Answer:
Potential difference is measured in volts
Explanation:
The standard metric unit on electric potential difference is the volt, abbreviated V and named in honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb.
Answer:
Your answer is A.) volts
Explanation:
A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is at a radial distance of 1.8 m from the central axis. Which is true?A- Boy and girl have zero tangential and angular accelerations.B- The girl has a larger angular acceleration than the boy.C- The boy has a larger tangential acceleration than the girl.D- The boy has a larger angular acceleration than the girl.E- The girl has a larger tangential acceleration than the boy.
Answer:
E) True. The girl has a larger tangential acceleration than the boy.
Explanation:
In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.
Angular and linear quantities are related
v = w r
a = α r
the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m
as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different
v₁ = w 1,2 (boy)
v₂ = w 1.8 (girl)
whereby
v₂> v₁
reviewing the claims we have
a₁ = α 1,2
a₂ = α 1.8
a₂> a₁
A) False. Tangential velocity is different from zero
B) False angular acceleration is the same for both
C) False. It is the opposite, according to the previous analysis
D) False. Angular acceleration is equal
E) True. You agree with the analysis above,
What's a line of best fit? Will give BRAINLIEST
A line of best fit expresses the relationship between the points.
Explanation:
It does not go through all the points but goes through most of them and it is like a hardrawn curve
A solid exerts a force of 500 N. Calculate the pressure exerted to the surface where area
of contact is 2000 cm2.
Answer:
2500 N/m²
Explanation:
Pressure: This can be defined as the force acting normally on a surface per unit area.
The expression for pressure is give as
P = F/A...................... Equation 1
Where P = pressure (N/m²), F = force (N), A = Contact area (m²)
Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.
Substitute into equation 1
P = 500/0.2
P = 2500 N/m²
Hence the pressure exerted to the surface is 2500 N/m²
A student has derived the following nondimensionally homogeneous equation: a=x/t2-vt+F/m where v is a velocity's magnitude , a is an acceleration's magnitude, t is a time, m is a mass, F is a force's magnitude , and x is a distance (or length). Which terms are dimensionally homogeneous? .
a) x/t
b) vt
c) a
d) F/m
Answer:
Letter C) and D) is the correct answer.
Explanation:
We know that the a is an acceleration's magnitude, so the units of a are m/s².
Now, let's analyze each terms. If we want that each term will be dimensionally homogeneous, all of them must have the same units of a.
[tex][\frac{x}{t}]=[\frac{m}{s}][/tex]
[tex][vt]=[m][/tex]
[tex][\frac{F}{m}]=[\frac{N}{kg}]=[kg\frac{m}{s^{2}kg}]=[\frac{m}{s^{2}}][/tex]
Therefore, the term F/m is the correct answer.
I hope it helps you!
We can see that a and F/M are dimensionally homogeneous.
In solving dimensions, we try to express a quantity in terms of the fundamental quantities;
MassLengthTimeFor the term a, its dimension is LT^-2
For the term F/m, its dimension is LT^-2
Hence, it follows that a and F/M are dimensionally homogeneous.
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Problem 3A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane
Answer:
5.1 rad/s
Explanation:
Mechanical energy of the system is conserved since no external work is done on the sphere.
[tex]mgh = mv^2/2 + I\omega^2/2[/tex]
Substituting v = ωr and I = 2 m r^2/5, we get,
=> [tex]mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2[/tex]
=> [tex]mgh = m\omega^2r^2/2 + m\omega^2r^2/5[/tex]
=> [tex]gh =\omega^2r^2/2+\omega^2r^2/5[/tex]
=> [tex]gh = 7\omega^2 r^2/10[/tex]
=> [tex]\omega r = (10gh/7)^{1/2}[/tex]
=> [tex]\omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2}[/tex] = 5.1 rad/s
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.653 kJ of heat. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Express your answer to three significant figures and include the appropriate units.
Answer:
ΔE = ‒0.271 kJ
Explanation:
Let's begin by listing out the given variables:
q = -0.653 kJ, w = 0.389 kJ
Using the formula ΔE = q + w
ΔE = -0.653 + 0.388
ΔE = (‒0.655 + 0.382) kJ
ΔE = ‒0.271 kJ
Therefore, the change in internal energy is -271 J or -0.271 kJ which implies that the system is exothermic
Which formation is one feature of karst topography?
Sinkholes formation is one feature of karst topography. The top of a cave falls if it develops large enough and its top extends near enough to the surface.
What is karst topography?Karst topography is a type of natural environment formed mostly by chemical weathering by water, resulting in caves, sinkholes, cliffs, and steep-sided hills known as towers.
The top of a cave falls if it develops large enough and its top extends near enough to the surface. Sinkholes are formed as a result of this, and they are one of the most distinguishing aspects of karst terrain.
When water absorbs carbon dioxide from the atmosphere and ground, it becomes carbonic acid.
Hence, sinkholes formation is one feature of karst topography
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Answer: A) Caves
Explanation:
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 4 s? (g = 10 m/s2)
Answer:
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
Explanation:
From the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = displacement
v = initial velocity = 0 (dropped with no initial speed)
t = time of flight = 4s
a = g = acceleration due to gravity = 10 m/s^2
Substituting the given values into equation 1;
d = 0(4) + 0.5(10 × 4^2)
d = 0.5(10×16)
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / -173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.
Explanation:
Complete the first and second sentences, choosing the correct answer from the given ones.
1. T = 100 K
[tex]^{\circ}C=K-273[/tex]
Put T = 100 K
[tex]T=100-273=-173^{\circ} C[/tex]
A temperature of 100 K corresponds on a Celsius scale to (-173 °C)
2. T = 50 °C
[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]
So, At 50 °C, it corresponds to a Kelvin scale of 323 K.
The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic field used to align the spins, with proportionality constant 5 hz/T. If the strength of the applied field is known to be 20 T plus or minus 3 T, which of the following correctly describes the uncertainty in the INVERSE frequency (1/frequency)?
A. 3/2000s
B. 3/5s
C. 1/15s
D. 1/4
Complete Question
The complete question is shown on the first uploaded image
Answer:
The uncertainty in inverse frequency is [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]
Explanation:
From the question we are told that
The value of the proportionality constant is [tex]k = 5 \frac{Hz }{T}[/tex]
The strength of the magnetic field is [tex]B = 20 \ T[/tex]
The change in this strength of magnetic field is [tex]\Delta B = 3 \ T[/tex]
The magnetic field is given as
[tex]B = \frac{k}{\frac{1}{w} }[/tex]
Where [tex]w[/tex] is frequency
The uncertainty or error of the field is given as
[tex]\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ][/tex]
The uncertainty in inverse frequency is given as
[tex]\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}[/tex]
[tex]\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }[/tex]
substituting values
[tex]\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }[/tex]
[tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]
A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity
a. The potential at the center of the sphere is zero.
b.The potential is lowest, but not zero, at the center of the sphere.
c. The potential at the center of the sphere is the same as the potential at the surface.
d. The potential at the center is the same as the potential at infinity.
e. The potential at the surface is higher than the potential at the center.
Answer:
a. FALSE
b. FALSE
c. TRUTH
d. FALSE
e. FALSE
Explanation:
To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:
[tex]V=\frac{Q}{4\pi \epsilon_o R}[/tex] inside the sphere
[tex]V'=\frac{Q}{4\pi \epsilon_o r}[/tex] for r > R (outside the sphere)
R: radius of the sphere
ε0: dielectric permittivity of vacuum
Q: charge of the sphere
As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.
Hence, you can conclude:
a. The potential at the center of the sphere is zero. FALSE
b.The potential is lowest, but not zero, at the center of the sphere. FALSE
c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH
d. The potential at the center is the same as the potential at infinity. FALSE
e. The potential at the surface is higher than the potential at the center. FALSE
A cylindrical shell of radius 7.00 cm and length 2.44 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 21.9 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.(a) Use approximate relationships to find thenet charge on the shell.
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from theaxis, measured radiallyoutward from the midpoint of the shell.
Answer:
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.
Consider a system of a cliff diver and the earth. The gravitational potential energy of the system decreases by 24,500 J as the diver drops to the water from a height of 44.0 m. Determine her weight in newtons.
Before she jumped from the cliff, her gravitational potential energy was
GPE = (her weight) x (height of the cliff) .
That's exactly the GPE that she lost on the way down to the water. So we can write
24,500 J = (her weight) x (44.0 m)
Divide each side by 44.0 m:
Her weight = 24,500 J / 44 m
Her weight = 556.8 Newtons
(about 125 pounds)
To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.7m. If the thrower takes 1.2s to complete one revolution, starting from rest, what will be the speed of the discus at release?
Answer:
4.437 m/s
Explanation:
Diameter of rotation d is 1.7 m
Radius of rotation = d/2 = 1.7/2 = 0.85 m
If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s
We convert to rad/s
Angular speed = 2 x pi x 0.83
= 2 x 3.142 x 0.83 = 5.22 rad/s
Speed is equal to the angular speed times the radius of rotation
Speed = 5.22 x 0.85 = 4.437 m/s
In the given case, the speed of the discus at release, If the thrower takes 1.2s to complete one revolution, starting from rest would be - 8.90 m/s.
Given:
diameter of the circle = 1.7 mradius f the circle would be = 1.7/2 = 0.85 m
time taken for one revolution t = 1.2 sThis rotation exercise can be treated using the rotation kinematics.
Angular acceleration:
θ = w₀ t + ½ α t²
t = 1.2 s to give a revolution (T = 2π rad) and with part of the rest the initial angular velocity is zero (wo = 0)
=> θ = 0 + ½ α t²
=> α = 2θ / t²
=> α= 2 × 2π / 1.2²
=> α = 4π = 8.7266 rad / s²
Let's calculate the angular velocity:
=> w = wo + α t
=> w = 0 + α t
=> w = 8.7266 × 1.2
=> w = 10.47192 rad / s
The relationship between linear and angular velocity is
=> r = d / 2
=> r = 1.7 / 2 = 0.85 m
=> v = w r
=> v = 10.47192 × 0.85
=> v = 8.90 m / s
Thus, the correct speed would be - 8.90 m/s
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A 1000-kg car is moving down the highway at 14m/s. What is the momentum?
Explanation:
Momentum = mass × velocity
p = mv
p = (1000 kg) (14 m/s)
p = 14000 kg m/s
The momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s
Given the data in the question
Mass of the car; [tex]g = 1000kg[/tex]Velocity of the car; [tex]v = 14m/s[/tex]Momentum; [tex]p = ?[/tex]Momentum is the product of the mass of a particle and its velocity.
Momentum = Mass × Velocity
[tex]P = m \ * \ v[/tex]
We substitute our given values into the equation
[tex]P = 1000kg \ * \ 14m/s\\\\P = 14000kg.m/s[/tex]
Therefore, the momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s
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Use the Lab screen to expand your ideas about what affects the landing location and path of a projectile. List any discoveries you made to identify additional things that affect the landing site of a projectile and/or path of a projectile. Next to each item, briefly explain why you think the motion of the projectile is affected..
Answer:
* air resistance.
*the direction of the rotation of the Earth
rotation of the thrown body
Explanation:
The projectile launch is described by the expressions
x-axis x = v₀ₓ t
y-axis y = [tex]v_{oy}[/tex] t - ½ gt²
When the things that affect this movement are analyzed, in order of importance we have:
* air resistance. This significantly changes the body's horizontal position, so it introduces a horizontal acceleration that is not contained in the equations.
* air resistance. At the height that the body reaches, since air resistance has the same direction as the gravity of gravity and therefore the relationship is more challenging.
* to a lesser extent the direction of launch, in the direction of the rotation of the Earth against. Since this creates an operational on the x and y axis that changes the initial assumption
* The possible rotation of the thrown body, since this rotation creates a lift that is not taken in the equations, this value is more noticeable the lighter the body, this effect has to keep the body longer in the air achieving more reach and height
A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.
Answer:
4.93 m
Explanation:
According to the question, the computation of the height is shown below:
But before that first we need to find out the speed which is shown below:
As we know that
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{5}{0.504}[/tex]
= 9.92 m/s
Now
[tex]v^2 - u^2 = 2\times g\times h[/tex]
[tex]9.92^2 = 2\times 9.98 \times h[/tex]
98.4064 = 19.96 × height
So, the height is 4.93 m
We simply applied the above formulas so that the height i.e H could arrive
wha is amplitde in sound
Answer:
The number of molecules displaced in a vibration makes the amplitude of a sound.
EXPLANATION ⛔
A 20 gram mass is suspended from meter rod at 20cm. The meter rod is balanced on 40cm mark. Weight of meter rod is
A. 0.4N
B. 0.6N
C. 6N
D. 60N
Answer:b
Explanation:I’m just trynna get more money dude
1. Calculate the centripetal force exerted on a 900kg900kg car that rounds a 600m600m radius curve on horizontal ground at 25.0m/s25.0m/s. 2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
Explanation:
It is given that,
Mass of a car is 900 kg
Radius of curve is 600 m
Speed of the car in the curve is 25 m/s
We need to find the centripetal force exerted on a car. The formula used to find the centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{900\times (25)^2}{600}\\\\F=937.5\ N[/tex]
So, the centripetal force exerted on a car is 937.5 N.
Static friction prevents the car from slipping. It means that the magnitude of centripetal force is balanced by the frictional force. So, the frictional force of 937.5 N is acting on the car.
A jeep starts from rest with a constant acceleration of 4m/s2.At the same time a car travels with a constant speed of 36km/h overtake and passes the jeep how far beyond the starting point will the jeep overtakes the car?
Answer:
25m
Explanation:
Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.
While the Jeep is accelerating to that speed, the car with that speed passes it.
Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.
This time is t = v/a; from Newton's Law of Motion:
a = V-U / t ; a-acceleration
V is final velocity = 36km/h
U is initial velocity 0 since the body starts from rest.
Hence t = 36000/3600 ÷ 4 = 2.5s
Note conversting from km/h to m/s we multiply by 1000/3600.
But the distance covered by the car while the Jeep just accelerates is
S = U × t = 10× 2.5 = 25m.
Note From Newton's law of Motion, distance for constant speed is defined as: U × t
Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.
The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A If the suitcase has a mass of 80.0 kg , how far can it be pushed across the level floor with 660 J of work
Answer:
Explanation:
The work required to push will be equal to work done by friction . Let d be the displacement required .
force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction
work done by force of friction
mg x μ x d = 660
80 x 9.8 x .272 x d = 660
d = 3 .1 m .
A ball is thrown straight up with an initial speed of 30 m/s. How long will it take to reach the top of its trajectory, and high will the ball go?
Answer:
About 3.06 seconds
Explanation:
[tex]v_f=v_o+at[/tex]
Since at the peak of its trajectory, the ball will have no velocity, you can write the following equation:
[tex]0=30+(-9.81)t\\\\-30=-9.81t\\\\t\approx 3.06s[/tex]
Hope this helps!
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying pan is only 0.400 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)? N
Answer:
normal force = 10 N
Explanation:
Given data
frictional force = 0.400 N
coefficient of kinetic friction = 0.04
Solution
we get here normal force that is express as
normal force = [tex]\frac{Frictional\ force}{coefficient\ of\ friction}[/tex] ............1
put here value and we will get value
normal force = [tex]\frac{0.400}{0.04}[/tex]
solve it we get
normal force = 10 N
PLS HELP,WILL GIVE BRAINLIEST + 30 POINTS
Describe how fractional distillation and cracking are used so that sufficient petrol is produced from crude oil to meet demand.
Answer:
☟
Explanation:
Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.
Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:
heated to 600-700°C
passed over a catalyst of silica or alumina
These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:
hexane → butane + ethene
C6H14 → C4H10 + C2H4
Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.
Fractional distillation of crude oil
Fractional distillation separates a mixture into a number of different parts, called fractions.
A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.
Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.
Oil fractions
The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.
As you go up the fractionating column, the hydrocarbons have:
lower boiling points
lower viscosity (they flow more easily)
higher flammability (they ignite more easily).
Other fossil fuels
Crude oil is not the only fossil fuel.
Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.
Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.
In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.
What is fractional distillation?Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.
Generally, the components have boiling points that differ by less than 25 °C from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.
The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.
Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.
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A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of the circle. (A) What is the speed of the stopper at the bottom of the circle? (HINT: Use energy conservation principles!) (10.2 m/s) (B) What is the tension in the string when the stopper is at the top of the circle? (0.27 N) (C) What is the tension in the string when the stopper is at the bottom of the circle?
Answer:
Explanation:
A )
At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy
1/2 m V² = mg x 2r + 1/2 mv²
m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top
1/2 V² = g x 2r + 1/2 v²
V² = g x 4r + v²
V² = 9.8 x 4 + 8²
V² = 103.2
V = 10.16 m/s
B )
If T be the tension at the top
Net downward force
= mg + T . This force provides centripetal force for the circular motion
mg +T = mv² / r
T = mv²/r -mg
= m ( v²/r - g )
= .005 ( 8²/1 -g )
= .005 x 54.2
= .27 N .
C ) At the bottom
Net force = T - mg , T is tension at the bottom , V is velocity at bottom
T-mg = mV²/r
T = m ( V²/r +g )
= .005 ( 10.16²/1 +9.8)
= .005 x 113
= .56 N .
Crystalline germanium (Z=32, rho=5.323 g/cm3) has a band gap of 0.66 eV. Assume the Fermi energy is half way between the valence and conduction bands. Estimate the ratio of electrons in the conduction band to those in the valence band at T = 300 K. (See eq. 10-11) Assume the width of the valence band is ΔΕV ~ 10 eV.
Answer:
= 8.2*10⁻¹²
Explanation:
Probability of finding an electron to occupy a state of energy, can be expressed by using Boltzmann distribution function
[tex]f(E) = exp(-\frac{E-E_f}{K_BT} )[/tex]
From the given data, fermi energy lies half way between valence and conduction bands, that is half of band gap energy
[tex]E_f = \frac{E_g}{2}[/tex]
Therefore,
[tex]f(E) = exp(-\frac{E-\frac{E_g}{2} }{K_BT} )[/tex]
Using boltzman distribution function to calculate the ratio of number of electrons in the conduction bands of those electrons in the valence bond is
[tex]\frac{n_{con}}{n_{val}} =\frac{exp(-\frac{[E_c-E_g/2]}{K_BT} )}{exp(-\frac{[E_v-E_fg/2}{K_BT} )}[/tex]
[tex]= exp(\frac{-(E_c-E_v}{K_BT} )\\\\=exp(\frac{-(0.66eV)}{(8.617\times10^-^5eV/K)(300K)} )\\\\=8.166\times10^-^1^2\approx8.2\times10^{-12}[/tex]
Which is the best description of the scientific theory
Explanation:
a scientific theory is a well substantiated explanation of some aspect of the nature world, based on a body of facts that have been repeatedly confirmed through observation and experiment. search fact-supported theories are not "guesses" but reliable account of the real world .
commune time to work ( physics) i need help pls :(
A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring
Answer:
a) The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex], b) The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].
Explanation:
This question is incomplete and complete version will be presented herein:
A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.
a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)
[tex]m\cdot g = k\cdot \Delta x[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]\Delta x[/tex] - Spring linear deformation, measured in meters.
Now, the spring constant is cleared in this equation and outcome is computed: ([tex]m = 0.573\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta x = 0.198\,m[/tex])
[tex]k = \frac{m\cdot g}{\Delta x}[/tex]
[tex]k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}[/tex]
[tex]k = 28.381\,\frac{N}{m}[/tex]
The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex].
b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Given that [tex]k = 28.381\,\frac{N}{m}[/tex] and [tex]m = 0.573\,kg[/tex], the angular frequency, measured in radians per second, of the block is:
[tex]\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }[/tex]
[tex]\omega = 7.038\,\frac{rad}{s}[/tex]
The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].
