Answer:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
[tex]E=\frac{F}{Q}[/tex]
But the force F
[tex]F= \frac{kQ1Q2}{r^2}[/tex]
But the electric field intensity due to a point charge Q at a distance r meters away is given by
[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
In a two-slit experiment, the slit separation is 3.34 ⋅ 10 − 5 m. The interference pattern is created on a screen that is 3.30 m away from the slits. If the 7th bright fringe on the screen is 29.0 cm away from the central fringe, what is the wavelength of the light?
Answer:
The wavelength is [tex]\lambda = 419 \ nm[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 3.34 *10^{-5} \ m[/tex]
The distance of the screen is [tex]D = 3.30 \ m[/tex]
The order of the fringe is n = 7
The distance of separation of fringes is y = 29.0 cm = 0.29 m
Generally the wavelength of the light is mathematically represented as
[tex]\lambda = \frac{y * d }{ n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}[/tex]
[tex]\lambda = 4.19*10^{-7}\ m[/tex]
[tex]\lambda = 419 \ nm[/tex]
There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is located 1.03 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
A particle with charge q and momentum p, initially moving along the x-axis, enters a region where a uniform magnetic field* B=(B0)(k) extends over a width x=L. The particle is deflected a distance d in the +y direction as it traverses the field. Determine the magnitude of the momentum (p).
Answer:
Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.
Explanation:
So, from the question, we are given that the charge = q, the momentum = p.
=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."
Momentum,P = mass × Velocity, v -----(1).
We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.
Radius,r = P × v / B0 -----------------------------(2).
Centrepetal force = q × B0 × v. ----------(3).
(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.
Equating Equation (2) and (3) gives;
P = B0 × q × r.
Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.
) Calculate current passing in an electrical circuit if you know that the voltage is 8 volts and the resistance is 10 ohms
Explanation:
Hey, there!
Here, In question given that,
potential difference (V)= 8V
resistance (R)= 10 ohm
Now,
According to the Ohm's law,
V= R×I { where I = current}
or, I = V/R
or, I = 8/10
Therefore, current is 4/5 A or 0.8 A.
(A= ampere = unit of current).
Hope it helps...
Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visible light with no greater than 1.0 mW total power. They're relatively safe because the eye's blink reflex limits exposure time to 250 ms.
Requried:
a. Find the intensity of a 1-mW class 2 laser with beam diameter 2.0 mm .
b. Find the total energy delivered before the blink reflex shuts the eye.
c. Find the peak electric field in the laser beam.
Answer:
a) 318.2 W/m^2
b) 2.5 x 10^-4 J
c) 1.55 x 10^-8 v/m
Explanation:
Power of laser P = 1 mW = 1 x 10^-3 W
exposure time t = 250 ms = 250 x 10^-3 s
If beam diameter = 2 mm = 2 x 10^-3 m
then
cross-sectional area of beam A = [tex]\pi d^{2} /4[/tex] = (3.142 x [tex](2*10^{-3} )^{2}[/tex])/4
A = 3.142 x 10^-6 m^2
a) Intensity I = P/A
where P is the power of the laser
A is the cros-sectional area of the beam
I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2
b) Total energy delivered E = Pt
where P is the power of the beam
t is the exposure time
E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J
c) The peak electric field is given as
E = [tex]\sqrt{2I/ce_{0} }[/tex]
where I is the intensity of the beam
E is the electric field
c is the speed of light = 3 x 10^8 m/s
[tex]e_{0}[/tex] = 8.85 x 10^9 m kg s^-2 A^-2
E = [tex]\sqrt{2*318.2/3*10^8*8.85*10^9}[/tex] = 1.55 x 10^-8 v/m
(a) The intensity of laser beam is [tex]318.2 \;\rm W/m^{2}[/tex].
(b) The total energy delivered before the blink reflex shuts the eye is [tex]2.5 \times 10^{-4} \;\rm J[/tex].
(c) The required value of peak electric field in the laser beam is [tex]1.55 \times 10^{-8} \;\rm V/m[/tex].
Given data:
The power of laser is, [tex]P=1 \;\rm mW = 1 \times 10^{-3} \;\rm W[/tex].
The exposure time is, [tex]t = 250\;\rm ms = 250 \times 10^{-3} \;\rm s[/tex].
The beam diameter is, [tex]d = 2 \;\rm mm = 2 \times 10^{-3} \;\rm m[/tex].
a)
The standard expression for the intensity of beam is given as,
I = P/A
Here, P is the power of the laser and A is the cross-sectional area of the beam. And its value is,
[tex]A =\pi /4 \times d^{2}\\\\A =\pi /4 \times (2 \times 10^{-3})^{2}\\\\A =3.142 \times 10^{-6} \;\rm m^{2}[/tex]
Then intensity is,
[tex]I = (1 \times 10^{-3})/(3.142 \times 10^{-6})\\\\I =318.2 \;\rm W/m^{2}[/tex]
Thus, the intensity of laser beam is [tex]318.2 \;\rm W/m^{2}[/tex].
(b)
The expression for the total energy delivered is given as,
E = Pt
Solving as,
[tex]E = 1 \times 10^{-3} \times (250 \times 10^{-3})\\\\E = 2.5 \times 10^{-4} \;\rm J[/tex]
Thus, the total energy delivered before the blink reflex shuts the eye is [tex]2.5 \times 10^{-4} \;\rm J[/tex].
(c)
The expression for the peak electric field is given as,
[tex]E = \sqrt{\dfrac{2I}{c \times \epsilon_{0}}}[/tex]
Solving as,
[tex]E = \sqrt{\dfrac{2 \times 318.2}{(3 \times 10^{8}) \times (8.85 \times 10^{9})}}\\\\E =1.55 \times 10^{-8} \;\rm V/m[/tex]
Thus, the required value of peak electric field in the laser beam is [tex]1.55 \times 10^{-8} \;\rm V/m[/tex].
Learn more about the laser intensity here:
https://brainly.com/question/24258754
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.
Answer:
a) k = 95.54 N / m, c = 19.55 , b) m₃ = 0.9078 kg
Explanation:
In a simple harmonic movement with friction, we can assume that this is provided by the speed
fr = -c v
when solving the system the angular value remains
w² = w₀² + (c / 2m)²
They give two conditions
1) m₁ = 1 kg
f₁ = 1.1 Hz
the angular velocity is related to frequency
w = 2π f₁
Let's find the angular velocity without friction is
w₂ = k / m₁
we substitute
(2π f₁)² = k / m₁ + (c / 2m₁)²
2) m₂ = 2 kg
f₂ = 0.8 Hz
(2π f₂)² = k / m₂ + (c / 2m₂)²
we have a system of two equations with two unknowns, so we can solve it
we solve (c / 2m)² is we equalize the expression
(2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁
k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)
k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)
a) Let's calculate
k = 4 π² (0.8² -1.1²) / (½ -1/1)
k = 39.4784 (1.21) / (-0.5)
k = 95.54 N / m
now we can find the constant of friction
(2π f₁) 2 = k / m₁ + (c / 2m₁)²
c2 = ((2π f₁)² - k / m₁) 4m₁²
c2 = (4ππ² f₁² - k / m₁) 4 m₁²
let's calculate
c² = (4π² 1,1² - 95,54 / 1) 4 1²
c² = (47.768885 - 95.54) 8
c² = -382.1689
c = 19.55
b) f₃ = 0.2 Hz
m₃ =?
(2πf₃)² = k / m₃ + (c / 2m₃) 2
we substitute the values
(4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²
1.579 = 95.54 / m₃ + 95.542225 / m₃²
let's call
x = 1 / m₃
x² = 1 / m₃²
- 1.579 + 95.54 x + 95.542225 x² = 0
60.5080 x² + 60.5080 x -1 = 0
x² + x - 1.65 10⁻² = 0
x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2
x = [1 ± 1.03] / 2
x₁ = 1.015 kg
x₂ = -0.015 kg
Since the mass must be positive we eliminate the second results
x₁ = 1 / m₃
m₃ = 1 / x₁
m₃ = 1 / 1.1015
A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of the magnetic field inside the solenoid
Answer:
1.37 ×10^-3 T
Explanation:
From;
B= μnI
μ = 4π x 10-7 N/A2
n= number of turns /length of wire = 1700/0.75 = 2266.67
I= 0.48 A
Hence;
B= 4π x 10^-7 × 2266.67 ×0.48
B= 1.37 ×10^-3 T
How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?
Answer:
[tex]y = 0.0394 \ m[/tex]
Explanation:
From the question we are told that
The distance of the screen is [tex]D = 2.20 \ m[/tex]
The distance of separation of the slit is [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]
The wavelength of light is [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]
Generally the condition for constructive interference is
[tex]dsin\theta = n * \lambda[/tex]
=> [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]
here n = 1 because we are considering the central diffraction peak
=> [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]
=> [tex]\theta = 1.0274 ^o[/tex]
Generally the width of central diffraction peak on a screen is mathematically evaluated as
[tex]y = D tan (\theta )[/tex]
substituting values
[tex]y = 2.20 * tan (1.0274)[/tex]
[tex]y = 0.0394 \ m[/tex]
A circular loop of wire of area 25 cm2 lies in the plane of the paper. A decreasing magnetic field B is coming out of the paper. What is the direction of the induced current in the loop?
Answer:
counterclockwise
Explanation:
given data
area = 25 cm²
solution
We know that a changing magnetic field induces the current and induced emf is express as
[tex]\epsilon = -N \frac{d \phi }{dt}[/tex] ..................................1
and we will get here direction of the induced current in the loop that is express by the Lens law that state that the direction of induces current is such that the magnetic flux due to the induced current opposes the change in magnetic flux due to the change in magnetic field
so when magnetic field decrease and point coming out of the paper.
so induced current in the loop will be counterclockwise
Can abnormality exist outside of a cultural context
At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W
Complete Question
At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W = 1 J/s)? Assume each fission reaction releases 200 MeV of energy.
Answer
a. Approximately [tex]5*10^{10}[/tex] fissions per second.
b. Approximately [tex]6*10^{12 }[/tex]fissions per second.
c. Approximately [tex]4*10^{11}[/tex] fissions per second.
d. Approximately [tex]3*10^{12}[/tex] fissions per second.
e. Approximately[tex]3*10^{14}[/tex] fissions per second.
Answer:
The correct option is d
Explanation:
From the question we are told that
The energy released by each fission reaction [tex]E = 200 \ MeV = 200 *10^{6} * 1.60 *10^{-19} =3.2*10^{-11} \ J /fission[/tex]
Thus to generated [tex]100 \ J/s[/tex] i.e (100 W ) the rate of fission is
[tex]k = \frac{100}{3.2 *10^{-11} }[/tex]
[tex]k =3*10^{12} fission\ per \ second[/tex]
c) If the ice block (no penguins) is pressed down even with the surface and then released, it will bounce up and down, until friction causes it to settle back to the equilibrium position. Ignoring friction, what maximum height will it reach above the surface
Answer:
y = 20.99 V / A
there is no friction y = 20.99 h
Explanation:
Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy
when the block of ice is submerged it is subjected to two forces its weight hydrostatic thrust
F_net= ∑F = B-W
the expression stop pushing is
B = ρ_water g V_ice
where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice
We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³
W = ρ-ice g V
F_net = (ρ_water - ρ_ ice) g V
this is the net force directed upwards, we can find the potential energy with the expression
F = -dU / dy
ΔU = - ∫ F dy
ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy
ΔU = - (ρ_water - ρ_ ice) g A y² / 2
we evaluate between the limits y = 0, U = 0, that is, the potential energy is zero at the surface
U_ice = (ρ_water - ρ_ ice) g A y² / 2
now we can use the conservation of mechanical energy
starting point. Ice depth point
Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2
final point. Highest point of the block
[tex]Em_{f}[/tex] = U = m g y
as there is no friction, energy is conserved
Em₀ = Em_{f}
(ρ_water - ρ_ ice) g A y² / 2 = mg y
let's write the weight of the block as a function of its density
ρ_ice = m / V
m = ρ_ice V
we substitute
(ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y
y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A
let's substitute the values
y = 0.913 / (1 - 0.913) 2 V / A
y = 20.99 V / A
This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction
V / A = h
where h is the height of the block
y = 20.99 h
When the current in a toroidal solenoid is changing at a rate of 0.0200 A/s, the magnitude of the induced emf is 12.7 mV. When the current equals 1.50 A, the average flux through each turn of the solenoid is 0.00458 Wb. How many turns does the solenoid have?
Answer:
[tex]N = 208 \ turns[/tex]
Explanation:
From the question we are told that
The rate of current change is [tex]\frac{di }{dt} = 0.0200 \ A/s[/tex]
The magnitude of the induced emf is [tex]\epsilon = 12.7 \ mV = 12.7 *10^{-3} \ V[/tex]
The current is [tex]I = 1.50 \ A[/tex]
The average flux is [tex]\phi = 0.00458 \ Wb[/tex]
Generally the number of turns the number of turn the solenoid has is mathematically represented as
[tex]N = \frac{\epsilon_o * I}{ \phi * \frac{di}{dt} }[/tex]
substituting values
[tex]N = \frac{ 12.7*10^{-3} * 1.50 }{ 0.00458 * 0.0200 }[/tex]
[tex]N = 208 \ turns[/tex]
Which statement belongs to Dalton’s atomic theory? Atoms have a massive, positively charged center. Atoms cannot be created or destroyed. Atoms can be broken down into smaller pieces. Electrons are located in energy levels outside of the nucleus.
Answer:
the correct statement is
* atoms cannot be created or destroyed
Explanation:
The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.
Based on the previous descriptive, the correct statement is
* atoms cannot be created or destroyed
Answer:
the Answer is b hope it help
Explanation:
A resistor and an inductor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the inductor is
Answer:
The voltage is equal to the batteries terminal voltage
Explanation:
Explanation:
5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be changed to raise the wave speed to 180 m/s?
Answer:
The tension on string when the speed was raised is 134.53 N
Explanation:
Given;
Tension on the string, T = 120 N
initial speed of the transverse wave, v₁ = 170 m/s
final speed of the transverse wave, v₂ = 180 m/s
The speed of the wave is given as;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
μ is mass per unit length
[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]
The final tension T₂ will be calculated as;
[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]
Therefore, the tension on string when the speed was raised is 134.53 N
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely immersed in water, the scale reads 18. 2 N. What are the volume and density of the block?
Answer:
7066kg/m³
Explanation:
The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:
Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)
= 7066kg/m³
Explanation:
Answer:
The volume of the block is 306 cm³
The density of the block is 7.07 g/cm³
Explanation:
Given;
weight of block in air, [tex]W_a[/tex] = 21.2 N
Weight of block in water, [tex]W_w[/tex] = 18.2 N
Mass of the block in air;
[tex]W_a = mg[/tex]
21.2 = m x 9.8
m = 21.2 / 9.8
m = 2.163 kg
mass of the block in water;
[tex]W_w = mg[/tex]
18.2 = m x 9.8
m = 18.2 / 9.8
m = 1.857 kg
Apply Archimedes principle
Mass of object in air - mass of object in water = density of water x volume of object
2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block
0.306 kg = 1000 kg/m³ x Volume of block
Volume of the block = [tex]\frac{0.306 \ kg}{1000 \ kg/m^3}[/tex]
Volume of the block = 3.06 x 10⁻⁴ m³
Volume of the block = 306 cm³
Determine the density of the block
[tex]Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3[/tex]
Scouts at a camp shake the rope bridge they have just crossed and observe the wave crests to be 9.70 m apart. If they shake the bridge twice per second, what is the propagation speed of the waves (in m/s)?
Answer:
The speed of the wave is 19.4 m/s
Explanation:
The wave's crest to crest distance (the wavelength of this rope's wave) λ= 9.70 m
The bridge is shaken twice, meaning that two wavelengths passed a given point on the rope per sec. The frequency of a wave is the amount of that wave that passes a given point in a second.
this means that the frequency f = 2 Hz
The speed of a wave = fλ = 9.70 x 2 = 19.4 m/s
W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively. Any energy not mentioned in the transformation is assumed to remain constant; if work is not mentioned, it is assumed to be zero.
1. Give a specific example of a system with the energy transformation shown.
W→ΔEth
2. Give a specific example of a system with the energy transformation shown.
a. Rolling a ball up a hill.
b. Moving a block of wood across a horizontal rough surface at constant speed.
c. A block sliding on level ground, to which a cord you are holding on to is attached .
d. Dropping a ball from a height.
Answer:
1) a block going down a slope
2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK
Explanation:
In this exercise you are asked to give an example of various types of systems
1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.
2)
a) rolling a ball uphill
In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact
W = ΔU + ΔK + ΔE
b) in this system work is transformed into internal energy
W = ΔE
c) There is no friction here, therefore the work is transformed into kinetic energy
W = ΔK
d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy
ΔU = ΔK
The velocity function (in meters per second) is given for a particle moving along a line. Find the total distance traveled by the particle during the given interval
Answer:
s=((vf+vi)/2)t vf is final velocity and vi is initial velocity
A thermos bottle works well because:
a. its glass walls are thin
b. silvering reduces convection
c. vacuum reduces heat radiation
d. silver coating is a poor heat conductor
e. none of the above
Answer:
A thermos bottle works well because:
A) Its glass walls are thin
Answer:
A thermos bottle works well because:
C
Vacuum reduces heat radiation
A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .Part A. Calcualte the coil's self-inductance.Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf froma to b or from b to a?
Complete Question
A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .
Part A. Calculate the coil's self-inductance.
Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.
Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?
Answer:
Part A
[tex]L = 0.000863 \ H[/tex]
Part B
[tex]\epsilon = 0.863 \ V[/tex]
Part C
From terminal a to terminal b
Explanation:
From the question we are told that
The number of turns is [tex]N = 590 \ turns[/tex]
The cross-sectional area is [tex]A = 6.20 cm^2 = 6.20 *10^{-4} \ m[/tex]
The radius is [tex]r = 5.0 \ cm = 0.05 \ m[/tex]
Generally the coils self -inductance is mathematically represented as
[tex]L = \frac{ \mu_o N^2 A }{2 \pi * r }[/tex]
Where [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]L = \frac{ 4\pi * 10^{-7} * 590^2 6.20 *10^{-4} }{2 \pi * 0.05 }[/tex]
[tex]L = \frac{ 2 * 10^{-7} * 590^2 6.20 *10^{-4} }{ 0.05 }[/tex]
[tex]L = 0.000863 \ H[/tex]
Considering the Part B
Initial current is [tex]I_1 = 5.00 \ A[/tex]
Current at time t is [tex]I_t = 3.0 \ A[/tex]
The time taken is [tex]\Delta t = 3.00 ms = 0.003 \ s[/tex]
The self-induced emf is mathematically evaluated as
[tex]\epsilon = L * \frac{\Delta I}{ \Delta t }[/tex]
=> [tex]\epsilon = L * \frac{ I_1 - I_t }{ \Delta t }[/tex]
substituting values
[tex]\epsilon = 0.000863 * \frac{ 5- 2 }{ 0.003 }[/tex]
[tex]\epsilon = 0.863 \ V[/tex]
The direction of the induced emf is from a to b because according to Lenz's law the induced emf moves in the same direction as the current
This question involves the concepts of the self-inductance, induced emf, and Lenz's Law
A. The coil's self-inductance is "0.863 mH".
B. The self-induced emf in the coil is "0.58 volts".
C. The direction of the induced emf is "from b to a".
A.
The self-inductance of the coil is given by the following formula:
[tex]L=\frac{\mu_oN^2A}{2\pi r}[/tex]
where,
L = self-inductance = ?
[tex]\mu_o[/tex] = permeability of free space = 4π x 10⁻⁷ N/A²
N = No. of turns = 590
A = Cross-sectional area = 6.2 cm² = 6.2 x 10⁻⁴ m²
r = radius = 5 cm = 0.05 m
Therefore,
[tex]L=\frac{(4\pi\ x\ 10^{-7}\ N/A^2)(590)^2(6.2\ x\ 10^{-4}\ m^2)}{2\pi(0.05\ m)}[/tex]
L = 0.863 x 10⁻³ H = 0.863 mH
B.
The self-induced emf is given by the following formula:
[tex]E=L\frac{\Delta I}{\Delta t}\\\\[/tex]
where,
E = self-induced emf = ?
ΔI = change in current = 2 A
Δt = change in time = 3 ms = 0.003 s
Therefore,
[tex]E=(0.000863\ H)\frac{2\ A}{0.003\ s}[/tex]
E = 0.58 volts
C.
According to Lenz's Law, the direction of the induced emf always opposes the change in flux that causes it. Hence, the direction of the induced emf will be from b to a.
Learn more about Lenz's Law here:
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When a monochromatic light of wavelength 433 nm incident on a double slit of slit separation 6 µm, there are 5 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.9 nm for the same double slit?
Answer:
The number of interference fringes is [tex]n = 3[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 433 \ nm = 433 *10^{-9} \ m[/tex]
The distance of separation is [tex]d = 6 \mu m = 6 *10^{-6} \ m[/tex]
The order of maxima is m = 5
The condition for constructive interference is
[tex]d sin \theta = n \lambda[/tex]
=> [tex]\theta = sin^{-1} [\frac{5 * 433 *10^{-9}}{ 6 *10^{-6}} ][/tex]
=> [tex]\theta = 21.16^o[/tex]
So at
[tex]\lambda_1 = 632.9 nm = 632.9*10^{-9} \ m[/tex]
[tex]6 * 10^{-6} * sin (21.16) = n * 632.9 *10^{-9}[/tex]
=> [tex]n = 3[/tex]
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 580 nm
Answer:
Explanation:
In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is
2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .
2 x 1.34 x t = 1 x 580
t = 216.42 nm .
Thickness must be 216.42 nm .
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°
A straight wire that is 0.56 m long is carrying a current of 2.6 A. It is placed in a uniform magnetic field, where it experiences a force of 0.24 N. The wire makes an angle of 900 with the magnetic field. What is the magnitude of the magnetic field
Answer:
0.165TeslaExplanation:
The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;
B is the magnetic field (in Tesla)
I is the current (in amperes)
L is the length of the wire (in meters)
∝ is the angle that the conductor makes with the magnetic field.
Given parameters
L = 0.56 m
I = 2.6A
F = 0.24N
∝ = 90°
Required
magnitude of the magnetic field (B)
Substituting the given values into the formula given above we will have;
F = BILsin∝
0.24 = B * 2.6 * 0.56 sin90°
0.24 = B * 2.6 * 0.56 (1)
0.24 = 1.456B
1.456B = 0.24
Dividing both sides by 1.456 will give;
1.456B/1.456 = 0.24/1.456
B ≈ 0.165Tesla
Hence the magnitude of the magnetic field is approximately 0.165Tesla
A solenoid inductor has an emf of 0.80 V when the current through it changes at the rate 10.0 A/s. A steady current of 0.20 A produces a flux of 8.0 μWb per turn.
Required:
How many turns does the inductor have?
Answer:
The number of turns of the inductor is 2000 turns.
Explanation:
Given;
emf of the inductor, E = 0.8 V
the rate of change of current with time, dI/dt = 10 A/s
steady current in the solenoid, I = 0.2 A
flux per turn, Ф = 8.0 μWb per
Determine the inductance of the solenoid, L
E = L(dI/dt)
L = E / (dI/dt)
L = 0.8 / (10)
L = 0.08 H
The inductance of the solenoid is given by;
[tex]L = \frac{\mu_o N^2 A}{l}[/tex]
Also, the magnetic field of the solenoid is given by;
[tex]B = \frac{\mu_o NI}{l}[/tex]
I is 0.2 A
[tex]B = \frac{\mu_oN(0.2)}{l} = \frac{0.2\mu_o N}{l}[/tex]
[tex]\frac{B}{0.2 } = \frac{\mu_o N}{l}[/tex]
[tex]L = \frac{\mu_o N^2 A}{l} \\\\L = \frac{\mu_o N }{l} (NA)\\\\L = \frac{B}{0.2} (NA)\\\\L = \frac{BA}{0.2} (N)[/tex]
But Ф = BA
[tex]L = \frac{\phi N}{0.2} \\\\\phi N = 0.2 L\\\\N = \frac{0.2 L}{\phi} \\\\N = \frac{0.2 *0.08}{8*10^{-6}}\\\\N = 2000 \ turns[/tex]
Therefore, the number of turns of the inductor is 2000 turns.
This question involves the concepts of magnetic flux, magnetic field, and inductance.
The inductor has "2000" turns.
The magnetic field due to an inductor coil is given as follows:
[tex]B=\frac{\mu_o NI}{L}\\\\[/tex]
where,
B = magnetic field
μ₀ = permeability of free space \
N = No. of turns
I = current = 0.2 A
L = length of inductor
Therefore,
[tex]\frac{\mu_oN}{L}=\frac{B}{0.2\ A}---------- eqn(1)[/tex]
Now, the inductance of a solenoid is given by the following formula:
[tex]E = L\frac{dI}{dt}\\\\L = \frac{E}{\frac{dI}{dt}}[/tex]
The inductance of solenoid can also be given using the following formula:
[tex]L = \frac{\mu_o N^2A}{L}[/tex]
comparing both the formulae, we get:
[tex]\frac{E}{\frac{dI}{dt}}= \frac{\mu_oN^2A}{L}\\\\E=\frac{dI}{dt}\frac{\mu_oN}{l}(NA)\\\\using\ eqn (1):\\\\E=\frac{dI}{dt}\frac{B}{0.2}(NA)\\\\[/tex]
where,
BA = magnetic flux = [tex]\phi[/tex] = 8 μWb/turn = 8 x 10⁻⁶ Wb/turn
N = No. of turns = ?
E = E.M.F = 0.8 volts
[tex]\frac{dI}{dt}[/tex] = rate of change in current = 10 A/s
Therefore,
[tex]0.8=(10)\frac{8\ x\ 10^{-6}}{0.2}N\\\\N=\frac{(0.8)(0.2)}{8\ x\ 10^{-5}}[/tex]
N = 2000 turns
Learn more about magnetic flux here:
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The attached picture shows the magnetic flux.
Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of the pipe is R/3
Answer:
v₂ = 9 v
Explanation:
For this exercise in fluid mechanics, let's use the continuity equation
v₁ A₁ = v₂ A₂
where v is the velocity of the fluid, A the area of the pipe and the subscripts correspond to two places of interest.
The area of a circle is
A = π R²
let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint
In this case v₁ = v and the area is
A₁ = π R²
in the second point
A₂= π (R / 3)²
we substitute in the continuity equation
v π R² = v₂ π R² / 9
v = v₂ / 9
v₂ = 9 v
When using science to investigate physical phenomena, which characteristic of the event must exist? predictable repeatable provable readable
Answer:
Not sure but I believe predictable.
Explanation:
Phenomena usually consists of :
- A history, a date in which the physical phenomenon has occurred.
- A source, a place or reason to why or where the physical phenomena has occured.
According to this, I want to say predictable.
It is not repeatable, there are one-time phenomenons that have occurred that scientists to this day still have not recorded again such as the Big Bang.
It is not provable. Most of the theories earlier scientists and historians have predicted have not yet been proved. Yet rather, somehow, they have been explored and investigated.
It is not readable. This is self explanatory, some things scientists investigate are not written down, nor read. It starts with a mental theory and then immediately goes to the next phase of investigation.
Given three resistors of different values, how many possible resistance values could be obtained by using one or more of the resistors?
Answer:
8 possible combinations
Assuming R 1, R 2 and R 3 be three different Resistance
1- all three in series
2-all three in parallel
3- R 1 and R 2 in series and parallel with R 3
4-R 1 and R 3 in series and parallel with R 2
5-R 2 and R 3 in series and parallel with R 1
6- R 1 and R 2
in parallel and series with R 3
7-R 1 and R 3 in parallel and series with R 2
8-R 2 and R 3 in V with R 1
