Answer:
The average density of the rod is 1.605 kg/m.
Explanation:
The average density of the rod is given by:
[tex] \rho = \frac{m}{l} [/tex]
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:
[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex] (1)
Using u = x+1 → du = dx → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]
[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]
Therefore, the average density of the rod is 1.605 kg/m.
I hope it helps you!
The average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Given data:
The length of rod is, L = 3 m.
The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, The expression for the average density is given as,
[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)
Using u = x+1
du = dx
u₁= x₁+1 = 0+1 = 1
and
u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]
Thus, we can conclude that the average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
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The charger for your electronic devices is a transformer. Suppose a 60 Hz outlet voltage of 120 V needs to be reduced to a device voltage of 3.0 V. The side of the transformer attached to the electronic device has 45 turns of wire.
How many turns are on the side that plugs into the outlet?
Answer:
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
Explanation:
The transformer turns ratio is given by the following equation:
V₁/V₂ = N₁/N₂
where,
V₁ = Voltage of outlet = 120 V
V₂ = Device Voltage = 3 V
N₁ = No. of turns on outlet side = ?
N₂ = No. of turns on side of device = 45
Therefore,
120 V/3 V = N₁/45
N₁ = (40)(45)
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
6. You push an object, initially at rest, across a frictionless floor with a constant force for a time interval t, resulting in a final speed of v for the object. You then repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v?
Answer:
t = t₀ / 2
Explanation:
In this exercise we must use Newton's second law
F = m a
a = F / m
now we can use kinematics
as in object part of rest (v₀ = 0)
v =a t₀
t₀ = v / a
these results are with the first experiment
now repeat the experiment, but F = 2F₀
a = 2F₀ / m = 2 a₀
v = 2 a₀ t
t = v / 2a₀
t = t₀ / 2
The time interval that is required to reach the same final speed (V) is equal to [tex]t=\frac{\Delta t}{2}[/tex].
Given the following data:
Initial speed = 0 m/s (since the object is at rest)Final speed = VTime = [tex]\Delta t[/tex]Speed = VTo find the time interval that is now required to reach the same final speed (V), we would apply Newton's Second Law of Motion:
Mathematically, Newton's Second Law of Motion is given by this formula;
[tex]F = \frac{M(V-U)}{t}[/tex]
Where:
F is the force.V is the final velocity.U is the initial velocity.t is the time.Substituting the given parameters into the formula, we have;
[tex]F = \frac{M(V-0)}{\Delta t}\\\\F = \frac{MV}{\Delta t}[/tex]
When the experiment is repeated, the magnitude of the force is doubled:
[tex]F = 2F[/tex]
Now, we can find the time interval that is required to reach the same final speed (V):
[tex]F = \frac{M(V-0)}{t}\\\\t=\frac{MV}{F}[/tex]
Substituting the value of F, we have:
[tex]t=\frac{MV}{2F} \\\\t=\frac{MV}{\frac{2MV}{\Delta t}} \\\\t=MV \times \frac{\Delta t}{2MV} \\\\t=\frac{\Delta t}{2}[/tex]
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g Two point sources emit sound waves of 1.0-m wavelength. The source 1 is at x = 0 and source 2 is at x = 2.0 m along x-axis. The sources, 2.0 m apart, emit waves which are in phase with each other at the instant of emission. Where, along the line between the sources, are the waves out of phase with each other by π radians?
Answer:
constructive interferencia 0, 1 , 2 m
destructive inteferencia 1/4, 3/4. 5/4, 7/4 m
Explanation:
This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression
Δr / λ = Φ / 2π
Δr = Φ/2π λ
let's apply this expression to our case
λ = 1 m
Δr = Φ 1 / 2π
We have constructive interference for angle of Φ = 0, 2π, ...
let's find the values where they occur
Φ Δr
0 0
2π 1
4π 2
Destructive interference occurs by Φ = π /2, 3π / 2, ...
Φ Δr
π/2 ¼ m
3π /2 ¾ m
5π /2 5/4 m
7π /2 7/4 m
What happens to the deflection of the galvanometer needle (due to moving the magnet) when you increase the number of loops
Answer:
If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil which will increase total induced electromotive force
Explanation:
galvanometer is an instrument that can detect and measure small current in an electrical circuit.
If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil. If it is move in a way into the coil,the needle deflect in that way and if it move in another way, it will deflect in the other way.
The total induced emf is equal to the emf induced in each loop by the changing magnetic flux, then multiplied by the number of loops and an increase in the number of loops will cause increase in the total induced emf.
One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 1 cm across, and you estimate that the distance from the window shade to the wall is about 4 m.
Estimate:
a. The average wavelength of the sunlight (in nm)
b. The diameter of the pinhole (in mm).
Given that,
Central maximum = 1 cm
Distance from the window shade to the wall =4 m
We know that,
The visible range of the sun light is 400 nm to 700 nm.
(a). We need to calculate the average wavelength
Using formula of average wavelength
[tex]\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}[/tex]
Put the value into the formula
[tex]\lambda_{avg}=\dfrac{400+700}{2}[/tex]
[tex]\lambda_{avg}=550\ nm[/tex]
(b). We need to calculate the diameter of the pinhole
Using formula for diameter
[tex]w=\dfrac{2.44\lambda L}{D}[/tex]
[tex]D=\dfrac{2.44\lambda L}{w}[/tex]
Put the value into the formula
[tex]D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}[/tex]
[tex]D=0.537\ mm[/tex]
Hence, (a). The average wavelength 550 nm.
(b). The diameter of the pinhole is 0.537 mm.
Calculate the answers to the appropriate number of significant
12.21 x 9.19 =
In a distant galaxy, whose light is just arriving from 10 billion light years away, our spectroscope should reveal that the most common element is
Answer:
In a distant galaxy, whose light is just arriving from 10 billion light years away, our spectroscope should reveal that the most common element is HELIUM
A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall. Find the potential energy of a particle due to this force when it is at a distance x from the wall, assuming the potential energy at the wall to be zero.
Answer:
it will be 10x
Explanation:
workdone(potential energy before it hits the wall)= horizontal force × distance
=10× x = 10x joules
A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall.The potential energy of a particle due to this force is 10x.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Given in the question a mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall the potential energy,
Work done (potential energy before it hits the wall)
= horizontal force × distance
=10× x = 10x joules
The potential energy of a particle due to this force is 10x.
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The atomic number of a nucleus increases during which nuclear reactions?
Answer:
Option (A) : Nuclear Fusion and Beta Decay (electron emission)
Answer:
A : Fusion followed by beta decay (electron emission)
Explanation:
Ap3x
Is the study of the moons places applied or pure science
Answer:
It is pure science
Explanation:
A basic knowledge for the discovery of unknown laws based on well controlled experiments and deductions from demonstrated facts or truths.
A force of 16.88 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20rad / (s ^ 2) . Calculate the rotational inertia of the wheel. A. 2.77 kg - m ^ 2 B. 0.73 kg - m ^ 2 C. 4.41 kg - m ^ 2 O. 4.78 kg - m ^ 2
Given.
force = 16.88 N is
radius = 0.340m
an angular acceleration = 1.20rad/s^2
the formula for torque is
F*r = I*a
where I is moment of inertia
16.88*.34 = I*1.2
I = 4.78Kg-m^2
so rotational inertia I = 4.78Kg-m^2
What will be the nature of the image formed from both a convex lens and a concave
lens of 20 centimeter focus distance, when the object is placed at a distance of
10 centimeters?
Answer:
Explanation:
Using the lens formula
1//f = 1/u+1/v
f is the focal length of the lens
u is the object distance
v is the image distance
For convex lens
The focal length of a convex lens is positive and the image distance can either be negative or positive.
Given f = 20cm and u = 10cm
1/v = 1/f - 1/u
1/v = 1/20-1/10
1/v = (1-2)/20
1/V = -1/20
v = -20/1
v = -20 cm
Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image
For concave lens
The focal length of a concave lens is negative and the image distance is negative.
Given f = -20cm and u = 10cm
1/v = 1/f - 1/u
1/v = -1/20-1/10
1/v = (-1-2)/20
1/V = -3/20
v = -20/3
v = -6.67 cm
Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image
The filament in the bulb is moving back and forth, first pushed one way and then the other. What does this imply about the current in the filament
Answer:
energy carried by the current is given by the pointyng vector
Explanation:
The current is defined by
i = dQ / dt
this is the number of charges per unit area over time.
The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.
But the energy carried by the current is given by the pointyng vector of the electromagnetic wave
S = 1 / μ₀ EX B
It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement
how many stars are in our solar system?
Answer:
there are over 100 billion stars in our galaxy.
A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave, in kHz
The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.
Given the following data:
Period = 645 μsNote: μs represents microseconds.
Conversion:
1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds
645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds
To find corresponding frequency of this sinewave, in kHz;
Mathematically, the frequency of a waveform is calculated by using the formula;
[tex]Frequency = \frac{1}{Period}[/tex]
Substituting the value into the formula, we have;
[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]
Frequency = 1550.39 Hz
Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);
Conversion:
1 hertz = 0.001 kilohertz
1550.39 hertz = X kilohertz
Cross-multiplying, we have;
X = [tex]0.001[/tex] × [tex]1550.39[/tex]
X = 155039 kHz
To 3 significant figures;
Frequency = 155 kHz
Therefore, the corresponding frequency of this sinewave, in kHz is 155.
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Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 55.3 kg, down a theta= 79.6º slope at constant acceleration a=-4.3 m/s2, as shown in Figure (here we assume the positive direction is going down the slope. So the given acceleration is a negative value, it means its direction is going up the slope, slowing down as it moving downward). So, the coefficient of friction between the sled and the snow is 0.100. How many Joules of work is done by the tension in the rope as the sled moves 2.1 m along the hill? Use g= 10 m/s2.
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.
Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.
From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex] (1)
Where:
[tex]W_{T}[/tex] - Work done by tension, in joules.
[tex]m[/tex] - Mass of the sled-victim system, in kilograms.
[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Travelled distance, in meters.
[tex]\theta[/tex] - Slope angle, in sexagesimal degrees.
[tex]a[/tex] - Net acceleration of the sled-victim system, in meters per square second.
If we know that [tex]\mu = 0.100[/tex], [tex]m = 55.3\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]s = 2.1\,m[/tex], [tex]\theta = 79.6^{\circ}[/tex] and [tex]a = -4.3\,\frac{m}{s^{2}}[/tex], then the work done by the tension in the rope is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex]
[tex]W_{T} = \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta -m\cdot a\cdot s[/tex]
[tex]W_{T} = (0.100)\cdot \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \cos 79.6^{\circ} + \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \sin 79.6^{\circ} - (55.3\,kg)\cdot \left(-4.3\,\frac{m}{s^{2}} \right) \cdot (2.1\,m)[/tex]
[tex]W_{T} = 1662.544\,J[/tex]
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
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Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is 4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?
Answer:
10.75 A
The current is in opposite direction since it causes a repulsion force between the wires
Explanation:
Force per unit length on the wires = 4.30×10^−5 N/m
distance between wires = 2.6 cm = 0.026 m
current through one wire = 0.52 A
current on the other wire = ?
Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as
[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]
where [tex]F/l[/tex] is the force per unit length on the wires
[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A
[tex]I_{1}[/tex] = current on the first wire = 0.520 A
[tex]I_{2}[/tex] = current on the other wire = ?
r = the distance between the two wire = 0.026 m
substituting the value into the equation, we have
4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] = [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]
4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]
[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A
The current is in opposite direction since it causes a repulsion force between the wires.
Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).
Image is missing, so i have attached it
Answer:
19.04 × 10⁻⁴ T in the +x direction
Explanation:
We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.
Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)
We are given;
I = 2.4 A
R = 5 cm = 0.05 m
μ₀ is a constant = 4π × 10⁻⁷ H/m
B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)
B = 9.6 × 10⁻⁴ T
To get the direction of the field from each wire, we will use Flemings right hand rule.
From the diagram attached:
We can say the field at P from the top wire will point up/right
Also, the field at P from the bottom wire will point down/right
Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.
If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.
Let the angle be θ
Thus;
sin(θ) = (1.3/2)/5
θ = sin⁻¹(0.13) = 7.47⁰
The x component of each field would be:
9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T
Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction
The magnitude of the magnetic field at the point P will be "9.6 × 10⁻⁴ T".
Magnetic fieldThe region of the environment close to something like a magnetic entity or a current-carrying body wherein this same magnetic forces caused by the body as well as a current might well be sensed.
According to the question,
Current, I = 2.4 A
Radius, R = 5 cm or,
= 0.05 m
Constant, μ₀ = 4π × 10⁻⁷ H/m
We know the relation,
The magnetic field, B = [tex]\frac{\mu_0 I}{2 \pi R}[/tex]
By substituting the values in the above relation, we get
= [tex]\frac{4 \pi\times 10^{-7}\times 2.4}{2 \pi\times 0.05}[/tex]
= 9.6 × 10⁻⁴ T
Thus the above answer is appropriate.
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A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sphere is then grounded as shown. Then the ground wire is removed. A is the inner surface and B is the outer surface. Which statement is correct
Explanation:
the missing figure in the Question has been put in the attachment.
Then from the figure we can observe that
the center of the sphere is positive, therefore, negative charge will be induced at A.
As B is grounded there will not be any charge on B
Hence the answer is A is negative and B is charge less.
An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance of 1 cm and there is vacuum in the region between the plates. The electron is initially found midway between the plates with a kinetic energy of 11.2 eV and with its velocity directed toward the negative plate. How close to the negative plate will the electron get if the potential difference between the plates is 100 V? (1 eV = 1.6 x 10-19 J)
Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:
[tex]F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}[/tex]
and since the electric field E in between parallel plates separated a distance d and under a potential difference [tex]\Delta V[/tex], is given by:
[tex]E=\frac{\Delta\,V}{d}[/tex]
then :
[tex]a=\frac{q\,\Delta V}{m\,d}[/tex]
We want to find when the particle reaches velocity zero via kinematics:
[tex]v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a[/tex]
We replace this time (t) in the kinematic equation for the particle displacement:
[tex]\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}[/tex]
Replacing the values with the information given, converting the distance d into meters (0.01 m), using [tex]\Delta V=100\,V[/tex], and the electron's kinetic energy:
[tex]\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J[/tex]
we get:
[tex]\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters[/tex]Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]
As a skydiver falls, his potential energy ___ and his kinetic energy __
increases,increases
increases,decreases
decreases,increases
decreases, decreases
Answer:
Hey there!
PE=mgh, so as height decreases, so does the potential energy.
KE=mv^2, so as velocity increases, kinetic energy increases.
Thus, the correct answer would be Decreases, Increases.
Let me know if this helps :)
Define the following, and give the letter which we will abbreviate them by:
Center of curvature:
Vertex:
Focal Point:
Radius of curvature:
Focal length:
Answer:
As in explanation.
Explanation:
A) Centre of Curvature: This is defined as the point in the center of the sphere from which the mirror was sliced. It is represented by the letter "C"
B) Vertex: It is defined as the point on the mirror's surface where the principal axis meets the mirror. It is represented by the letter A.
C) Focal Point: This is defined as the Midway point between the vertex and the center of curvature. It is represented by the letter "F"
D) Radius of Curvature: This is defined as the distance from the vertex to the center of curvature. It is represented by the letter "R"
E) Focal Length: This is defined as the distance from the mirror to the focal point. It's represented by the letter "f"
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
Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
Answer:
The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Explanation:
Given;
intensity of light, I = 1 kW/m²
The radiation pressure of light is given as;
[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]
I kW = 1000 J/s
The energy flux density = 1000 J/m².s
The speed of light = 3 x 10⁸ m/s
Thus, the radiation pressure of the light is calculated as;
[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]
Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
13. A sinusoidal wave of frequency f is traveling along a stretched string. The string is brought to rest, and a second traveling wave of frequency 2f is established on the string. What is the wavelength of the second wave?
Answer:
It will be half that if the first wave
Explanation:
Because the wave speed remains the same, the result of doubling the frequency is that the wavelength is half as large as it
A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.220 s, during which it produces an average 0.520 W from an average 3.00 V.
A. How much charge moves through the lamp (C)?
B. Find the capacitance (F).
C. What is the resitance of the lamo?
Answer:
A. 0.0374C
B. 0.012F
C. 18 ohms
Explanation:
See attached file
A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation. With what period does it now oscillate?
Answer:
The new time period is [tex]T_2 = 3.8 \ s[/tex]
Explanation:
From the question we are told that
The period of oscillation is [tex]T = 5 \ s[/tex]
The new length is [tex]l_2 = 0.76 \ m[/tex]
Let assume the original length was [tex]l_1 = 1 m[/tex]
Generally the time period is mathematically represented as
[tex]T = 2 \pi \sqrt{ \frac{ I }{ mgh } }[/tex]
Now I is the moment of inertia of the stick which is mathematically represented as
[tex]I = \frac{m * l^2 }{12 }[/tex]
So
[tex]T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }[/tex]
Looking at the above equation we see that
[tex]T \ \ \ \alpha \ \ \ l[/tex]
=> [tex]\frac{ T_2 }{T_1} = \frac{l_2}{l_1}[/tex]
=> [tex]\frac{ T_2}{5} = \frac{0.76}{1}[/tex]
=> [tex]T_2 = 3.8 \ s[/tex]
Object A, with heat capacity CA and initially at temperature TA, is placed in thermal contact with object B, with heat capacity CB and initially at temperature TB. The combination is thermally isolated. If the heat capacities are independent of the temperature and no phase changes occur, the final temperature of both objects is
Answer:
d) (CATA + CBTB) / (CA + CB)
Explanation:
According to the given situation, the final temperature of both objects is shown below:-
We assume T be the final temperature
while m be the mass
So it will be represent
m CA (TA - T) = m CB (T - TB)
or we can say that
CATA - CA T = CB T - CBTB
or
(CA + CB) T = CATA + CBTB
or
T = (CA TA + CBTB) ÷ (CA + CB)
Therefore the right answer is d
The final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
The given parameters;
heat capacity of object A = CAinitial temperature of object A = TAheat capacity of object B = CBinitial temperature of object B = TBThe final temperature of both objects is calculated as follows;
heat lost by object A is equal to heat gained by object B
[tex]mC_A (T_A - T) = mC_B(T- T_B)\\\\C_AT_A-C_AT = C_BT - C_BT_B\\\\C_BT+C_AT = C_AT_A+ C_BT_B\\\\T(C_B + C_A) = C_AT_A+ C_BT_B \\\\T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex]
Thus, the final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
Learn more here:https://brainly.com/question/17163987
How do for-profit and nonprofit fitness centers compare with each other?
For-profit centers offer more luxury services than nonprofit centers.For-profit centers offer more luxury services than nonprofit centers. , ,
For-profit centers have lower membership fees than nonprofit centers.For-profit centers have lower membership fees than nonprofit centers. , ,
For-profit centers place more emphasis on weight lifting than nonprofit centers.For-profit centers place more emphasis on weight lifting than nonprofit centers. , ,
For-profit centers are more common in low-income communities than nonprofit centers.
Comparison of For-profit and Non-profit Fitness Centers
This is how they compare with each other: For-profit centers offer more luxury services than nonprofit centers.
For-profit fitness centers are fitness centers that offer their services to cover their costs and make some profits. They are business entities that generate income for their shareholders. Their fees are based on cost plus profit. This approach enables them to remain sustainable as business ventures.
They are not like non-profit fitness centers, which may offer their services for free or at cost. Sometimes, the nonprofit centers may not even cover their fixed and operational costs. They are usually located in low-income communities, unlike for-profit fitness centers, which are mainly located in rich-income communities.
Considering their clientele, for-profit centers provide modern and better facilities to satisfy their rich-income clients.
Thus, these for-profit centers offer more luxury services than their non-profit counterparts.
For more information about how for-profit centers base their charges, see https://brainly.com/question/21302488.
The primary difference between a barometer and a manometer is
A. a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.
B. a barometer uses mercury, while a manometer can use any liquid. a barometer is used to measure atmospheric pressure, and a manometer is used to measure absolute pressure.
C a barometer reads in mm, while a manometer reads in Pa.
D a barometer can measure either positivee or negative pressure, while a manometer only
E positive pressure. measures
Answer:
a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.
Explanation:
A barometer measures air pressure at any locality with sea level as the reference.
However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure
The algebraic sum of these two values gives the absolute pressure.