Answer:
The direction of the force is towards the East.
Explanation:
Using the right hand rule, the force on the current carrying conductor is east.
In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.
In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled
Answer:
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
Explanation:
The diffraction of a slit is given by the expressions
a sin θ = m λ
where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.
sin θ = m λ / a
If this equation
a ’= 2 a
we substitute
2 a sin θ'= m λ
sin θ'= (m λ / a) 1/2
sin θ ’= sin θ / 2
We can use trigonometry to find the width
tan θ = y / L
as the angle is small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
y ’/ L = y/L 1/2
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment
Answer: Speed = [tex]3.10^{-31}[/tex] m/s
Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:
[tex]p_{f} = p_{i}[/tex]
Relativistic momentum is calculated as:
p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]
where:
m is rest mass
u is velocity relative to an observer
c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)
Initial momentum is zero, then:
[tex]p_{f}[/tex] = 0
[tex]p_{1}-p_{2}[/tex] = 0
[tex]p_{1} = p_{2}[/tex]
To find speed of the heavier fragment:
[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]
[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]
[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]
[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]
[tex]u_{1} = 3.10^{-31}[/tex]
The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.
The ancient Greek Eratosthenes found that the Sun casts different lengths of shadow at different points on Earth. There were no shadows at midday in Aswan as the Sun was directly overhead. 800 kilometers north, in Alexandria, shadow lengths were found to show the Sun at 7.2 degrees from overhead at midday. Use these measurements to calculate the radius of Earth.
Answer:
The radius of the earth is [tex]r = 6365.4 \ km[/tex]
Explanation:
From the question we are told that
The distance at Alexandria is [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]
The angle of the sun is [tex]\theta = 7.2 ^o[/tex]
So we want to first obtain the circumference of the earth
So let assume that the earth is circular ([tex]360 ^o[/tex])
Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many [tex](7.2 ^o)[/tex] are in [tex]360^o[/tex]
i.e [tex]N = \frac{360}{7.2}[/tex]
=> [tex]N = 50[/tex]
With this value we can evaluate the circumference as
[tex]c = 50 * 800[/tex]
[tex]c = 40000 \ km[/tex]
Generally circumference is mathematically represented as
[tex]c = 2\pi r[/tex]
[tex]40000 = 2 * 3.142 * r[/tex]
=> [tex]r = 6365.4 \ km[/tex]
A small helium-neon laser emits red visible light with a power of 5.40 mW in a beam of diameter 2.30 mm.
Required:
a. What is the amplitude of the electric field of the light? Express your answer with the appropriate units.
b. What is the amplitude of the magnetic field of the light?
c. What is the average energy density associated with the electric field? Express your answer with the appropriate units.
d. What is the average energy density associated with the magnetic field? Express your answer with the appropriate units.
E) What is the total energy contained in a 1.00-m length of the beam? Express your answer with the appropriate units.
Answer:
A. 990v/m
B.330x10^-8T
C.2.19x10^-6J/m³
D.1.45x10^-11J
Explanation:
See attached file
light of wavelength 550 nm is incident on a diffraction grating that is 1 cm wide and has 1000 slits. What is the dispersion of the m = 2 line?
Answer:
The dispersion is [tex]D = 2.01220 *10^{5} \ rad/m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ = 550 *10^{-9} \ n[/tex]
The width of the grating is[tex]k = 1\ cm = 0.01 \ m[/tex]
The number of slit is N = 1000 slits
The order of the maxima is m = 2
Generally the spacing between the slit is mathematically represented as
[tex]d = \frac{k}{N}[/tex]
substituting values
[tex]d = \frac{ 0.01}{1000}[/tex]
[tex]d = 1.0 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]d\ sin(\theta ) = m * \lambda[/tex]
substituting values
[tex]1.0 *10^{-5} sin (\theta) = 2 * 550 *10^{-9}[/tex]
[tex]\theta = sin^{-1} [\frac{ 2 * 550 *10^{-9}}{ 1.0 *10^{-5}} ][/tex]
[tex]\theta = 6.315^o[/tex]
Generally the dispersion is mathematically represented as
[tex]D = \frac{ m }{d cos(\theta )}[/tex]
substituting values
[tex]D = \frac{ 2 }{ 1.0 *10^{-5} cos(6.315 )}[/tex]
[tex]D = 2.01220 *10^{5} \ rad/m[/tex]
How can I solve this?
I managed to find Part A, but I got stuck trying to find Part B and C
Answer:
Parte B : 31.18º , Parte C: 31.17º
Explanation:
Parte B: The angle between the glass and the water before it enters the water is going to be equal to the value of the angle when it enters the glass , 27.13º.
Using the formula n1 senθ1 = n2 senθ2 , where n1=1.51 , θ1=27.13º, n2=1.33 , it gives us θ2=31.18º.
Parte C: n1= 1 , θ1=43.5º, n2=1.33
Using the same formula : n1 senθ1 = n2 senθ2 , it gives us θ2= 31.17º.
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:
1.
(a)
P
center
Figure 1
A ball is released at point P with a tangential velocity of 5 ms to move in a circular track in a
vertical plane as shown in the Figure 1. Can the ball reach the highest point of the circular track
of radius 1.0 m? Give reasons. (4 marks]
Answer:
No.
Explanation:
Given the following :
Velocity (V) of ball = 5m/s
Radius = 1m
Can the ball reach the highest point of the circular track
of radius 1.0 m?
The highest point in the track could be considered as the diameter of the circle :
Radius = diameter / 2;
Diameter = (2 * Radius) = (2*1) = 2
Maximum height which the ball can reach :
Using the relation :
Kinetic Energy = Potential Energy
0.5mv^2 = mgh
0.5v^2 = gh
0.5(5^2) = 9.8h
0.5 * 25 = 9.8h
12.5 = 9.8h
h = 12.5 / 9.8
h = 1.2755
h = 1.26m
Therefore maximum height which can be reached is 1.26m.
Since h < Diameter
A light beam has a wavelength of 330 nm in a material of refractive index 1.50. In a material of refractive index 2.50, its wavelength will be In a material of refractive index 2.50, its wavelength will be:_________
a. 495 nm .
b. 330 nm .
c. 220 nm .
d. 198 nm .
e. 132 nm .
Answer:
The wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm
d. 198 mm
Explanation:
Refractive index is given by;
[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}}[/tex]
where;
[tex]\lambda_{vacuum}[/tex] is the wavelength of the light beam in vacuum
[tex]\lambda_{medium}[/tex] is the wavelength of the beam in a material
[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}} \\\\\lambda_{vacuum} = \mu *\lambda _{medium}\\\\\ the \ wavelength \ of \ the \ light \ beam \ is \ constant \ in \ a \ vacuum\\\\ \mu_1 *\lambda _{medium}_1 = \mu_2 *\lambda _{medium}_2\\\\\lambda _{medium}_2 = \frac{ \mu_1 *\lambda _{medium}_1 }{ \mu_2} \\\\\lambda _{medium}_2 =\frac{1.5*330}{2.5} \\\\\lambda _{medium}_2 = 198 \ mm[/tex]
Therefore, the wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm.
d. 198 mm
Light of wavelength 520 nm is used to illuminate normally two glass plates 21.1 cm in length that touch at one end and are separated at the other by a wire of radius 0.028 mm. How many bright fringes appear along the total length of the plates.
Answer:
The number is [tex]Z = 216 \ fringes[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 520 \ nm = 520 *10^{-9} \ m[/tex]
The length of the glass plates is [tex]y = 21.1cm = 0.211 \ m[/tex]
The distance between the plates (radius of wire ) = [tex]d = 0.028 mm = 2.8 *10^{-5} \ m[/tex]
Generally the condition for constructive interference in a film is mathematically represented as
[tex]2 * t = [m + \frac{1}{2} ]\lambda[/tex]
Where t is the thickness of the separation between the glass i.e
t = 0 at the edge where the glasses are touching each other and
t = 2d at the edge where the glasses are separated by the wire
m is the order of the fringe it starts from 0, 1 , 2 ...
So
[tex]2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=> [tex]2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=>
[tex]m = 215[/tex]
given that we start counting m from zero
it means that the number of bright fringes that would appear is
[tex]Z = m + 1[/tex]
=> [tex]Z = 215 +1[/tex]
=> [tex]Z = 216 \ fringes[/tex]
I need help on Weight vs mass.
The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more-approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater than 100 kilograms. In common usage, the mass of an object is often referred to as its weight, though these are in fact different concepts and quantities. In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity. In other words, an object with a mass of 1.0 kilogram weighs approximately 9.81 newtons. Weight and mass are considered to be the same quantities. But many people tend to misuse these terms in their daily conversations. The main difference between weight and mass is that weight is the force of gravity by which the earth attracts towards it whereas mass is the amount of matter in an object.
which is example of radiation
Answer:
Ultraviolet light from the sun.
Explanation:
This is an example of radiation.
Answer:
X-Ray
Explanation:
x-Ray is an example of radiation.
The angle of the resultant vector is equal to
the inverse tangent of the quotient of the x-component divided by the y-component of the resultant vector
the inverse cosine of the quotient of the y-component divided by the x-component of the resultant vector.
the inverse cosine of the quotient of the x-component divided by the y-component of the resultant vector.
the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
The angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
To find the angle of a resultant vector, the vector must be resolved into y-component and x-component.
The y-component of a vector is the product of the magnitude of the vector and the sine of the angle of the vector to the horizontal. The x-component of a vector is the product of the magnitude of the vector and the cosine of the angle of the vector to the horizontal.The angle of this resultant vector is also known as the direction of the vector.
Mathematically, the direction of a resultant vector is given as;
[tex]\theta = tan^{-1} (\frac{R_y}{R_x} )\\\\where;\\\\\theta \ is \ the \ direction \ of \ the \ resultant \ vetcor\\\\R_y \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ y - direction\\\\R_x \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ x - direction[/tex]
Therefore, the angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
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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
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
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The attached picture shows the magnetic flux.
A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exerts a resistive force with magnitude proportional to the square of the speed with k
Answer:
The velocity is 40 ft/sec.
Explanation:
Given that,
Force = 3200 lb
Angle = 30°
Speed = 64 ft/s
The resistive force with magnitude proportional to the square of the speed,
[tex]F_{r}=kv^2[/tex]
Where, k = 1 lb s²/ft²
We need to calculate the velocity
Using balance equation
[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]
Put the value into the formula
[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]
Put the value of k
[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]
[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]
At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]
So, [tex]1600-v^2=0[/tex]
[tex]v=\sqrt{1600}[/tex]
[tex]v=40\ ft/sec[/tex]
Hence, The velocity is 40 ft/sec.
At a department store, you adjust the mirrors in the dressing room so that they are parallel and 6.2 ft apart. You stand 1.8 ft from one mirror and face it. You see an infinite number of reflections of your front and back.(a) How far from you is the first "front" image? ft (b) How far from you is the first "back" image? ft
Answer:
a) 3.6 ft
b) 12.4 ft
Explanation:
Distance between mirrors = 6.2 ft
difference from from the mirror you face = 1.8 ft
a) you stand 1.8 ft in front of the mirror you face.
According to plane mirror rules, the image formed is the same distance inside the mirror surface as the distance of the object (you) from the mirror surface. From this,
your distance from your first "front" image = 1.8 ft + 1.8 ft = 3.6 ft
b) The mirror behind you is 6.2 - 1.8 = 4.4 ft behind you.
the back mirror will be reflected 3.6 + 4.4 = 8 ft into the front mirror,
the first image of your back will be 4.4 ft into the back mirror,
therefore your distance from your first "back" image = 8 + 4.4 = 12.4 ft
Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors
Answer:
it is going to D. all of these are resistors
A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the highest point, as a wave passes. If the ripples decreaseto 4.7 cm, by what factor does thebug's maximum KE change?
Answer:
factor that bug maximum KE change is 0.52284
Explanation:
given data
vertical distance = 6.5 cm
ripples decrease to = 4.7 cm
solution
We apply here formula for the KE of particle that executes the simple harmonic motion that is express as
KE = (0.5) × m × A² × ω² .................1
and kinetic energy is directly proportional to square of the amplitude.
so
[tex]\frac{KE2}{KE1} = \frac{A2^2}{A1^2}[/tex] .............2
[tex]\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}[/tex]
[tex]\frac{KE2}{KE1}[/tex] = 0.52284
so factor that bug maximum KE change is 0.52284
The factor does the bug's maximum KE change should be considered as the 0.52284.
Calculation of the factor:Since
vertical distance = 6.5 cm
ripples decrease to = 4.7 cm
So here we apply the given formula
KE = (0.5) × m × A² × ω² .................1
here,
kinetic energy is directly proportional to square of the amplitude.
So,
= 4.7^2/ 6.5^2
= 0.52284
hence, The factor does the bug's maximum KE change should be considered as the 0.52284.
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The resistor used in the procedures has a manufacturer's stated tolerance (percent error) of 5%. Did you results from Data Table agree with the manufacturer's statement? Explain.
Resistor Measured Resistance
100 99.1
Answer:
e% = 0.99% this value is within the 5% tolerance given by the manufacturer
Explanation:
Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by
e% = | R_nominal - R_measured | / R_nominal 100
where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors
let's apply this formula to our case
R_nominal = 10 kΩ = 10000 Ω
R_measured = 100 99 Ω
e% = | 10000 - 10099.1 | / 10000 100
e% = 0.99%
this value is within the 5% tolerance given by the manufacturer
A current-carrying loop of wire lies flat on a horizontal tabletop. When viewed from above, the current moves around the loop in a counterclockwise sense. For points on the tabletop outside the loop, the magnetic field lines caused by this current
a. circle the loop in a counterclockwise direction.
b. point straight up. point straight down.
c. circle the loop in a clockwise direction.
Answer: i dont do physics yet lol
Explanation:
Experiments are performed with ultracold neutrons having velocities of 7.54 m/s. (a) What is the wavelength (in nm) of such a neutron
Answer:
λ = 52.5 nm
Explanation:
De Broglie's duality principle states that all matter has wave and particle characteristics, being related by the expression
p = h / λ
where the moment
p = mv
λ = h / mv
let's calculate
λ = 6.63 10⁻³⁴ / 1.675 10⁻²⁷ 7.54
λ = 5.25 10⁻⁸ m
Let's reduce anm
λ = 5.25 10⁻⁸ m (10⁹ nm / 1m)
λ = 52.5 nm
1. What does the acronym LASER stand for? What characteristic of a laser makes it suitable for today's experiment?
Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.
Explanation:
Answer:
Light amplification by stimulated emission of radiation
A hot cup of coffee is placed on a table. Which will happen because of conduction? Answer options with 4 options A. The temperature of the coffee will decrease while the temperature of the table decreases. B. The temperature of the coffee will increase while the temperature of the table increases. C. The temperature of the coffee will decrease while the temperature of the table increases. D. The temperature of the coffee will increase while the temperature of the table decreases.
Answer:
C.Explanation:
C. The temperature of the coffee will decrease while the temperature of the table increases.
Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.
Complete Question
An emf is induced in a conducting loop of wire 1.12m long as its shape is.
changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.
Answer:
The induced emf is [tex]\epsilon = 0.0863 \ V[/tex]
Explanation:
From the question we are told that
The time taken is [tex]\Delta t = 0.125 \ s[/tex]
The magnitude of the magnetic field is B = 0.504 T
The length of the loop wire is [tex]l = 1.12 \ m[/tex]
Generally the circumference of the wire when in circular form is
[tex]C = 2 \pi r[/tex]
=> [tex]l = 2 \pi r[/tex]
=> [tex]r =[/tex][tex]\frac{l}{2 \pi}[/tex]
=> [tex]r =[/tex][tex]\frac{1.12}{2 * 3.142}[/tex]
=> [tex]r =[/tex][tex]0.1782 \ m[/tex]
Now the area of the wire as a circle is
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.1782)^2[/tex]
=> [tex]A = 0.0998 \ m^2[/tex]
The length of one side of the square is
[tex]b = \frac{l}{4}[/tex]
[tex]b = \frac{1.12}{4}[/tex]
[tex]b = 0.28 \ m[/tex]
Now the area of the wire as a square is
[tex]A_s = b^2[/tex]
=> [tex]A_s =(0.28 )^2[/tex]
[tex]A_s = 0.0784 \ m^2[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = \frac{B * [A - A_s ]}{\Delta t }[/tex]
=> [tex]\epsilon = \frac{0.504 * [0.0998 - 0.0784 ]}{0.125 }[/tex]
=> [tex]\epsilon = 0.0863 \ V[/tex]
A Van de Graaff generator produces a beam of 2.02-MeV deuterons, which are heavy hydrogen nuclei containing a proton and a neutron.
A) If the beam current is 10.0 μA, how far apart are the deuterons?
B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.
Answer:
A) The distance of the deuterons from one another = 2.224× 10⁻⁷ m
B) The electrical force of repulsion among them shows a small effect in beam stability.
Explanation:
Given that:
A Van de Graaff generator produces a beam of 2.02-MeV deuterons
If the beam current is 10.0 μA, the distance of the deuterons from one another can be determined by using the concept of kinetic energy of the generator.
[tex]\mathtt{K.E = \dfrac{1}{2}mv^2}[/tex]
2 K.E = mv²
[tex]\mathtt{v^2 = \dfrac{2 K.E }{m}}[/tex]
[tex]\mathtt{v =\sqrt{ \dfrac{2 K.E }{m}}}[/tex]
so, v is the velocity of the deuterons showing the distance of the deuterons apart from one another.
[tex]\mathtt{v =\sqrt{ \dfrac{2 (2.02 \ MeV) \times \dfrac{10^6 \ eV}{ 1 \ MeV} \times \dfrac{1.60 \times 10^{-19} \ J }{1 \ eV} }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]
[tex]\mathtt{v =\sqrt{ \dfrac{6.464 \times 10^{-13} \ J }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]
v = 13911611.49 m/s
v = 1.39 × 10⁷ m/s
So, If the beam current is 10.0 μA.
We all know that:
[tex]I = \dfrac{q}{t}[/tex]
[tex]t = \dfrac{q}{I}[/tex]
[tex]\mathtt{ t = \dfrac{1.6 * 10 ^{-19} \ C}{10.0 * 10^{-6} \ A}}[/tex]
t = 1.6 × 10⁻¹⁴ s
Finally, the distance of the deuterons from one another = v × t
the distance of the deuterons from one another = (1.39 × 10⁷ m/s × 1.6 × 10⁻¹⁴ s)
the distance of the deuterons from one another = 2.224× 10⁻⁷ m
B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.
The electrical force of repulsion among them shows a small effect in beam stability. This is because, one nucleus tends to put its nearest neighbor at potential V = (k.E × q) / r = 7.3e⁻⁰³ V. This is very small compared to the 2.02-MeV accelerating potential, Thus, repulsion within the beam is a small effect.
How does a negative ion differ from an uncharged atom of the same
element?
O A. The ion has a greater number of protons.
B. The ion has fewer protons.
O C. The ion has a greater number of electrons.
O D. The ion has fewer neutrons.
Answer:
C if it is a negitive ion it has more electrons because protons determine what element it is
An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.440 mm.
1. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
2. What would be the angular position of the second-order, two-slit, interference maxima in this case?
3. Let the slits have a width 0.310 mm . In terms of the intensity I0 at the center of the central maximum, what is the intensity at the angular position of θ1?
4. What is the intensity at the angular position of θ2?
Answer:
1) θ = 0.00118 rad, 2) θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4) I / Io = 0.216
Explanation:
In the double-slit interference phenomenon it is explained for constructive interference by the equation
d sin θ = m λ
1) the first order maximum occurs for m = 1
sin θ = λ / d
θ = sin⁻¹ λ / d
let's reduce the magnitudes to the SI system
λ = 520 nm = 520 10⁻⁹ θ = 0.00118 radm
d = 0.440 mm = 0.440 10⁻³ m ³
let's calculate
θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)
θ = sin⁻¹ (1.18 10⁻³)
θ = 0.00118 rad
2) the second order maximum occurs for m = 2
θ = sin⁻¹ (m λ / d)
θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)
θ = 0.00236 rad
3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains
I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )
where the function sinc = sin x / x
and b is the width of the slits
we caption the values
x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)
x = 2.21
I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²
remember angles are in radians
I / I₀ = cos² (3.0945) [0.363] 2
I / I₀ = 0.9978 0.1318
I / I₀ = 0.1738
4) the maximum second intensity is
I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)
x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)
x = 4.41
I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²
I / Io = cos² 6.273 0.216
I / Io = 0.216
.
A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant when it has turned through 0.40 radian, what is the magnitude of the total linear acceleration of a point on the rim (radius = 13 cm)?
a. 0.31 m/s^2
b. 0.27 m/s^2
c. 0.35 m/s^2
d. 0.39 m/s^2
e. 0.45 m/s^2
Answer:
The magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
Explanation:
The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.
The radial acceleration is given by;
[tex]a_t = ar[/tex]
where;
a is the angular acceleration and
r is the radius of the circular path
[tex]a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2[/tex]
Determine time of the rotation;
[tex]\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\[/tex]
Determine angular velocity
ω = at
ω = 1.6 x 0.707
ω = 1.131 rad/s
Now, determine the radial acceleration
[tex]a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2[/tex]
The magnitude of total linear acceleration is given by;
[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2[/tex]
Therefore, the magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
Calculate the focal length (in m) of the mirror formed by the shiny bottom of a spoon that has a 3.40 cm radius of curvature. m (b) What is its power in diopters? D
Answer:
The power of the mirror in diopters is 58.8 D
Explanation:
Given;
radius of curvature of the spoon, R = 3.4 cm = 0.034 m
The focal length of a mirror is given by;
[tex]f = \frac{R}{2} \\\\f = \frac{0.034}{2} \\\\f = 0.017 \ m[/tex]
The focal length of the mirror is 0.017 m
(b) The power of the mirror is given by;
[tex]P = \frac{1}{f}[/tex]
where;
P is the power of the mirror
f is the focal length
[tex]P = \frac{1}{f}\\\\P= \frac{1}{0.017}\\\\P = 58.8 \ D[/tex]
Thus, the power of the mirror in diopters is 58.8 D