac{0.0018 \times 9.8}{0.012} \\\\K=\frac{0.01764}{0.012}[/tex]
Spring constant, K = 1.47 N/m.
For the electrostatic force:
[tex]F = ke\\\\F = 1.47 \times 0.08[/tex]
F = 0.01176 Newton.
Coulomb's law of electrostatic force.
Mathematically, the charge in an electric field is given by this formula:
[tex]q = \sqrt{\frac{F}{k} } \times r[/tex]
Substituting the given parameters into the formula, we have;
[tex]q = \sqrt{\frac{0.01176 }{8.99 \times 10^9} } \times 0.048\\\\q=\sqrt{1.3228 \times 10^{-12}} \times 0.048\\\\q=1.1502 \times 10^{-6} \times 0.048\\\\q= 5.521 \times 10^{-8}\;C[/tex]
Note: 1 nC = [tex]1 \times 10^{-9}\;C[/tex]
Charge, q = 55.21 nC.
Read more on electric field here: https://brainly.com/question/14372859