Answer:
convection is the answer
Explanation:
good luck on your assignment <3
how much work does an ideal battery with a 3.50 v do to an electron that passes through the battery from the positive to the negative terminal?
An ideal battery with a 3.50 V, does a work of 3.50 J on an electron passing through the battery from the positive to the negative terminal.
What is work?Work can be defined as the energy transfer that occurs when an object is moved through a distance by a force that is applied to it. A positive work indicates that energy is transferred to the system from the surroundings, and a negative work indicates that energy is transferred from the system to the surroundings.
Voltage can be defined as the electric potential energy per unit charge of an electric field. The unit of voltage is the volt (V), and it is the energy per charge that must be imparted to move a unit charge from the negative to the positive terminal of an electric circuit or to move an electron from a point of low potential to a point of high potential.
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what is the kinetic energy, in joules, of each ejected electron when light of 258.0 nm strikes the metal surface?
When the light of 258.0 nm strikes the metal surface, each ejected electron has a kinetic energy of 4.80 eV.
To calculate the kinetic energy, we use the formula:
Kinetic Energy (KE) = hc/λ, where h is Planck's constant (6.626×10⁻³⁴ Js), c is the speed of light (2.998x10⁸ m/s) and λ is the wavelength of the light (258.0 nm).
Therefore,
KE = (6.626x10⁻³⁴ Js)(2.998x10⁸ m/s) / (2.58x10^-7 m)
= 7.69x10⁻¹⁹ J = 4.80eV, where (1eV = 1.6 x 10⁻¹⁹ J)
Thus, each ejected electron has a kinetic energy of 4.80 eV or 7.69x10⁻¹⁹ J when the light of 258.0 nm strikes the metal surface.
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Blocks A (mass 3.50 kg) and B (mass 10.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 9.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of A. Find the maximum energy stored in the spring bumpers and the velocity of each block at the time of the collision
The total energy that can be stored in the spring bumpers is 43.8 J, or KE = 43.8.
What is the formula for energy capacity?The battery's power capacity is the amount of energy it can hold. Its power is commonly stated in Watt-hours (the symbol Wh) (the symbol Wh). A Watt-hour is equal to the voltage (V) and current (Amps) that a battery can produce for a specific period of time (generally in hours). Voltage * Amps * hours = Wh.
Block A's momentum before to the impact can be calculated using the formula p1 = m1v1 = (3.50 kg)(9.00 m/s) = 31.5 kgm/s.
Block B's initial momentum is p2 = m2v2 = 0, indicating that it is at rest.
Prior to the collision, the system's total momentum was equal to 31.5 kgm/p1 + p2.
[tex]p1 + p2 = (m1 + m2)v[/tex]
[tex]31.5 kgm/s = (3.50 kg + 10.00 kg) * v[/tex]
[tex]31.5 kgm/s = 13.50 kg * v[/tex]
[tex]v = 31.5 kg*m/s / 13.50 kg = 2.33 m/s[/tex]
The kinetic energy of block A before the collision is given by KE1 = ([tex]1/2)m1v1^2 = (1/2)(3.50 kg)(9.00 m/s)^2 = 141.8[/tex] J
The kinetic energy of block B before the collision is KE2 = [tex](1/2)m2v2^2 = 0[/tex]
The total kinetic energy before the collision is KE1 + KE2 = 141.8 J
[tex]ΔKE = KEf - KEi = (1/2)(m1 + m2)v^2 - KE1 - KE2[/tex]
[tex]ΔKE = (1/2)(3.50 kg + 10.00 kg)(2.33 m/s)^2 - 141.8 J - 0[/tex]
ΔKE = 43.8 J[tex]31.5 kgm/s = 13.50 kg * v[/tex]
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a 23.9 a current flows in a long, straight wire. find the strength of the resulting magnetic field at a distance of 58.3 cm from the wire.
The magnetic field at a distance of 58.3 cm from a long, straight wire carrying a 23.9 A current, the strength of the resulting magnetic field can be found using the equation B = μ0*I/2π*r, where B is the magnetic field strength, μ0 is the permeability of free space, I is current, and r is the distance.
Therefore, the strength of the magnetic field at 58.3 cm from the wire is B = 4π * 10-7 * 23.9/2π * 58.3 = 0.0067 N/Amp.
The magnetic field strength due to the current in the wire is caused by the current producing a magnetic field, which is a result of moving electric charges (electrons) in the wire. The strength of the magnetic field depends on the magnitude of the current and the distance from the wire.
As the current increases, the magnetic field strength increases; likewise, as the distance from the wire increases, the magnetic field strength decreases. The direction of the magnetic field can be determined using the right-hand rule.
The strength of the magnetic field can be used to calculate the force on a moving charged particle, F = q * v * B, where q is the charge of the particle, v is its velocity, and B is the magnetic field strength. By using this equation, the force acting on a charged particle due to the magnetic field at 58.3 cm from the wire can be found.
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q101 is a local radio station operating at 101.7 mhz. a. what is the wavelength of their radio waves?
The wavelength of q101 is equal to 2.946 meters.
The wavelength of a radio wave is determined by the frequency, and for q101 the frequency is 101.7 MHz.
The formula for calculating wavelength is: wavelength = speed of light (3 x 10^8 m/s) divided by the frequency (101.7 MHz).
The wavelength of a radio wave is the distance from the crest of one wave to the crest of the next, and the frequency is the number of waves passing a point in a second.
As the frequency increases, the wavelength decreases, and vice versa.
Since q101 is operating at 101.7 MHz, its wavelength is much shorter than a station operating at a lower frequency, such as the FM station 88.3 MHz, which has a wavelength of 3.41 meters.
The wavelength is also important in antenna design. An antenna needs to be designed according to the specific wavelength of the station in order to pick up the signal. In the case of q101, a 2.946 meter antenna is needed.
q101 is a local radio station operating at 101.7 MHz, and its wavelength is 2.946 meters. The wavelength is determined by the frequency, and is also important in antenna design.
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the force of gravity on the side of the earth facing the moon is the force of gravity acting on the center of the earth group of answer choices greater than smaller than equal to
The force of gravity on the side of the earth facing the moon is greater than the force of gravity acting on the center of the earth.
This is because of the gravitational attraction between the earth and the moon.
The moon’s gravity pulls on the side of the earth that is closer to it, resulting in a larger gravitational force on that side than on the center of the earth. The size of the force on the side of the earth is slightly more than double that at the center, due to the inverse square law.
Thus, the force of gravity at the side of the earth facing the moon is greater than the force of gravity acting on the center of the earth.
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Complete question:
The force of gravity on the side of the earth facing the moon is the force of gravity acting on the center of the earth
greater than
smaller than
equal to
1st attempt rolling-circle replication of plasmids proceeds choose one: in opposite directions from multiple origin sites. in one direction from multiple origin sites. in one direction from a single fixed origin. in opposite directions from a single fixed origin.
Based on the given options, the correct answer is: "in one direction from a single fixed origin."
The 1st attempt rolling-circle replication of plasmids proceeds in one direction from a single fixed origin.
This process involves the initiation of DNA replication from a specific origin site on the plasmid.
The replication then proceeds in a circular direction, generating multiple copies of the plasmid.
Overall, plasmids are small, circular pieces of DNA that are separate from the chromosome.
They replicate independently of the chromosome and can carry genes that provide a selective advantage to the cell, such as antibiotic resistance.
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Do any of the force pairs suggested in Question 5 not produce an acceleration? If so which one(s).
A. A skier uses her ski poles to start moving downhill
B. A boat propeller spins rapidly in the water
C. A baseball player hits a pitched ball with a bat
D. A party balloon contains rapidly moving helium atoms
All of the given options produce an acceleration that are force pairs suggested in Question 5.
When a skier uses her ski poles to start moving downhill then the ski poles exert a backward force on the ground while the ground exerts a forward force on poles and produces acceleration.
Similarly in case B. when a boat propeller spins rapidly in the water the propeller exert a backward force on the water while the water exerts a forward force on propeller and produces acceleration.
In case C. when a baseball player hits a pitched ball with a bat the bat exert a backward force on the ball while the ball exerts a force away from bat and produces acceleration.
In case D. when a party balloon contains rapidly moving helium atoms the helium atoms exert an outward force on the balloon while the balloon exerts an inward force on helium atoms and produces acceleration.
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star y appears much brighter than star z when viewed from earth, but is found to actually give off much less light. assign a set of possible values for the apparent and absolute magnitudes of these stars that would be consistent with the information given in the previous statement. explain your reasoning.
Star Y appears to be much brighter than star Z when viewed from Earth, but is found to actually give off much less light.
This could be due to a number of factors, such as the distance of the stars from Earth, their relative sizes, and other characteristics. To be consistent with this statement, the apparent magnitude (m) of star Y should be lower than that of star Z, while the absolute magnitude (M) of star Y should be higher than that of star Z.
For example, if star Y has an apparent magnitude of -2 and an absolute magnitude of +2, and star Z has an apparent magnitude of 0 and an absolute magnitude of -2, this would be consistent with the information given.
This is because star Y appears brighter than star Z, since its apparent magnitude is lower, but star Y gives off less light since its absolute magnitude is higher.
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if you place a positive test charge at the origin, would the test charge be at a point of stable equilibrium?
If you place a positive test charge at the origin, it will not be at a point of stable equilibrium. In fact, it will be at a point of unstable equilibrium.
The reason for this is that there will be no other charges in the system that will stabilize the position of the test charge.
Let's explore this idea further.
First, let's define what we mean by equilibrium. An equilibrium point is a point where the net force on an object is zero. This means that if we place an object at an equilibrium point, it will remain there unless a force is applied to it. There are two types of equilibrium points: stable and unstable.
A stable equilibrium point is one where if an object is displaced slightly from that point, it will experience a force that will push it back toward the equilibrium point. An unstable equilibrium point is one where if an object is displaced slightly from that point, it will experience a force that will push it further away from the equilibrium point.
In the case of a positive test charge at the origin, there are no other charges in the system that will exert a force on the test charge that will push it toward the origin. In fact, any slight displacement of the test charge from the origin will cause it to experience a force that will push it further away from the origin. Therefore, the equilibrium point at the origin is an unstable equilibrium point.
In summary, if you place a positive test charge at the origin, it will not be at a point of stable equilibrium. Instead, it will be at a point of unstable equilibrium where any slight displacement will cause it to experience a force that will push it further away from the origin.
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What evidence is there to explain how the temperature of the blocks can be measured?
Answer:
in addition infrared thermometers don't measure metal surfaces particularly well anyway (metals typically have a low emissivity). Measuring electrical resistance is better.
what is the maximum speed with which a 1200- kg car can round a turn of radius 85.0 m on a flat road if the coefficient of static friction between tires and road is 0.60?
When you want to find the maximum speed with which a 1200- kg car can round a turn of radius 85.0 m on a flat road if the coefficient of static friction between tires and road is 0.60, you can use the following formula: v = sqrt(μrg).
Where:v represents the maximum speed with which the car can round a turn r is the radius of the turn g is the acceleration due to gravity, andμ is the coefficient of frictionIn this case, the mass of the car is 1200 kg and the radius of the turn is 85.0 m, while the coefficient of friction is 0.60.
To find the acceleration due to gravity, we can use the value 9.81 m/s². Therefore:v = sqrt(0.60 * 9.81 m/s² * 85.0 m) = 23.7 m/sTherefore, the maximum speed with which the 1200-kg car can round a turn of radius 85.0 m on a flat road if the coefficient of static friction between tires and road is 0.60 is approximately 23.7 m/s.
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In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A rider is towed at a constant speed by a rope that is at an angle of 19 ∘
from horizontal. The tension in the rope is 1500 N. The force of the sail on the rider is 30∘
from horizontal
We may use trigonometry to address this issue by dividing the forces into their horizontal and vertical components.
...... 'S,""" '
T horizontal equals Tension * cos(19°)
T vertical = 1437.61 N
Then, we may determine the tension force's vertical component:
T vertical equals Tension * Sin(19°)
T horizontal = 484.94 N
We can now calculate the horizontal component of the sail's force on the rider:
F horizontal is equal to F sail * cos(30°).
vertical = 25.98 N
Last but not least, we may determine the vertical component of the sail's force on the rider:
F vertical is F sail times sin(30°).
F horizontal = 14.99 N
The net horizontal force must be zero since the rider is not accelerating in the horizontal direction. In light of this, the horizontal component of the tension force and the horizontal component
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find an expression for the magnitude of the force on the lower book by the upper book in the elevator situation. is it equal to the weight of the upper book? should it be?
Yes, the magnitude of the force on the lower book by the upper book in the elevator situation is equal to the weight of the upper book. This can be expressed as F = mg,
What is magnitude expression?The expression for the magnitude of the force on the lower book by the upper book in the elevator situation is;
F(lower on upper) = (m(lower) + m(upper))g
Where F(lower on upper) is the magnitude of the force on the lower book by the upper book,
m(lower) is the mass of the lower book,
m(upper) is the mass of the upper book and
g is the acceleration due to gravity.
Therefore, the magnitude of the force on the lower book by the upper book is not equal to the weight of the upper book since it takes into account the mass of the lower book as well.
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a 5100-pound vehicle is driven at a speed of 30 miles per hour on a circular interchange of radius 100 feet. to keep the vehicle from skidding off course, what frictional force must the road surface exert on the tires?
The frictional force must the road surface exert on the tires is 58.667 ft / s.
Weight of the vehicle W = 5600 lb
Speed v = 40 miles/h
Radius of circular interchanger = 100 feet.
mass of the vehicle m= w/g = 5600 lb / 32 ft/s2
= m = 175 lb s2 / ft.
Speed of the vehicle V = ds/dt.
V = 40 miles/h 1mile = 5280ft
=40 x 5280 ft / 3600 S
V = 58.667 ft / s
Also curvature k = 1/r = 100ft.
when a vehicle in moving along a circular track, the tyres have a tendancy to slip outwards So to avoid skidding the surface exerts frictional force on the times towards the cente
frictional force F = m x normal component of acceleration
= m x an.
where a_N = k (ds/dt)^2 = kv^2.
F = mk v^2.
Frictional force is a force that opposes the relative motion or tendency of motion between two surfaces in contact. It arises due to the roughness and irregularities present on the surfaces in contact.Static frictional force is the force that prevents two objects from moving relative to each other when a force is applied to them. It is always equal and opposite to the applied force until the maximum value of static frictional force is reached.
Kinetic frictional force is the force that opposes the motion of two surfaces sliding over each other. It is generally less than the maximum static frictional force. The magnitude of frictional force depends on various factors such as the nature of the surfaces in contact, the normal force acting between them, the temperature, and the presence of any lubricants.
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Complete Question: -
A 5600-pound vehicle is driven at a speed of 40 miles per hour on a circular interchange of radius 100 feet. To keep the vehicle from skidding off course, what frictional force must the road surface exert on the tires? (Round your answer to one decimal place.)
a reservoir behind a dam is 15 m deep. what is the pressure a. at the base of the dam? b. 5.0 m from the top of the dam?
a. The pressure at the base of the dam is 147.15 kPa.
b. The pressure 5.0 m from the top of the dam is 98.1 kPa.
The pressure at the base of the dam can be calculated using the formula:
P = ρgh
where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth of the liquid.
Assuming the density of water is 1000 kg/m³ and acceleration due to gravity is 9.81 m/s², the pressure at the base of the dam is:
P = 1000 x 9.81 x 15
P = 147,150 Pa or 147.15 kPa
Therefore, the pressure at the base of the dam is 147.15 kPa.
b. To calculate the pressure 5.0 m from the top of the dam, we can use the formula:
P = ρgh
where h is the depth of the liquid from the surface to the point where we want to calculate the pressure. In this case, h = 15 - 5 = 10 m.
Using the same values for density and acceleration due to gravity, the pressure at 5.0 m from the top of the dam is:
P = 1000 x 9.81 x 10
P = 98,100 Pa or 98.1 kPa
Therefore, the pressure 5.0 m from the top of the dam is 98.1 kPa.
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calculate the centripetal acceleration, in m/s2, at the tip of a 3.50-meter-long helicopter blade that rotates at 300 rev/min.
To calculate the centripetal acceleration in m/s2 at the tip of a 3.50-meter-long helicopter blade that rotates at 300 rev/min, the given values should be converted into suitable units.
Then, we can use the following formula:Centripetal acceleration = (angular velocity)2 (radius)The conversion factor for rpm (rev/min) to rad/s is 2π/60 radians/second.
Therefore,Angular velocity = (300 rev/min)(2π/60) = 31.42 rad/sRadius = 3.50 centripetal acceleration = (31.42 rad/s)2 (3.50 m)= 3476 m/s2Therefore, the centripetal acceleration at the tip of a 3.50-meter-long helicopter blade that rotates at 300 rev/min is 3476 m/s2.
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the generator of a car idling at 1300 rpm produces 14.9 v . part a what will the output be at a rotation speed of 2100 rpm , assuming nothing else changes? express your answer to three significant figures and include the appropriate units.
The output of the car generator at a rotation speed of 2100 rpm would be 24.1 V, assuming that nothing else changes.
To calculate the output of the generator, we can use the formula:-
Output Voltage = (RPM/1300) x 14.9 V
We know that the generator is rotating at 2100 rpm, so the output voltage is:-
Output Voltage = (2100/1300) x 14.9 V= 24.09 V (rounded to three significant figures)
Therefore, the output voltage of the car generator at a rotation speed of 2100 rpm is 24.1 V (rounded to one decimal place), assuming nothing else changes. The appropriate unit of voltage is volts (V).
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water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. the density of water is 1 000 kg/m3. determine its average velocity. multiple choice question. 20 m/s 200 m/s 0.02 m/s 2 m/s 0.2 m/s
Option D: 2 m/s is the average velocity of the water flowing through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s.
According to the question:
cross-sectional area of the pipe = 0.002m²
Mass flowrate = 4 kg/s
Density of water = 1000 kg/m³
We are asked to find, average velocity =?
Average velocity is the net or total displacement covered by a body in a given time. The mass flow rate divided by the pipe's cross-sectional area and density ratio is the formula for calculating a fluid's average velocity.
As a result, the water's average flow rate through the pipe is provided by:
v = m / (ρ × A)
where, v is the average velocity, m is the mass flow rate, ρ is the density of water, and A is the cross-sectional area of the pipe. Substituting the values in the above equation we get:
v = 4 / (1000 × 0.002)
v = 2m/s
Therefore, the average velocity of water flowing through a pipe of cross-sectional area of 0.002m² is 2m/s.
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Correct question is:
Water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. The density of water is 1 000 kg/m3. Determine its average velocity. Multiple choice question.
20 m/s
200 m/s
0.02 m/s
2 m/s
0.2 m/s
Part 1: A cello string vibrates in its fundamental mode with a frequency of 303 1/s. The vibrating segment is 42.7 cm long and has a mass of 1.04 g. Find the tension in the string. Answer in units of N.
Part 2: Find the frequency of the string when it vibrates in two segments. Answer in units of 1/s.
Answer:
Part 1:
The frequency of a vibrating string in its fundamental mode is given by:
f = (1/2L) √(T/μ)
where L is the length of the string, T is the tension in the string, and μ is the linear mass density (mass per unit length) of the string.
In this problem, f = 303 1/s, L = 42.7 cm = 0.427 m, and μ = m/L, where m is the mass of the vibrating segment. Substituting these values into the formula, we get:
303 1/s = (1/2 × 0.427 m) √(T/(1.04 g/0.427 m))
303 1/s = (1/2 × 0.427 m) √(T/0.00243 kg/m)
303 1/s = 0.0949 √T
T = (303 1/s / 0.0949)^2 × 0.00243 kg/m
T = 4.29 N
Therefore, the tension in the string is 4.29 N.
Part 2:
When a string vibrates in two segments, it is vibrating in its second harmonic or first overtone, which has two segments of equal length vibrating in opposite directions. The frequency of the second harmonic is given by:
f = (1/L) √(T/μ) × 2
where L, T, and μ have the same meaning as in Part 1. Substituting the values we found in Part 1, we get:
f = (1/0.427 m) √(4.29 N / 0.00243 kg/m) × 2
f = 712.7 1/s
Therefore, the frequency of the string when it vibrates in two segments is 712.7 1/s.
if you rub a balloon against your head, then electrons from the atoms that make up your hair get transferred to the balloon. the balloon becomes negatively charged and your hair becomes positively charged. what happens if you place balloon by hair?
When you rub a balloon against your head, electrons from the atoms in your hair are transferred to the balloon. This causes the balloon to become negatively charged, while your hair becomes positively charged. If you then place the balloon near your hair, the negative charge of the balloon will be attracted to the positive charge of your hair, causing the two to stick together. This phenomenon is known as electrostatic attraction.
The attraction of the negative charge of the balloon to the positive charge of your hair creates a strong force that causes the two objects to stick together. This force is known as the electrostatic force of attraction. It is the same force that makes two magnets stick together when their poles are placed near each other. The attraction between the balloon and your hair will remain until the charge on the balloon is dissipated by contact with another object.
To demonstrate this force of attraction, you can try rubbing the balloon against your head and then holding it near your hair. You will notice that the balloon will become attracted to your hair and will stick to it. You can also experiment with other materials that become charged when rubbed together, such as a cloth and a comb.
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Unknown Material A
Mass of metal
Mass of water
50 g д
100g
Temp. Of Unknown
90 deg * C
2 * o deg * c 23. 5 deg * c
Temp. Of Water
Temp at Equilibrium
C of water
4. 18H * g deg * C
C meTAL = C wateR M waree Delta T water m neTAL Delta T metAL
C of unknown metal
The specific heat of other liquid is 2.09 J/g°C.
Using the formula Q = mcΔT, where Q is amount of heat lost, m is the mass of liquid, c is specific heat of liquid, and ΔT is the change in temperature, we can set up two equations for each liquid.
For water, Q = (50 + 30)g × 4.18 J/g°C × (30 - 25)°C = 1045 J
For other liquid, Q = 100g × c × (30 - 25)°C = 500c J
Since both liquids lost same amount of heat, took same amount of time to cool, we can set these two equations equal to each other and solve for c: 1045 J = 500c J, c = 2.09 J/g°C
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--The correct Question is, A mass of 50 g of a certain metal at 150 degree C is immersed in 100 g of water at 11 degree C. The final temperature is 20 degree C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2Jg −1 K−1 .--
why does hydrogen, which is abundant in the sun's atmosphere, have relatively weak spectral lines, while calcium, which is not abundant, has very strong spectral lines? (assume the spectrum is observed on the surface of the earth.)
Hydrogen has relatively weak spectral lines, while calcium, which is not abundant, has very strong spectral lines because hydrogen is a comparatively lighter element, whereas calcium is much heavier than hydrogen.
In the sun's atmosphere, hydrogen is more prevalent and spread over a larger area, while calcium is less frequent, making it more concentrated, and hence they have more intense spectral lines.Spectral lines are unique to every element, and their patterns are utilized to identify elements present in any given compound. The intensity of spectral lines is determined by the concentration of the element. The more concentrated the element, the more intense its spectral lines will be.
Calcium has a more massive atomic structure than hydrogen, which explains why its spectral lines are more concentrated than hydrogen's. As a result, hydrogen's spectral lines are more dispersed, making them weaker in contrast. Thus, hydrogen, which is abundant in the sun's atmosphere, has relatively weak spectral lines, while calcium, which is not abundant, has very strong spectral lines.
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A load of 46 N attached to a spring hanging
vertically stretches the spring 5.5 cm. The
spring is now placed horizontally on a table
and stretched 13 cm.
What force is required to stretch it by this
amount?
Answer in units of N.
Answer:
108.8N
Explanation:
We can use Hooke's Law to solve for the force required to stretch the spring. Hooke's Law states that the force (F) applied to a spring is directly proportional to the amount it is stretched or compressed (x), as long as the spring does not exceed its limit of proportionality. Mathematically, this can be expressed as:
F = kx
where k is the spring constant, which depends on the properties of the spring and is measured in units of N/m (newtons per meter).
To solve for the force required to stretch the spring by 13 cm, we first need to find the spring constant. We can use the information given in the problem to do this. When the load of 46 N was attached to the spring and it was stretched 5.5 cm, we can write:
46 N = k (5.5 cm) (1)
To convert the units of length to meters, we divide both sides by 100:
46 N = k (0.055 m)
Solving for k, we find:
k = 46 N ÷ 0.055 m = 836.36 N/m
Now we can use Hooke's Law again to solve for the force required to stretch the spring by 13 cm. Since the spring is now horizontal, we need to convert the displacement from vertical to horizontal. We can assume that the spring is stretched in a straight line, so the displacement is the same for both orientations. Therefore, we can write:
F = kx
F = (836.36 N/m) (0.13 m)
F ≈ 108.8 N
Therefore, the force required to stretch the spring by 13 cm is approximately 108.8 N.
an electron starts from rest a distance of 42 cm from a fixed point charge of 0.128 how fast will electron be moving when it is very far away
The speed of the electron when it is very far away from the point charge of 0.128 depends on the amount of energy it has gained from the electric field. As the electron moves closer to the charge, the electric field gets stronger and the electron accelerates. By the time the electron reaches a distance of 42 cm from the point charge, it has gained enough energy from the electric field to reach a velocity of 8.97 x 106 m/s.
As the electron moves away from the point charge, the strength of the electric field decreases and the electron starts to decelerate. Eventually, the electric field will become so weak that the electron reaches a point where its speed stops decreasing and stabilizes. This point is referred to as the “asymptote”, and the speed of the electron at this point is known as the “asymptotic velocity”.
The asymptotic velocity of the electron can be calculated using the formula: V asymptotic = (2q/m)1/2, where q is the charge of the electron and m is its mass.
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does the propagation speed of the transmitted wave depend on the propagation speed of the incident wave
Yes, the propagation speed of the transmitted wave does depend on the propagation speed of the incident wave.
When a wave passes through a different medium, the speed of the wave changes due to a change in the medium’s properties such as density, elasticity, and permeability.
There are different types of waves including mechanical waves and electromagnetic waves. A mechanical wave, also called a traveling wave, requires a medium to travel.
Examples of mechanical waves include water waves, sound waves, and seismic waves. Electromagnetic waves, on the other hand, do not require a medium to travel.
Examples of electromagnetic waves include radio waves, X-rays, and light waves.
A mechanical wave's speed is determined by the medium's properties, whereas electromagnetic wave's speed is determined by a universal constant which is the speed of light in vacuum.
If the wave passes from one medium to another, the wave's velocity changes, and the wavelength changes as well. The frequency of the wave, however, does not change when it enters a different medium.
The speed of the wave is slower when it passes from a denser medium to a lighter medium. In this case, the transmitted wave has a lower speed compared to the incident wave because it travels at a slower rate.
When the wave passes from a lighter medium to a denser medium, the transmitted wave has a higher speed than the incident wave because it travels at a faster rate.
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radio waves are diffracted by large objects such as buildings, whereas light is not noticeably diffracted. why is this?
The reason why radio waves are diffracted by large objects, whereas light is not noticeably diffracted is the wavelength of light is much smaller than the wavelength of radio waves.
Thus, the correct answer is the wavelength of light is much smaller than the wavelength of radio waves (B).
The wаvelength of rаdio wаves being much lаrger thаn light, hаs а size compаrаble to those of buildings, hence diffrаct from them. Both rаdio аnd light wаves аre electromаgnetic wаves, just in different wаvelength rаnges. The wаvelength of visible light is typicаlly аround the 400-700 nm rаnge. Rаdio wаves on the other hаnd, often wаve wаvelengths of а few meters long.
For а wаve to diffrаct аround аn object, the size of the object must be on the sаme order of the wаvelength of the wаve. Hence, rаdio wаves diffrаct through buildings becаuse rаdio wаves hаve much lаrger wаvelength thаn light wаves.
Your question is incomplete, but most probably your options were
a. Radio waves are unpolarized, whereas light is plane polarized.
b. The wavelength of light is much smaller than the wavelength of radio waves.
c. Light is coherent and radio waves are usually not coherent.
d. Radio waves are coherent and light is usually not coherent.
Thus, the correct option is B.
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a refrigerator with a cop of 3.0 accepts heat from the refrigerated space at a rate of 10 kw. determine the power consumed.
The power consumed is 3.33 kW.
The coefficient of performance (COP) of a refrigerator is defined as the ratio of the heat extracted from the refrigerated space to the work done by the compressor. In other words, it's a measure of how much cooling effect the refrigerator can produce for a given amount of electrical energy input.
Here, the rate at which the refrigerator accepts heat from the refrigerated space is 10 kW.
COP of the refrigerator is 3.0.
The power consumed by the refrigerator can be calculated using the following formula:
Power consumed = Heat absorbed / Coefficient of Performance
Power consumed = 10 kW / 3.0 = 3.33 kW
Therefore, the power consumed by the refrigerator is 3.33 kilowatts.
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what is the relationship between velocity of moving body and the force acting on it?
Answer:
The relation between the momentum of a body and the force acting on it is that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force acting.
a little aluminum boat with a mass of 14.5 g has a volume of 450 cm3 . the boat is placed in a small pool of water and carefully filled with pennies. if each penny has a mass of 2.5 g, what is the minimum number of pennies needed to make the boat sink?
The boat is placed in a small pool of water and carefully filled with pennies. The minimum number of pennies needed to make the boat sink is 181 pennies.
To solve the given problem, you need to apply the Archimedes principle, which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object.
A little aluminum boat with a mass of 14.5 g has a volume of 450 cm³. The density of aluminum is 2.70 g/cm³. The mass of water displaced by the boat is the same as the mass of the boat. The mass of water displaced by the boat is given by the product of the volume of the boat and the density of water, which is 1 g/cm³. The mass of water displaced by the boat is then:
Mass of water displaced by the boat = Volume of the boat × Density of water
= 450 cm³ × 1 g/cm³
= 450 g
Since the buoyant force on the boat is equal to the weight of the water displaced by the boat, the buoyant force on the boat is 450 g.
For the boat to sink, the weight of the pennies added to the boat must be greater than 450 g. Each penny has a mass of 2.5 g.
Let's assume that the minimum number of pennies needed to make the boat sink is n. Then the total mass of pennies is 2.5n g. For the boat to sink, the total mass of pennies must be greater than 450 g.
Hence, we have the inequality:2.5n > 450
Dividing both sides of the inequality by 2.5, we get:
n > 180
The minimum number of pennies needed to make the boat sink is 181 pennies.
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