Anything parallel to the horizon is called horizontal.
What is the period of the motion?The period of the object's motion is defined as the time for the object to complete one full cycle. Being a time, the period is measured in units such as seconds, milliseconds, days or even years. The standard metric unit for period is the second. An object in periodic motion can have a long period or a short period.Given that,The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.“Period” is rooted in the Greek words “peri” and “hodos” (periodos) meaning “around” and “way/path.” This eventually turned into the Latin “periodus” meaning “recurring cycle.” Use of the English term “period” to describe menstruation began in the early 1800s (1).A mass m suspended by a wire of length L is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15º. The period of a simple pendulum is T=2π√Lg T = 2 π L g , where L is the length of the string and g is the acceleration due to gravity.
The mass attached to the spring, m = 1.5 kg
The amplitude of the spring, A = 0.3 m
The spring constant of the spring, k = 85 N/m
We need to find the period of the mass's motion. The time period in spring mass system is given by :
[tex]T=2\pi \sqrt{\frac{M}{K} }[/tex]
[tex]T=2\pi \sqrt{} \frac{1.5}{85}[/tex]
[tex]T=0.83s[/tex]
So, the period of the mass's motion is 0.83 seconds.
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Answer: 0.96s
Explanation:
Khan
A diffraction grating 20.0 mm wide has 6000 rulings. Light of wavelength 589 nm is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of theta at which maxima appear on a distant viewing screen?
6000 rulings are present on a 20.0 mm broad diffraction grating. The maximum value of theta at which maxima show on a distant viewing screen for light with a wavelength of 589 nm is 62.1 °, followed by 45.0 ° and 32.0 °.
Adjacent rulings are separated by a distance of d = 20.0mm/6000 = 0.00333mm = 3.33m. Let dsin equal m (0, 1, 2,...). The greatest value of corresponds to m=5, which results in = sin 1(m/d)= sin 1 (5(0.589m)/3.33m)=62.1 ° b) because m/d>1 for m6. The second-largest value of is equal to m=4, which results in the expression = sin 1 (m/d) = sin 1 (4(0.589 m)/ 3.33 m)= 45.0° c. When m=3—the third-largest value of —is used, the result is = sin 1 (m/d) = sin 1 (3(0.589 m)/3.33 m) = 32.0°.
Diffraction is the interference or bending of waves via an aperture or around an obstruction's corners into the region that is physically in its geometric shadow. In essence, the diffracting component or aperture becomes a second source of the wave. Francesco Maria Grimaldi, an Italian physicist, was the first to precisely record the occurrences of the phenomena in 1660 and also coined the name "diffraction".
An unlimited number of locations (three shown) along length d that project phase contributions from the wavefront cause a continuously varying intensity () to be created on the license plate.
The diffraction phenomenon is explained by the classical physics idea known as the Huygens-Fresnel principle, which views each point in a propagating wave front as a group of distinct spherical wavelets.
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The half-life for a 200 gram sample of radioactive element X is 5 days. How much of element X remains after 15 days have passed?
A. 200 g - no change
B. 25 g
C. 50 g
D. 12.5 g
Answer:
B.) 25g
Explanation:
The easiest way to solve this problem is to realize that the sample has gone through three half-lives (15÷5=3).
Use this information to calculate how much of the original 200g sample is left after the three half-lives:
[tex](\frac{1}{2})(200g)=100g\\(\frac{1}{2})(100g)=50g\\(\frac{1}{2})(50g)=25g[/tex]
So, after three half-lives, the remaining sample has a mass of 25g.
find the finel angular velocity (w) when the intial angular velocity(w) is 5rad/s the Angular acceleration is 2rad/s² and the time of motion is 5sec
Answer:
15 rad/s
Explanation:
Given:
α = 2 rad/s²
ω₀ = 5 rad/s
t = 5 s
Find: ω
ω = αt + ω₀
ω = (2 rad/s²) (5 s) + (5 rad/s)
ω = 15 rad/s
when an object of unknown mass is attached to an ideal spring with force constant 130 n/m, it is found to vibrate with a frequency of 7.25 hz.
When a massless item is fastened to an ideal spring with a force constant of 130 n/m, it is discovered that it vibrates at a frequency of 7.25 hz. The motion lasts 0.138 seconds.
What does a spring's force constant mean?In accordance with Hooke's law, the force necessary to compress or enlarge a spring is directionally related to or proportionate to the length it is stretched to. K is used to indicate the force constant.
What exactly is a vibration's frequency?The rate of vibrations and oscillations is called frequency, and it is expressed in hertz (Hz) units. To distinguish and ascertain vibrational patterns, frequencies are used. Therefore, a faster-vibrating atom would be seen as having a greater frequency than a slower-vibrating one.
Briefing:T= 1/f = 1/7.25 Hz = 0.138s
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The given question is incomplete. The complete question is:
When an object of unknown mass is attached to an ideal spring with force constant 130 n/m, it is found to vibrate with a frequency of 7.25 hz. What is the period of the motion?
what is the force on the electron, magnitude, and direction? (i.e. when it is in the plane of the page as stated above.) justify your answer.
Faraday's left hand rule gives us the direction of the force on the particles, then [tex]F^{\prime}=B q v\\[/tex] gives the size of the force.
What is direction?
Direction is characterized as the course that anything follows, the route that must be taken to go to a particular location, the direction in which something is beginning to take shape, or the direction you are facing. When you turn right instead of left, that is an illustration of direction.
If we use Faraday's left hand rule on each particle, we can determine the effect of the magnetic field on the particle. The diagram shows the direction of the fields and the "current" so we can apply the rule.
For the electron, the magnetic field will exert a force directed into the plane of the paper (or if travelling on the earth, this will be towards the earth).
For the proton, the magnetic field will exert a force out of the paper (or if travelling on the earth, this will be away from the earth).
The force on a particle of charge q moving in a magnetic field, B with a velocity v is given by:
[tex]F=B q v[/tex]
Both proton and electron have a charge of [tex]1.6 \times 10^{-19} C\\[/tex]
For the electron
[tex]$$F=\left(190 \times 10^{-3}\right) \cdot\left(1.6 \times 10^{-19}\right) \cdot 575000=1.748 \times 10^{-14} N$$[/tex]
For the proton
[tex]$$F=\left(54.2 \times 10^{-6}\right) \cdot\left(1.6 \times 10^{-19}\right) \cdot 130000=1.127 \times 10^{-18} N$$[/tex]
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A student applies a force F at an angle of θ below the horizontal to a 50 kg desk, which causes the desk to move across the room at a constant speed.
Angle θ= 20 Degrees
Force= 200 N
Draw and label an appropriate force vector diagram for the desk.
What is the weight of the desk?
What is the magnitude of the normal force acting on the desk?
What is the magnitude of the friction force acting on the desk?
The magnitude of the weight, normal and friction forces are 490N, 455N and 16.66N respectively.
The force applied by the student at the angle of 20 degrees has a magnitude of 200N.
The desk is moving across the room with a constant speed.
The mass of the desk is Kg.
The weight of the desk is given by,
W = Mg
Where,
g is acceleration due to gravity.
Putting values,
W = 50 x 9.8
W = 490N.
The magnitude of the normal force is given by,
N = WcosA
Where,
A is the angle with x-axis.
Putting values,
N = 490 x cos(20)
N = 455N.
The magnitude of the Friction force is given by,
F = W x sin(20)
F = 490 x 0.34
F = 16.66N.
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Two blocks, with masses indicated in the figure above, are at rest on a horizontal surface and connected by a string of negligible mass and a compressed spring. There is negligible friction between the blocks and the surface. The string is cut, and the spring pushes the blocks away from each other. Which of the following statements are true about the motion of the blocks immediately after the string is cut? select two answers.
1. The velocity of the center of mass of the two-block system is zero.
2. The magnitude of the acceleration of the left block is greater than that of the right block.
The center of mass, or balancing point, of a distribution of mass in space is the only location where the total of the weighted relative positions of the distributed mass equals zero. In order to accelerate linearly without also accelerating angularly, a force might be applied at this moment. Using the center of mass as a reference point simplifies calculations in mechanics quite a bit. To imagine how an object would move, one can imagine a hypothetical location where the object's total mass is concentrated. To apply Newton's equations of motion to a particular object, the center of mass is the particle equivalent of that thing.
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aphasia
Due to an automobile accident, Jenny suffered damage to her cerebral cortex in Broca's area. Jenny is most likely to experience:
Due to an automobile accident, Jenny suffered damage to her cerebral cortex in Broca's area. Jenny is most likely to experience: aphasia.
Aphasia is the inability to understand or form language due to damage to certain areas of the brain The main causes are stroke and head trauma. Although the prevalence is difficult to determine, aphasia due to stroke is estimated at 0.1-0.4% in the Global North. Aphasia can also be the result of a brain tumor, brain infection, or neurodegenerative disease (such as dementia).
To be diagnosed with aphasia, communication after an acquired brain injury Speech or language must be significantly impaired in one (or more) of the four aspects of Alternatively, in the case of progressive aphasia, it must have decreased significantly over a short period of time. The four dimensions of communication are auditory comprehension, verbal expression, reading and writing, and functional communication.
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large redshifts move the positions of spectral lines to longer wavelengths and change what can be observed from the ground. for example, suppose a quasar has a redshift of z
Laboratory experiments on Earth have examined that each element in the periodic table emits photons only at definite wavelengths (determined by the excitation state of the atoms).
These photons are manifest as either emission or absorption lines in the spectrum of an astronomical object, and by observing the position of these spectral lines, we can detect which components are present in the object itself or along the line of sight.
although, when astronomers observe spectral lines in extragalactic objects (such as galaxies and quasars), they find that the wavelength of the notice spectral lines is different from the laboratory experiments.
In most cases, the wavelength of the spectral lines is longer and thus are shifted toward the red end of the spectrum they are redshifted.
There are many explanations for this redshift phenomenon.
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the following items describe observational characteristics that could indicate that an object is either a white dwarf or a neutron star. match each characteristic to the correct object.
When a white dwarf acquires enough mass for its carbon interior to start fusing, a white dwarf supernova results.
The white dwarf entirely explodes as the star's fusion starts nearly immediately throughout the entire star. When iron is produced during fusion in a star's core, a "massive stellar supernova" results. If a white dwarf in a near binary system acquires enough mass to surpass the "white dwarf limit (1.4 solar masses)," it will erupt as a supernova. The absence of hydrogen characteristics in type I supernovae's spectra at maximum luminosity is a crucial identifying trait. The remnant core of a low-mass star known as a white dwarf is protected from the force of gravity by electron degeneracy pressure. The mass of the Sun is often condensed into a body no bigger than Earth in a white dwarf.
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Filed sdi as269len insbude sil velfs ovom
What distance will a bus go if it is traveling at a rate of 50 km/hr for 3.8
hours?
A. 190.0 kilometers
B. 53.8 kilometers
C. 46.2 kilometers
D. 13.2 kilometers
The distance the bus will go, if it is travelling at a rate of 50 km/hr for 3.8 hours is 190 kilometers (Option A)
How do I determine the distance the bus will go?We know that speed is defined as distance travelled per unit time i.e
Speed = Distance travelled / time
Cross multiply
Distance travelled = Speed × time
With the above formula, we can calculate the distance the bus will go. Details below:
Speed = 50 km/hr Time = 3.8 hoursDistance travelled =?Distance travelled = Speed × time
Distance travelled = 50 × 3.8
Distance travelled = 190 kilometers
Thus, we can conclude that the distance the bus will go is 190 kilometers (Option A)
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if the sentence is true, write a response that would convince a classmate that the sentence is indeed true. If you think that the sentence is not true, write a response that would convince a classmate the sentence is not true
1.A 10 cm radius Styrofoam sphere is submerged in water. So is a 10 cm radius lead sphere. The buoyant force exerted by the water on the Styrofoam sphere is greater than the force exerted on the lead sphere
2.Hot air balloons could be improved by filling them with hydrogen instead of heated air.
A 10 cm radius Styrofoam sphere is submerged in water. So is a 10 cm radius lead sphere. The buoyant force exerted by the water on the Styrofoam sphere is greater than the force exerted on the lead sphere is true.
2.Hot air balloons could be improved by filling them with hydrogen instead of heated air is ta true statement.
How do you prove the statements above?Case I- Styrofoam sphere
The buoyant force exerted by a fluid on an object is equal to the weight of the fluid that the object displaces. This means that the buoyant force exerted on the Styrofoam sphere and the lead sphere will be determined by the amount of water that each sphere displaces when it is submerged.
The Styrofoam sphere has a lower density than water, so it will displace a greater volume of water than the lead sphere, which has a higher density than water. This means that the buoyant force exerted on the Styrofoam sphere will be greater than the force exerted on the lead sphere.
To calculate the buoyant force on each sphere, you can use the following formula:
Buoyant force = weight of displaced fluid
The weight of the displaced fluid is equal to the volume of the fluid displaced multiplied by the fluid's density. You can use the formula for the volume of a sphere (V = 4/3πr^³) to calculate the volume of water displaced by each sphere.
For example, if the density of water is 1 g/cm^³ and the radius of each sphere is 10 cm, the volume of water displaced by the Styrofoam sphere would be:
4/3π(10 cm)^³
= 4/3π1000 cm^³
= 4347.19 cm^³
The weight of the displaced water would be:
4347.19 cm^³ * 1 g/cm^³
= 4347.19 g.
The buoyant force on the Styrofoam sphere would therefore be 4347.19 g.
The volume of water displaced by the lead sphere would be the same as the Styrofoam sphere (4347.19 cm^3), but the weight of the displaced water would be greater, due to the higher density of lead.
The weight of the displaced water for the lead sphere would be:
4347.19 cm^³ * 11.34 g/cm^³
= 49359.93 g,
The buoyant force on the lead sphere would be:
49359.93 g.
Therefore, the buoyant force on the Styrofoam sphere would be less than the force on the lead sphere, as stated in the original statement.
Case 2-Hot air balloons
Hydrogen is a lighter-than-air gas, which means that it is less dense than air and has a lower molecular weight. This makes it a potential alternative to heated air as a lifting gas for hot air balloons.
One advantage of using hydrogen as a lifting gas is that it is more buoyant than air, so a hot air balloon filled with hydrogen would be able to lift a greater payload than a balloon filled with air. This could make it possible to design hot air balloons that are larger or carry more passengers or cargo.
However, there are also some significant drawbacks to using hydrogen as a lifting gas for hot air balloons. One concern is the flammability of hydrogen, which is highly explosive and can ignite easily. This makes it potentially dangerous to use in a hot air balloon, as any leaks or accidents could result in a fire or explosion.
Overall, while hydrogen has some potential advantages as a lifting gas for hot air balloons, the risks and costs associated with its use may make it less practical than other options.
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sphere a has mass m and is moving with velocity v. it makes a head-on elastic collision with a stationary sphere b of mass 2m. after the collision their speeds (va, vb) are:
When a ball of mass m elastically collides with mass 2m, momentum and kinetic energy are conserved.
We know that,
Momentum Conservation:
mv+ 0= m(va)+ 2m(vb)......(a)
m(v)=m(va+2vb)
=> v= va+ 2vb
We know that,(a+b)2=a2+b2+2ab
Then,
(va+2ab)2=va2+4ab2+4(va×vb)
Squaring both sides gives v2= va2+4vb2+4(va×vb).....(1)
Kinetic conservation of energy
mv2=mva2+2mvb2
v2=va2+2vb2.....(2
Subtracting 2 from 1 gives
vb=-2va
Substituting into (a), we get
v a=-v/3
v b=2v/3
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Aspace station has a mass M and orbits Earth in a circular orbit at a height above Earth's surface. Space Station Earth A payload of mass m, where m < M is delivered to the space station. Soon after, the space station's orbit is adjusted so that it is 50 km farther away from Earth's surface than before. Which of the following best describes the effects of these changes on Earth's gravitational field strength at the space station's new location? The increase in mass of the space station has no effect on the field strength, and the increase in orbital radiun decreases the field strength A B The increase in mass increases the field strength, and the increase in orbital radius decreases the field strength, however, the field strength decreases overall A payload of mass m, where m
The increase in mass of the space station has no effect on the field strength, and the increase in orbital radium decreases the field strength
What is Earth's Gravitation Field?
Earth's gravitation field is the force any object experiences due to the gravitational pull by earth. It's formula is given below where G is the force, m is the mass of earth and r is radius of distance.
As seen through the formula, the gravitation field strength is only depend on earth's mass which is constant and radius of the distance. Hence, the increase in mass of the space station has no effect on the field strength, and the increase in orbital radium decreases the field strength.
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Elcft parallelepiped is in an electric field which to the left of the dashed line; has the value Eleft =< -55.37,0,0 > N/C,and Eright =< 55.37,0,0 > N/C to the right of the dashed line_ The top and bottom of the parellelepiped are rectangles Iying in the = plane, and measure /1 by l2, as shown: The left and right faces are rectangles inclined by an angle 0 = 83.489 from the € axis, and measure lz by l3. The values Of L1,l2, and l3 are 23.82 cm, 11.27 cm, and 32.48 cm, respectively. What charge is contained inside the parallelepiped?
There is a net flux going into the surface, the parallelepiped must have a net charge (negative charge). Additionally, without an external field, all lines would point at the slab.
Six parallelograms come together to form a three-dimensional shape called a parallelepiped (the term rhomboid is also sometimes used with this meaning). It is comparable to a parallelogram in the same way that a cube is comparable to a square. There are three similar descriptions of a parallelepiped. A prism with a parallelogram-shaped base, a hexahedron with three pairs of parallel faces, and a polyhedron with six faces, each of which is a parallelogram.
There are pairs of parallel and congruent faces. The parallelepiped has twelve (12) edges and eight (8) vertices. The edges of the parallelepiped can be split into three (3) sets of four (4) edges each, where the edges in each set are all parallel and the same length. The parallelepiped is also a zonohedron.
A parallelepiped is a three-dimensional object having six faces and a parallelogram-like design. It has 8 vertices, 6 faces, and 12 edges. The parallelepiped is also known by the names rhombic, cuboid, and cube.
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A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 10 kPa and 10 MPa, with a turbine inlet temperature of 600 degree C. The rate of heat transfer in the boiler is 800 kJ/s. Disregarding the pump work, the power output of this plant is: A. 284 kW B. 800 kW C. 508 kW D. 243 kW E. 335 kW
Option E; The pump work, the power output of this plant is 335KW
Pressure. at turbine. inlet P1 = 10 Mpa.
Enthalpy at turbine inlet = h1=3625.84 KJ/kg.
Entropy at turbine inlet S1 =6.90 KJ/kg-K.
It is given that process performed acceording to ideal rankine cycle which is "Reversible adiabatic expansion' in case of turbine wor., Since it is reversible adiabatic expansion, entropy at turbine inlet and entropy at turbine outlet is constant. So quality at turbine outlet is evaluated by using this relation. S1 = S2.
Pressure at turbine outlet = 10 Kpa.
Enthalphy of steam = hg =2675.1 KJ/kg.
Enthaphy of liquid=hf =417.51 KJ/kg
Entropy of steam = Sg =7.3589 KJ/kg-K.
Entropy of liquid =Sf =1.302 KJ/kg-K.
We know S1 = S2 =6.090=1.302+x(7.3589 - 1.302).
Quality at turbine exit x =0.79.
Enthalphy at exit = h2 = 417.51+ 0.79*(2675.1 - 417.51) =2201 KJ/kg-K.
heat transfered= mass*(enthlphy at exit of condenser - enthalphy at turbine inlet), Since pump work negleted
At exit of condenser, steam gets converted into water at 10 Kpa pressure whose enthalphy is hf =417.51 KJ/kg.
800 = m(3625.84 - 417.51).
mass flow rate of steam = 0.24 Kg/sec.
Turbine work the power output of this plant is = m(h1 - h2) =0.24(3625.84 - 2201) = 335 KW.
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A 20 g ball of clay traveling east at 2.5 m/s collides with a 25 g ball of clay traveling north at 2.0 m/s .
What is the speed of the resulting 45 g ball of clay?
What is the direction of the resulting ball of clay?
The final velocity of the two balls after collision is 1.7 m/s and the direction is 45⁰.
The given parameters;
mass of the first clay ball, m₁ = 20 g = 0.02 kg
initial velocity of the first clay ball, u₁ = 3 m/s
mass of the second clay ball, m₂ = 30 g = 0.03 kg
initial velocity of the second ball, u₂ = 2 m/s
The initial momentum of the fist ball is calculated as follows;
P₁ = m₁u₁
P₁ = (0.02)(3)
P₁ = 0.06 kg.m/s
The initial momentum of the second ball is calculated as follows
P₂ = m₂u₂
P₂ = (0.03)(2)
P₂ = 0.06 kg.m/s
The resultant initial momentum of the two balls is calculated as follows;
Apply the principle of conservation of momentum, to determine the final velocity of the two balls;
The direction of the two ball's velocity is calculated as follows;
Thus, the final velocity of the two balls after collision is 1.7 m/s and the direction is 45⁰To know more about speed direction visit:
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A 50 gg mass rotates in a vertical plane--call it the xy-plane with the y-axis pointing up--at the end of a 75-cmcm-long, massless, rigid rod. The other end of the rod is attached to a frictionless pivot at the origin.
What is the gravitational torque about the pivot when the mass is 60 ∘∘ above the +x-axis? Give your answer using unit vectors.
The gravitational torque about the pivot when the mass is 60 ∘∘ above the +x-axis is (0.1838 Nm)k
What is torque?Torque is the turning effect of a force which causes rotation. It is given by τ = F × d where
F = force and d = perpendicular distance of force from point of action.What is the gravitational torque about the pivot when the mass is 60° above the +x-axis?Since a 50 g mass rotates in a vertical plane--call it the xy-plane with the y-axis pointing up--at the end of a 75-cm-long, massless, rigid rod.
The gravitational torque is given by τ = F × d where
F = component of weight perpendicular to rod = -(mgcosФ)j where m = mass, g = acceleration due to gravityФ = angle between rod and x axis and d = length of rodSo, τ = F × d
= -(mgcosФ)j × di
= mgdcosФ(-j × i)
= -mgdcosФ(j × i)
= -mgdcosФ(-k)
= (mgdcosФ)k
Given that
m = 50 g = 0.05 kgg = 9.8 m/s²,d = 75 cm = 0.75 m and Ф = 60°Substituting the values of the variables into the equation, we have
τ = (mgdcosФ)k
= (0.05 kg × 9.8 m/s² × 0.75 m × cos60°)k
= (0.05 kg × 9.8 m/s² × 0.75 m × cos60°)k
= (0.3675 kgm²/s² × 0.5)k
= (0.18375 kgm²/s²)k
= (0.18375 Nm)k
≅ (0.1838 Nm)k
So, the torque is (0.1838 Nm)k
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A uniform plane wave with parallel polarization is propagating in a lossless dielectric medium (l is € 1), and is incident obliquely onto a plane boundary with another lossless dielectric medium (u 2, E 2). (30%) (a) Derive the Snell's laws of reflection and refraction in terms of the refractive indices and angles. (b) Derive the general expressions for the reflection and the transmission coefficients of the EM field. Find such expressions in terms of the refractive indices if both media are nonmagnetic, i.e., M F M 240. (c) Derive the general expression for the Brewster angle. Find such an expression in terms of the refractive indices if both media are nonmagnetic. (d) Prove that, under the condition of no reflection, the sum of the Brewster angle and the angle of refraction is 90° if both media are nonmagnetic. (e) For total reflection to occur, find the condition for the medium property and the critical angle.
Transmission coefficients are used in physics and electrical engineering when considering wave propagation in discontinuous media. The snells law is n₁sinθ1 = n₂sinθ2
The transmission coefficient describes the amplitude intensity or total power of the transmitted wave relative to the incident wave. The transmission coefficient is defined as the ratio of the transmitted particle flux to the incident particle flux and depends on the incident energy.
The sum of the reflected and transmitted energy must equal the total incident energy, so the transmission coefficient is calculated simply by subtracting the reflection coefficient. The ratio of the reflected wave amplitude to the incident wave amplitude is called the reflection coefficient.
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muscle mass in a newborn constitutes % of total body weight, while muscle mass in an adult constitutes % of total body weight.
Muscle mass in a newborn constitutes 50% of total body weight, while muscle mass in an adult constitutes 70 % of total body weight.
For the execution of daily tasks, muscle strength is crucial. There is a growing understanding that physical fitness and overall health are related. Ages 20 to 39: 75 to 89 percent for men, 63 to 75.5 percent for women. These are the usual ranges for muscle mass. Men aged 40–59: 73–86%; women aged 40–59: 62–73.5%. Men aged 60 to 79: 70 to 84 percent; women aged 60 to 72.5%. To be deemed healthy, a lean muscle percentage should be between 70% and 90%. Your body fat percentage so ranges from 10% to 30%. Body fat levels in athletes typically range from 7 to 22%. Men typically have lean mass in the 80–90% range, whereas women typically have lean mass in the 70–85% range.
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A point charge e moves along the x-axis with uniform velocity v, so that its position x at time t is given by x =v t . Derive an explicit expression for the potential φ(t ) at the point (0, b, 0). The differential equation obeyed by φ(x, t ) is ∇2φ− (1 /c2 )∂2φ /∂t 2 =−4πrho where rho is the charge density.
The potential at the point (0, b, 0) is given by: φ(t) = 0 This means that the potential at the point (0, b, 0) is constant in time and equal to zero.
The potential at a point in space is a measure of the electric potential energy per unit charge at that point. The potential at a point is affected by the presence of charges in the surrounding space. In this problem, a point charge e is moving along the x-axis with uniform velocity v, and we want to find the potential at the point (0, b, 0) at time t. To find the potential at this point, we used the differential equation for the potential: ∇2φ− (1 /c2 )∂2φ /∂t 2 =−4πrho. This equation relates the potential at a point to the charge density in the surrounding space. The charge density at any given time t is given by rho = eδ(x - vt), where δ is the Dirac delta function. This expression represents a point charge located at x = vt. Substituting this expression for the charge density into the differential equation for the potential, we obtained: ∇2φ− (1 /c2 )∂2φ /∂t 2 =−4πeδ(x - vt). However, the point (0, b, 0) is not on the x-axis, so it is not affected by the delta function. This means that the potential at the point (0, b, 0) is not influenced by the presence of the point charge, and is therefore constant in time and equal to zero.
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show an example of the method described above. ii. why might your calculated value for the half-life of barium-137 from this method differ from the accepted value?
By monitoring a sample's activity as it decays, Ba-137m can be calculated. Utilizing AktivLab, the radioisotope Ba-137half-life, m's which was isolated from Cs-137, is found.
How is the experiment's half-life determined?After that, the half-life is calculated using the basic concept of activity, which is the sum of the radionuclide decay constant,, and the number of radioactive atoms present, N. By using the formula = ln2/T1/2, one may solve for and obtain the half-life.
Describe half-life in detail.?A half-life is the duration needed for something to reduce in quantity by half. The phrase is most frequently used in reference to radioactive decay, which takes place when unstable atomic particles shed energy. There are 29 elements that can travel through this, according to known facts.
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A taxi company is trying to decide whether to purchase brand A or brand B tires for its fleet of taxis. To estimate the difference in the two brands, an experiment is conducted using 12 of each brand. The tires are run until they wear out. The results are Brand A: \overline{x}_1 = 36. 300 x
1
=36.300 kilometers, s_1s 1
= 5000 kilometers. Brand B: \overline{x}_2 = 38. 100 x
2
=38.100 kilometers, s_2 = 6100s 2
=6100 kilometers. Compute a 95% confidence interval for \mu A − \mu BμA−μB assuming the populations to be approximately normally distributed. You may not assume that the variances are equal.
A 95% confidence interval for μA-μB is 0.025
If we want to make a 95% confidence interval estimate for an unknown population mean. This means that there is a 95% probability that the confidence interval will have the true population mean.
Thus, P( [sample mean] - margin of error < μ < [sample mean] + margin of error) = 0.95.
We have,
n1 = 12
s1 = 5000
mean x1 = 36300
n2= 12
s2 =6100
mean x2 =38100
We have to find 95%confidence interval for μA- μB
from 95% = 100 (1-α%)
1-α = 0.95
α = 1-0.95 =0.05
α/2 = 0.025
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A student releases the 4kg bob of a 0.47m long
pendulum from the position shown.
What is the period of the harmonic motion?
S
What is the frequency of the harmonic motion?
Hz
How many cycles would the pendulum make in
45s?
6
(You may answer with a decimal)
6°
Projectile launch is a kinematics application to the movement of objects near the earth's surface.
How to find Projectile launch?Since the acceleration is constant, the time it takes for the body to go up is equal to the time it takes to go down; thus, the time it takes to go up is half of the total time.
tu = ttotal/2
tu = 4.70 / 2
tu = 2.35 s
The height the ball reach
The point of maximum height where the vertical velocity is zero
vy = voy - gt
0 = voy - gt
voy = gt
= 9.8 x 2.35
= 23 m / s
By using the equation.
v²y = v²oy - 2gy
0 = v²oy - 2gy
y = v²oy/2g
y = 23²/2 x 9.8
y = 27 m
To find the initial velocity,
Initial vertical velocity = 23 m / s
vx = x/t
The throw range is 44m in time of 4.60 s
vₓ = 44 / 4.60
v₀ = [tex]\sqrt{v^{2}ox + v^{2}oy }[/tex]
v₀ = [tex]\sqrt{9.56^{2}+23^{2} }[/tex]
v₀ = 24.9 m / s
The launch angle is:
Let's use trigonometry
tan θ = voy/vox
θ = tan⁻¹ voy/vox
θ = tan⁻¹ 23/9.56
θ = 67.4°
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A runner first runs a displacement A of 3.20 km due south, and then a second displacement Bthat points due east.(a) The magnitude of the resultant displacement A + B is 5.18 km. What is the magnitude (in m) of B?____kmWhat is the angle that A + Bmakes relative to due south? (Your answer must be a positive number from 0 to 180 degrees).Is this angle east or west of south?(b) Consider a situation where the runner still runs a displacement B due east, and we find that the vector A − B has a magnitude of 5.18 km. In this situation, what is the magnitude (in km) of B?_____kmWhat is the angle that A − B makes relative to due south? (Your answer must be a positive number from 0 to 180 degrees).Is this angle east or west of south?
The magnitude of the resultant displacement A + B is 5.18 km. Displacement of A is 3.2 km. The magnitude of B is 4.07 km.
It can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A + B as the hypotenuse, 3.2 km as the opposite and the displacement B as the adjacent.
By using Pythagoras theorem, the adjacent side of the triangle can be found.
(5.18)² = (3.20)² + B²
26.83 = 10.24 + B²
B² = 26.83 - 10.24
B² = 16.59
B = √16.59 = 4.07 km
Thus, the magnitude of B is 4.07 km.
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umama wants to enable user experience virtualization (ue-v) by configuring group policy settings. two important settings that she needs to configure are the setting storage location and settings template catalog location. which of the following is true of this scenario?
Correct. When new user profiles are created, the default profile is utilised. Umama wants to configure Group Policy settings to enable User Experience Virtualization (UE-V).
UEV policy settings: What are they?You can enable or disable User Experience Virtualization using this Group Policy setting (UE-V). Only versions of UE-V 2.x and earlier are affected by this setting. Use the Enable UE-V setting in Wind ows 10, version 1607, for UE-V. The location of the settings template catalogue and the location of the settings storage location are two crucial settings that she must set up.
Microsoft User Experience Virtualization: What is it?A programme called Micro soft UE-V (User Experience Virtualization) enables users to switch between Wind ows devices while keeping their current operating system (OS) and application settings.
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sand dunes form when ? ? sand build up in one place
Sand dunes form when the mound grows in circumference and height due to subsequent sand build up in one place by wind.
What is sand dunes?A sand dune is a mass of loose sand grains that have accumulated due to wind movement. The mound will eventually get so massive that it will collapse under its own weight to form a sand dune as it increases in height and circumference as a result of additional sand deposits. A set of sand dunes is referred to as a sand dune complex or sand dune system by geologists.
On the other hand, depending on the quantity of vegetation present, vast clusters of sand dunes are referred to as sand seas or sand dune fields. Ergs are another name for sand seas.
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Your question is incomplete, but most probably the question was:
sand dunes form when _______ sand build up in one place by wind.
Part B
The general kinematic equations of motion for vertical displacement can also be simplified significantly. Write the simplified equation for y in this case. Explain how you have been able to simplify it from this general displacement equation:
The general vertical displacement equation for object moving upwards is written as h = v₀t - ¹/₂gt².
What is the vertical displacement of an object?The vertical displacement of an object is the height travelled by the object.
The general kinematic equations of motion for vertical displacement is written as follows;
y = y₀ + v₀t + ¹/₂at²
where;
y₀ is initial vertical position of the objectv₀ is the initial vertical velocity of the objecta is the acceleration of the objectt is the time of motionIf the object is moving upwards, the new kinematic equation becomes;
h = 0 + v₀t - ¹/₂gt²
h = v₀t - ¹/₂gt²
where;
h is the vertical displacement of the objectv₀ is the initial vertical velocity of the objectg is the acceleration due to gravityt is the time of motion of the objectFor the simplified equation above, the acceleration of the object changes to acceleration due to gravity since the object is moving against gravity in the negative direction.
Thus, the general kinematic equations of motion for vertical displacement is written in terms of acceleration due to gravity in negative direction since it is moving upwards.
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apply the loop rule to loop 2 (the smaller loop on the right). sum the voltage changes across each circuit element around this loop going in the direction of the arrow. remember that the current meter is ideal.express the voltage drops in terms of vb , i2 , i3 , the given resistances, and any other given quantities.
The voltage drops in terms of vb , i2 , i3 , the given resistances, and any other given quantities.ΣΔV = 0 = I3 ⋅ R3 - I2 ⋅ R2.
When applying Kirchhoff's second rule the loop rule we need to identify closed loops and decide whether to loop clockwise or counterclockwise. For example, in Figure 3 the loop was traversed in the same direction as the current.
Loop 1 is the full loop and Loop 2 is the small loop on the right. To apply the loop rule, add all those voltage changes. Kirchhoff's Second Law, also known as Kirchhoff's Voltage Law states that the sum of all voltages around the closed loop of any circuit must be zero. This is a result of charge conservation and energy conservation.
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the two weights are released from rest. the coefficient of kinetic friction between the horizontal surface and the 5 lb weight is 0.18
The final velocity of the weights will be 3.31 m/s
Weight = 5 lb
Coefficient of kinetic friction (µ) = 0.18
let T be tension
[tex]m_{1}[/tex]*g-T =[tex]m_{1}[/tex]*a
[tex]m_{1}[/tex] =10 lb = 4.53 kg
T-(µ)[tex]m_{2}[/tex]g =[tex]m_{2}[/tex]a
Adding both equations, we have
([tex]m_{1}[/tex]-k[tex]m_{2}[/tex])g =[tex]m_{2}[/tex] a
a = (4.53-0.2*2.26)*10/2.26 =18.04 [tex]m/s^{2}[/tex]
Using the equation of motion, we have,
[tex]v^{2} -u^{2} =2as[/tex]
Since the body starts from rest, the initial velocity, that is, u is zero.
So, u = 0 m/s
Putting this value in the above equation to calculate the final velocity, we have,
[tex]v^{2}[/tex] = 2as
v = [tex]\sqrt{(2*18.04*0.304) }[/tex]= 3.31 m/s
Thus, the final velocity of the weights will be 3.31 m/s
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