Answer:
its potential energy decreases and its electric potential decreases.
Explanation:
Let's consider a radial field for simplicity. We have:
- The electric potential of the field is given by:
[tex]V=\frac{kQ}{r}[/tex]
where
k is the Coulomb's constant
Q is the charge source of the field
r is the distance from Q
We see that the electric potential decreases as we move away from the source. If we consider a positive charge q moving in the direction of the electric field, this charge q will move away from the charge Q (because the field lines generated by the positive particle Q point away from the particle), so the electric potential will decrease.
- The potential energy of the moving charge is given by
[tex]U=qV[/tex]
where q is the magnitude of the charge. As we said previously, V is decreasing while the charge is moving in the direction of the field, so since U is directly proportional to V, U will decrease as well.
1) If a pendulum clock keeps perfect time at the base of a mountain, will it also keep perfect time when it is moved to the top of the mountain? Explain.
2)if a grandfather clock is running slow, how could we adjust the length of the pendulum to correct the time?
3)Explain (a) how it is possible for a large force to produce only a small, or even zero, torque, and (b) how it is possible for a small force to produce a large torque.
1) No
Explanation:
The period of a pendulum is given by
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
where
L is the length of the pendulum
g is the acceleration due to gravity
The value of g, acceleration due to gravity, is not exactly the same in all locations of the Earth. In fact, its value is given by
[tex]g=\frac{GM}{r^2}[/tex]
where G is the gravitational constant, M is the Earth's mass, and r the distance of the point from the Earth's center. This means that at the top of a mountain, r is slightly larger than at the Earth's surface, so the value of g is slightly smaller at the top of the mountain, and therefore the period of the pendulum will also be different (it will be slightly longer than at Earth's surface).
2) We need to decrease the length of the pendulum
Again, the period is given by
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex] (1)
If the clock is running slow, it means that its period T' is slightly longer than the expected period T: so, we need to shorten the period.
From eq.(1), we see that the period is proportional to the square root of the length of the pendulum, L: therefore, if the length increases the period increases, and if the length decreases, the period will decreases.
Here we want to shorten the period: therefore, according to the equation, we need to decrease the length of the pendulum.
3)
The torque of a force applied is given by
[tex]\tau = Fd sin \theta[/tex]
where
F is the magnitude of the force
d is the distance between the point of application of the force and the pivot point
[tex]\theta[/tex] is the angle between the direction of the force and d
So we have:
a) If we have a large force F, it is possible to produce a small torque by decreasing d, so by applying the force really close to the pivot, or by decreasing [tex]\theta[/tex], which means applying the force as more parallel as possible to d. The torque will be even zero if d=0 (force applied at the pivot point) or if [tex]\theta=0^{\circ}[/tex] (force parallel to d)
b) if we have a small force F, it is possible to produce a large torque by increasing d, so by applying the force really far to the pivot, or by increasing [tex]\theta[/tex], which means applying the force as more perpendicular as possible to d.
If a pendulum clock is accurate at the base of a mountain, it may not be accurate at the top because gravity is weaker at higher altitudes. To adjust a slow grandfather clock, shorten the pendulum. Torque depends on both force and its application point, so large forces can produce small torques and small forces can produce large torques.
1) If a pendulum clock keeps perfect time at the base of a mountain, it may not keep perfect time at the top due to a slight variation in the acceleration due to gravity. A pendulum clock's timing depends on gravity, and at higher altitudes, gravity is slightly weaker which would affect the clock's accuracy.
2) If a grandfather clock is running slow, the length of the pendulum needs to be shortened to increase the rate at which the pendulum swings, thereby correcting the time.
3) (a) A large force can produce a small or zero torque if the force is applied directly along the line of action of the pivot point, as torque is a measure of rotational force and depends on both the amount of force applied and its distance from the pivot.
(b) A small force can produce a large torque if applied at a large distance from the pivot point, as torque is the product of the force and the perpendicular distance from the pivot - thus, even a small force can create significant torque if applied far enough away from the pivot.
For a pendulum clock moved to a location with greater gravity, you would need to shorten the pendulum to maintain the same period and keep correct time.
In summer, the pendulum of a clock generally expands due to heat, which makes the clock run slower, while in winter, it would contract and make the clock run faster.
What controls the net direction of molecules?
Molecular diffusion is the process of mass transfer as a result of a concentration difference in a mixture.
This process is related to the thermal movement of the particles and their velocity, which is a function of the temperature, the viscosity of the fluid and the size of the particles. This is how diffusion controls the net direction of the molecular flow from a region of higher concentration to one of lower concentration, as long as there is a difference (or gradient) of concentration.
Which phenomenon can only be explained by assuming that light is quantized
Light can be considered as a wave or as particles, in this context Einstein proposed that light behaves like a stream of particles called photons with an energy, in order to correctly explain the photoelectric effect.
This fenomenom consists in the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.
So, if we consider light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.
This means the photoelectric effect can only be explained based on the corpuscular model of light, that is, light is quantized.
A drunk guy walking on a narrow lane takes 8 steps forward and 6 steps backward and so on.
Each step is 1m long and he requires 1 second to travel. The time that the drunk guy takes to fall in a pit 18m away from the starting point is
a) 18s
b) 126s
c) 78s
d) 62s
Answer: 126 secs
Explanation: he gains 2 steps every time and it takes 9 2 steps to reach 18 meters so what u do is take that 9 and multiply it to 14 (8+6) giving u ur answer of 126 secs.
Why does temperature decrease with increasing altitude in the troposphere
In the troposphere, the temperature generally decreases with altitude. The reason is that the troposphere's gases absorb very little of the incoming solar radiation. Instead, the ground absorbs this radiation and then heats the tropospheric air by conduction and convection.
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A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if the upper electrode is 17.8 V more positive than the lower electrode. The density of the oil is 885kg/m3. Part A What is the droplet's mass? Express your answer to two significant figures and include the appropriate units. m m = nothing nothing Request Answer Part B What is the droplet's charge? Express your answer to two significant figures and include the appropriate units. q q = nothing nothing Request Answer Part C Does the droplet have a surplus or a deficit of electrons? How many? Does the droplet have a surplus or a deficit of electrons? How many? deficit 9 electrons surplus 9 electrons surplus 16 electrons deficit 7 electrons
A) [tex]2.4\cdot 10^{-16}kg[/tex]
The radius of the oil droplet is half of its diameter:
[tex]r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m[/tex]
Assuming the droplet is spherical, its volume is given by
[tex]V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3[/tex]
The density of the droplet is
[tex]\rho=885 kg/m^3[/tex]
Therefore, the mass of the droplet is equal to the product between volume and density:
[tex]m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg[/tex]
B) [tex]1.5\cdot 10^{-18}C[/tex]
The potential difference across the electrodes is
[tex]V=17.8 V[/tex]
and the distance between the plates is
[tex]d=11 mm=0.011 m[/tex]
So the electric field between the electrodes is
[tex]E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m[/tex]
The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:
[tex]qE=mg[/tex]
So, from this equation, we can find the charge of the droplet:
[tex]q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C[/tex]
C) Surplus of 9 electrons
The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is
[tex]q=-1.5\cdot 10^{-18}C[/tex]
we can think this charge has made of N excess electrons, so the net charge is given by
[tex]q=Ne[/tex]
where
[tex]e=-1.6\cdot 10^{-19}C[/tex] is the charge of each electron
Re-arranging the equation for N, we find:
[tex]N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9[/tex]
so, a surplus of 9 electrons.
What causes charges to move in a circuit?
Charges move in a circuit due to the presence of an electrical field created by a voltage difference. The electrical field exerts forces on charged particles, causing them to accelerate and move through the circuit.
Explanation:Circuit charges refer to the movement of electric charge (usually electrons) through an electrical circuit. In a closed circuit, charges flow due to voltage (potential difference), creating an electric current. This flow of charges powers electrical devices and is described by Ohm's law, which relates current, voltage, and resistance in the circuit.
Charges move in a circuit due to the presence of an electrical field created by a voltage difference. The electrical field exerts forces on charged particles, causing them to accelerate and move through the circuit. As charges move, they lose potential energy and gain kinetic energy, traveling from an area of higher potential to an area of lower potential.
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Charges move in a circuit primarily due to the electrical field created by a voltage difference. The electrical field forces the charges, often free electrons, to accelerate, creating an electrical current and eventually reaching a constant 'drift velocity'. The presence and strength of a magnetic field can also affect the flow of charges.
Explanation:In a circuit, charges move due to an electrical field created by a voltage difference, such as a battery. This voltage difference exerts forces on the free electrons, causing them to accelerate and thus creating an electrical current. The exact rate at which these charges flow, meaning the amount of charge per unit of time, is influenced by factors such as the voltage applied, the state and type of the material (conductor or insulator), and the strength of the electrical field.
In the case of a conducting material, which is often the type of material used in circuits, the electrical field forces charge to flow and lose kinetic energy in the process until it reaches a constant velocity, known as the 'drift velocity'. This is essentially an equilibrium state where charges are constantly moving due to the force provided by the electrical field, but their speed or kinetic energy does not increase due to interactions with atoms and free electrons, similar to an object falling and reaching its terminal velocity.
It's also important to acknowledge here the role of a magnetic field, which can also impact the movement of charges in a conductor, causing a change in the direction of the electron flow, and hence, influencing the current within the circuit.
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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the 1.00 s? (d) With the angular acceleration unchanged, through what additional angle (rad) will the disk turn during the next 1.00 s?
With constant angular acceleration [tex]\alpha[/tex], the disk achieves an angular velocity [tex]\omega[/tex] at time [tex]t[/tex] according to
[tex]\omega=\alpha t[/tex]
and angular displacement [tex]\theta[/tex] according to
[tex]\theta=\dfrac12\alpha t^2[/tex]
a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of
[tex]21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}[/tex]
b. Under constant acceleration, the average angular velocity is equivalent to
[tex]\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2[/tex]
where [tex]\omega_f[/tex] and [tex]\omega_i[/tex] are the final and initial angular velocities, respectively. Then
[tex]\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}[/tex]
c. After 1.00 s, the disk has instantaneous angular velocity
[tex]\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}[/tex]
d. During the next 1.00 s, the disk will start moving with the angular velocity [tex]\omega_0[/tex] equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle [tex]\theta[/tex] according to
[tex]\theta=\omega_0t+\dfrac12\alpha t^2[/tex]
which would be equal to
[tex]\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}[/tex]
Please help on this one?
Answer: d
Explanation:
When the core of a star like the Sun uses up its supply of hydrogen for fusion, the core begins to ________.
Answer:
Shrink and heat
Explanation:
Hope my answer has helped you!
Answer:
Shrink and heat up
Explanation:
When the core of a star like the Sun uses up its supply of hydrogen for fusion, the core begins to contract or shrink up which leads to release of energy from the core.
The released energy starts to heat up core until it has gotten to the point where the core is hot enough to able start up the fusion of hydrogen into another element (helium).
The layers on the outside of the sun absorbs the released the energy and starts to enlarge or swell up and the sun develops a very high luminosity which means it start to shine brighter and brighter.
As the outside large swells up, due to the absorption of the released energy, it start to become cool thereby causing a low surface temperature.
It is at the stage that the sun becomes a red giant.
Scientists have recently discovered one the smallest stars on record. How long will this star last
compared to a larger star formed at the same time?
O A. The smaller star will last longer than the larger star because it has a smaller chance of getting hit by an
asteroid
O B. The smaller star will last longer than the larger star because it burns less fuel.
O C. The larger star will last longer than the smaller star because it uses fuel more efficiently.
O D. The larger star will last longer than the smaller star because it attracts more nebulae
Answer:
The correct answer is B.
Explanation:
Stars start their lives being dense clouds of gas and dust. After a star is formed, it begins to burn hydrogen and transform it into helium. Once the hydrogen is depleted, new stages of nuclear combustion begin, such as burning helium to obtain heavier elements. This stage is shorter in larger stars since the greater the mass of a star, the greater the temperature in its nucleus and the greater the rate of fusion of hydrogen into helium, with which the fuel is depleted more quickly.
Have a nice day!
Scientists have recently discovered one the smallest stars on record. How long will this star last compared to a larger star formed at the same time is B. The smaller star will last longer than the larger star because it burns less fuel.
What is the starsThe length of time a star can live is mainly decided by how much mass it has. Smaller stars weigh less and therefore use up their fuel more slowly than bigger stars. This means that smaller stars can live longer because they use fuel at a slower rate.
The amount of fuel, usually hydrogen, in a star determines how long it can keep producing energy through nuclear fusion. Smaller stars have cooler centers and lower pressure, so they can burn fuel more slowly and efficiently. As a result, they can live much longer than bigger stars.
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The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure force on the earth? Give your answers in (a) newtons (b) as a fraction of the sun's gravitational force on the earth
The radiation pressure force on the Earth due to the Sun's light is approximately 1.09 x 109 Newtons, and it is roughly 3.1 x 10-14 the Sun's gravitational force on Earth.
Explanation:The intensity of sunlight reaching the earth is given as 1360 W/m2. The pressure due to radiation can be calculated using the formula P = E/c, where P is pressure, E is energy and c is the speed of light. From this formula, we find that the radiation pressure force on Earth is roughly 4.53 x 10-6 N/m2.
(a) In Newtons:
The radiation pressure in Newtons will depend on the cross-sectional area exposed to sunlight. For the entire Earth, we would use the cross-sectional area of a circle with radius equals to Earth's radius (6.37 x 106 m). This gives us an approximate radiation pressure force of 1.09 x 109 Newtons.
(b) As a fraction of the Sun's gravitational force:
The gravitational force (Fg) between Earth and the Sun can be calculated using the formula Fg = GMm/r2, with G being the gravitational constant, M the mass of the Sun, m the mass of Earth and r the distance between Earth and the Sun. This gives an approximate Fg of 3.54 x 1022 Newtons. Hence, the radiation force is roughly 3.1 x 10-14 the Sun's gravitational force on Earth.
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The radiation pressure force on the Earth can be calculated using the intensity of sunlight and the surface area of the Earth. The force is approximately 1.09858256 × 10^7 N and is about 1.37 × 10^-38 times the Sun's gravitational force on the Earth.
Explanation:The radiation pressure force on the Earth can be calculated using the formula:
Force = Intensity × Area
Given that the intensity of sunlight reaching the Earth is 1360 W/m2 and assuming all the sunlight is absorbed, we can calculate the force:
Force = 1360 W/m2 × Area
The area of the Earth can be calculated using the formula for the surface area of a sphere:
Area = 4πr2
Where r is the radius of the Earth. Plugging in the value of the radius of the Earth, we can solve for the force:
Area = 4π(6371 km)Area = 4π(6.371 × 10^6 m)Area = 4π(6.371 × 10^6 m)Area = 4(3.14159)(6.371 × 10^6 m)Area = 4(20.0979 × 10^6 m)Area = 80.3916 × 10^6 mArea = 803.916 × 10^4 m2Force = 1360 W/m2 × 803.916 × 10^4 m2Force = 1098.58256 × 10^4 WForce = 1.09858256 × 10^7 WForce = 1.09858256 × 10^7 N(a) The radiation pressure force on the Earth is approximately 1.09858256 × 10^7 N.
(b) To find the force as a fraction of the Sun's gravitational force on the Earth, we need to calculate the gravitational force between the Sun and the Earth. Using Newton's law of gravitation:
Force = G × (mass of the Sun)(mass of the Earth) / (distance between the Sun and the Earth)^2
The mass of the Sun is about 2 × 10^30 kg. The mass of the Earth is about 6 × 10^24 kg. The average distance between the Sun and the Earth is about 1.5 × 10^11m. Plugging these values into the formula, we can calculate the gravitational force:
Force = (6.6743 × 10^-11 N m^2 / kg^2) × (2 × 10^30 kg)(6 × 10^24 kg) / (1.5 × 10^11m)^2Force = (6.6743 × 10^-11 N m^2 / kg^2) × (1.2 × 10^55 kg^2) / (2.25 × 10^22 m^2)Force = 8.00876033 × 10^44 N m^2 / kg^2The radiation pressure force as a fraction of the Sun's gravitational force on the Earth can be calculated as:
Fraction = (Radiation Pressure Force) / (Gravitational Force)
Fraction = (1.09858256 × 10^7 N) / (8.00876033 × 10^44 N)Fraction = 1.37 × 10^-38(b) The radiation pressure force on the Earth is approximately 1.37 × 10^-38 times the Sun's gravitational force on the Earth.
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A coin completes 18 spins in 12 seconds. The centripetal acceleration of the edge of the coin is 2.2 m/s2. The radius of the coin is ____-blank
Answer:
0.025 m
Explanation:
The coin completes 18 revolutions in 12 seconds. The angular velocity of the coin is:
[tex]\omega = \frac{18 rev}{12 s} \cdot (2\pi \frac{rad}{rev})=9.42 rad/s[/tex]
The centripetal acceleration of the edge of the coin is given by
[tex]a=\omega^2 r[/tex]
where we have
a = 2.2 m/s^2 is the acceleration
r is the radius of the coin
Solving for r, we find
[tex]r=\frac{a}{\omega^2}=\frac{2.2 m/s^2}{(9.42 rad/s^2)^2}=0.025 m[/tex]
Answer: 0.025 m or 2.5m
Explanation:
The number of spins given in 12 seconds = 18
So the number of spins in 1 second will be
= 18 / 12 = 1.5rps
The above result is the frequency because it is the number of spins in one second
Let f be frequency
f = 1.5 rps
Using centripetal acceleration given
Ca = 2.2 m/s^2
m is the radius of the given coin
Using the formula
Centripetal acceleration, = m x w^2
Knowing that w = 2πf
2.2 = m x ( 2 x 3.14 x 1.5) ^2
2.2 = m x 88.7364
m = 0.025 m
What causes convection currents to occur?
Answer:
Convection currents are caused by an uneven temperature within something.
For example, within the earth, Convection currents occur when a reservoir of fluid is heated at the bottom, and allowed to cool at the top.. Heat causes the fluid to expand, decreasing its density. If there is cooler material on top, it will be more compact and therefore, will sink to the bottom. The heated material will rise to the top.
What is the brightest star in the night sky
the brightest star in the night sky is Sirius A
The brightest star in the night sky is Sirius A.
Sirius A is the brightest star that we see in the night sky. It is a binary system with a white dwarf orbiting a main-sequence star. Despite common misconceptions, Sirius is the actual brightest star, not Polaris.
Which statement is true?Which statement is true? Gas atoms subjected to electricity emit bright lines. Gas atoms absorb white light and emit a continuous spectrum of light. Gas atoms under high pressure emit dark lines against a continuous spectrum. Gas atoms heated to a high temperature emit white light. Gas atoms subjected to electricity do not emit a spectrum.
Answer:
Gas atoms subjected to electricity emit bright lines.
Explanation:
Gas atoms subject to electricity emit bright lines. This can be seen in fluorescent lamps, for example.
Fluorescent lamps work by ionizing atoms of argon gas and mercury vapor. After receiving an electric current, ionization occurs, at that moment the atoms are accelerated by the potential difference established between the lamp terminals and emit bright light and electromagnetic waves when they return to the natural state.
Answer:
A. Gas atoms subjected to electricity emit bright lines.
Explanation:
What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?
Answer: [tex]6.268(10)^{-16}J[/tex]
The kinetic energy of an electron [tex]K_{e}[/tex] is given by the following equation:
[tex]K_{e}=\frac{(p_{e})^{2} }{2m_{e}}[/tex] (1)
Where:
[tex]K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}[/tex]
[tex]p_{e}[/tex] is the momentum of the electron
[tex]m_{e}=9.11(10)^{-31}kg[/tex] is the mass of the electron
From (1) we can find [tex]p_{e}[/tex]:
[tex]p_{e}=\sqrt{2K_{e}m_{e}}[/tex] (2)
[tex]p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}[/tex]
[tex]p_{e}=2.091(10)^{-24}\frac{kgm}{s}[/tex] (3)
Now, in order to find the wavelength of the electron [tex]\lambda_{e}[/tex] with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:
[tex]\lambda_{e}=\frac{h}{p_{e}}[/tex] (4)
Where:
[tex]h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s}[/tex] is the Planck constant
So, we will use the value of [tex]p_{e}[/tex] found in (3) for equation (4):
[tex]\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}[/tex]
[tex]\lambda_{e}=3.168(10)^{-10}m[/tex] (5)
We are told the wavelength of the photon [tex]\lambda_{p}[/tex] is the same as the wavelength of the electron:
[tex]\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m[/tex] (6)
Therefore we will use this wavelength to find the energy of the photon [tex]E_{p}[/tex] using the following equation:
[tex]E_{p}=\frac{hc}{lambda_{p}}[/tex] (7)
Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum
[tex]E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}[/tex]
Finally:
[tex]E_{p}=6.268(10)^{-16}J[/tex]
The energy of a photon is about 6.3 × 10⁻¹⁶ Joule
[tex]\texttt{ }[/tex]
Further explanationThe term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :
[tex]\large {\boxed {E = h \times f}}[/tex]
E = Energi of A Photon ( Joule )
h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )
f = Frequency of Eletromagnetic Wave ( Hz )
[tex]\texttt{ }[/tex]
The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.
[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]
[tex]\large {\boxed {E = qV + \Phi}}[/tex]
E = Energi of A Photon ( Joule )
m = Mass of an Electron ( kg )
v = Electron Release Speed ( m/s )
Ф = Work Function of Metal ( Joule )
q = Charge of an Electron ( Coulomb )
V = Stopping Potential ( Volt )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
kinetic energy of an electron = Ek = 15 eV = 2.4 × 10⁻¹⁸ Joule
Asked:
energy of photon = E = ?
Solution:
Firstly , we will use the formula of kinetic energy:
[tex]Ek = \frac{1}{2}mv^2[/tex]
[tex]v^2 = \frac{2Ek}{m}[/tex]
[tex]v = \sqrt{ \frac{2Ek}{m}}[/tex]
[tex]\texttt{ }[/tex]
Next , we will use the formula of The Broglie's Wavelength:
[tex]\lambda = \frac{h}{mv}[/tex]
[tex]\lambda = \frac{h}{m\sqrt{2Ek/m}}[/tex]
[tex]\lambda = \frac{h}{\sqrt{2mEk}}[/tex]
[tex]\texttt{ }[/tex]
[tex]E = h f[/tex]
[tex]E = h \frac{c}{\lambda}[/tex]
[tex]E = h \frac{c}{\frac{h}{\sqrt{2mEk}}}[/tex]
[tex]E = c\sqrt{2mEk}[/tex]
[tex]E = 3 \times 10^8 (\sqrt{2(9.11 \times 10^{-31} \times 2.4 \times 10^{-18}}[/tex]
[tex]E \approx 6.3 \times 10^{-16} \texttt{ Joule}[/tex]
[tex]\texttt{ }[/tex]
Learn morePhotoelectric Effect : https://brainly.com/question/1408276Statements about the Photoelectric Effect : https://brainly.com/question/9260704Rutherford model and Photoelecric Effect : https://brainly.com/question/1458544[tex]\texttt{ }[/tex]
Answer detailsGrade: College
Subject: Physics
Chapter: Quantum Physics
[tex]\texttt{ }[/tex]
Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light
A constant voltage is applied across a circuit. If the resistance in the circuit is doubled, what is the effect on the power dissipated by the circuit?
a.The power dissipated is quadrupled
b.The power dissipated is reduced by a factor of 2.
c.The power dissipated is reduced by a factor of 4
d. The power dissipated is doubled.
e. The power dissipated remains constant.
Answer:
b.The power dissipated is reduced by a factor of 2.
Explanation:
The power dissipated in the circuit is given by
[tex]P=\frac{V^2}{R}[/tex]
where
V is the voltage
R is the resistance
In this problem:
- The voltage V is kept constant
- The resistance is doubled, so R' = 2R
Therefore, the new power dissipated is
[tex]P'=\frac{V^2}{R'}=\frac{V^2}{2R}=\frac{1}{2}\frac{V^2}{R}=\frac{1}{2}P[/tex]
so, the power dissipated is reduced by a factor of 2.
Final answer:
When resistance is doubled and voltage is constant, the power dissipated by the circuit is halved. The power is proportional to the inverse of resistance, so doubling resistance reduces power by a factor of 2.
Explanation:
When the resistance in a circuit is doubled while keeping the voltage constant, the current in the circuit according to Ohm's Law (V = IR) will be halved, because I = V/R. The power dissipated by the circuit can be calculated using the formula P = V2/R. If the resistance is doubled, the new power dissipated becomes Pnew = V2/(2R), which is half the original power. Since the original power is Porig = V2/R, by doubling the resistance, the power is effectively reduced by a factor of 2, not 4.
Thus, the correct answer is: b. The power dissipated is reduced by a factor of 2.
When is the angular momentum of a system constant?
a. When the total kinetic energy is constant.
b. When no net external force acts on the system.
c. When the linear momentum and the energy are constant.
d. When no net torque acts on the system.
e. When the moment of inertia is constant.
The angular momentum of a system constant when no net torque acts on the system. The correct option is d.
When there is no external torque operating on a system in a net way, there will be no change in the system's angular momentum. The principle of the conservation of angular momentum is one of the most fundamental principles in all of physics. The rotational motion of an object or a system of objects can be described using a vector quantity known as the angular momentum of the system.
When there is no net external torque acting on a system, the total angular momentum of the system will not change over time; this property will be known as the system's inertia. This indicates that if an item or system is originally at rest or has a particular angular momentum, it will keep that angular momentum unless an external torque is applied to it and causes it to rotate in the opposite direction.
The other options (a,b,c, and e) do not guarantee continuous angular momentum. Even if some of those conditions could result in particular outcomes for the system, the conservation of angular momentum requires that the system have no net external torque.
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1.Work is only being done when what happens to an object?
A) the object does not move
B) the object disappears without explanation
C) it moves in the same direction as the force
D) The object moves in the opposite direction of the force
2.Ben and Jerry, arch rivals, decide to have a weight lifting contest during PE. Ready, set, go! Ben and Jerry both lift a 250 kg barbell 10 times over their heads. They are approximately the same height and lift the barbell the same distance in the air. It takes Ben 5 seconds to complete 10 lifts; it takes Jerry 25 seconds to complete his 10 lifts.
Which statement is MOST accurate regarding the weightlifting contest?
A) Ben did more work than Jerry.
B) Ben has more power than Jerry.
C) Ben and Jerry have the same power.
D) Ben does more work and is more powerful than Jerry.
3.Which energy source can be found on the electromagnetic spectrum?
A) sound energy
B) chemical energy
C) UV light energy
D) mechanical energy
4. Sam blew up a balloon and rubbed it on his head. Then he stuck the balloon on the wall. This was all the result of electrostatics or the transfer of charge. The appearance of negative charge on a balloon is the result of its gaining electrons. And these electrons must come from somewhere; in this case, from Sam's hair. Electrons are transferred in any charging process. In the case of charging by friction, they are transferred between the two objects being rubbed together. Prior to the charging, both objects are electrically neutral and the situation obeys the law of conservation of charge. How does the law of conservation of charge explain Sam and his balloon?
A) Friction creates charge that collected on the balloon.
B) The balloon gained electrons; Sam's hair gained protons.
C) If the balloon acquires charge, Sam's hair loses charge.
D) The net charge before is zero; the net charge after is negative.
1.
C) it moves in the same direction as the force
D) The object moves in the opposite direction of the force
Work is defined as follows:
[tex]W=Fdcos \theta[/tex]
where
F is the magnitude of the force applied on an object
d is the displacement of the object
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
As we see from the formula, when d=0 (the object does not move), then W=0 (no work done). While we have work done if the object is moving. In particular, we have two situations:
- The object moves in the same direction as the force: [tex]\theta=0^{\circ}[/tex], so [tex]cos \theta= 1[/tex] and the work done is positive
- The object moves in the opposite direction of the force: [tex]\theta=180^{\circ}[/tex], so [tex]cos \theta= -1[/tex] and the work done is negative
2. B) Ben has more power than Jerry.
Power is defined as:
[tex]P=\frac{W}{t}[/tex] (1)
where
W is the work done
t is the time taken
Work is the product of force and displacement. Since Ben and Jerry lift the same mass (250 kg), they apply the same force, and since they lift the barbell the same distance in air the same number of times (10), the displacement is also the same: so, they did the same work.
However, Ben did it in less time (5 s) then Jerry (25 s): looking at eq.(1), we see that less time means more power, so Ben has more power than Jerry.
3. C) UV light energy
The electromagnetic spectrum is the classification of all electromagnetic waves according to their different wavelength.
From shortest to longest wavelength, we have:
Gamma rays
X-rays
UV (ultraviolet)
Visible light
Infrared radiation
Microwaves
Radio waves
So, we see that UV light energy (ultraviolet radiation) is a type of electromagnetic waves.
4. C) If the balloon acquires charge, Sam's hair loses charge
The law of conservation of charge applied to this case states that the total charge on Sam hair and the balloon before must be equal to the total charge after: no charge can be created or destroyed in the process, but only moved from one object to the other.
In this example, Sam rubs the balloon on his head: electrons are transferred from Sam's hair to the balloon's surface. Therefore, we can say that the balloon has acquired negative charge, while Sam's hair has lost negative charge.
A runner moves 2.88 m/s north. She accelerates at 0.350 m/s^2 at a -52.0 angle. At the point in the motion where she is running directly east, what is Δx?
Answer:
Δx = 11.7 and Δy = 15
The question requires the use of kinematics and vector decomposition to calculate the horizontal displacement of a runner when she changes direction from the north to the east due to acceleration at a given angle.
The student is asking about the projection motion of a runner moving north, who accelerates at an angle. The question focuses on calculating the horizontal displacement (denoted as Δx) when the runner is running directly east. To solve this, one would have to break down the acceleration vector into its northward and eastward components and then use kinematic equations to determine the eastward displacement from the point of initial velocity to the point where the northward velocity component reaches zero and the runner is moving directly east.
Which of these stars has the coolest surface temperature?A. A starB. F starC. G starD. K star
Answer:
D. K star
Explanation:
Stars are classified into different groups according to their peak wavelength and their surface temperature.
In particular, we have the following group of stars, which correspond to the following surface temperatures:
Group O - Temperature > 25,000 K
Group B - Temperature 11,000 - 25,000 K
Group A - Temperature 7,500 - 11,000 K
Group F - Temperature 6,000 - 7,500 K
Group G - Temperature 5,000 - 6,000 K
Group K - Temperature 3,500 - 5,000 K
Group M - Temperature < 3,500 K
So among the options given, the star with the coolest surface temperature is star in group K.
The coolest surface temperature among the options is a K star. Ait depends on the surface temperature. The correct option is option (D).
Stars are categorized into different spectral classes based on their surface temperature.
The spectral classes are labeled with letters, starting from the hottest to the coolest: O, B, A, F, G, K, and M. So, a K star has a cooler surface temperature compared to an F star, a G star, and an A star.
Therefore, the correct option is option (D) K star has the coolest temperature.
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Please help on this one?
Answer:
D. An image that is smaller than the object and is behind the mirror
A 282 kg bumper car moving 3.50 m/s collides with a 155 kg bumper car moving -1.38 m/s. Afterwards the 282 kg car moves at 1.10 m/s. Find the velocity of the 155 kg car afterwards?
Answer:
4.03 m/s
Explanation:
Initial momentum = final momentum
(282 kg) (3.50 m/s) + (155 kg) (-1.38 m/s) = (282 kg) (1.10 m/s) + (115 kg) v
v = 4.03 m/s
We use the conservation of momentum to find the velocity of the 155 kg bumper car after the collision. Substituting the given values, we solve to get the final velocity as 2.99 m/s. The final velocity of the 155 kg car is 2.99 m/s.
The question is about a collision between two bumper cars and finding the velocity after the collision. We can use the principle of conservation of momentum which states that the total momentum before the collision is equal to the total momentum after the collision.
Let's define the variables:
Mass of first car, m₁ = 282 kgInitial velocity of first car, u₁ = 3.50 m/sMass of second car, m₂ = 155 kgInitial velocity of second car, u₂ = -1.38 m/sFinal velocity of first car, v₁ = 1.10 m/sFinal velocity of second car, v₂ = ?Using the conservation of momentum:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Substituting the given values into the equation:
(282 kg)(3.50 m/s) + (155 kg)(-1.38 m/s) = (282 kg)(1.10 m/s) + (155 kg)v₂
Simplifying:
987 kg·m/s - 213.9 kg·m/s = 310.2 kg·m/s + 155 kg·v₂
773.1 kg·m/s = 310.2 kg·m/s + 155 kg·v₂
Subtracting 310.2 kg·m/s from both sides:
462.9 kg·m/s = 155 kg·v₂
Solving for v₂:
v₂ = 462.9 kg·m/s / 155 kg = 2.99 m/s
Therefore, the final velocity of the 155 kg car after the collision is 2.99 m/s.
Two hundred grams of liquid A is at a temperature of 100◦C. One hundred grams of liquid B is at a temperature of 0◦C. When the two liquids are mixed, the final temperature is 50◦C. Which material has a higher specific heat? 1. The specific heat of A is greater than that of B. 2. The specific heat of A is great than B if its viscosity is greater. 3. The specific heats of A and B are equal. 4. The specific heat cannot be defined when fluids are mixed in this way. 5. The specific heat of B is greater than that of A. "correct"
Answer:
The specific heat of B is greater than that of A.
Explanation:
The amount of heat lost by A is:
q = m C ΔT
q = (200 g) Ca (100°C - 50°C)
q = 10,000 Ca
The amount of heat gained by B is:
q = (100 g) Cb (50°C - 0°C)
q = 5,000 Cb
Since the heat lost by A = heat gained by B:
10,000 Ca = 5,000 Cb
2 Ca = Cb
So the heat capacity of B is double the heat capacity of A. Answer 5.
Taking into account the definition of calorimetry, the correct answer is option 5: The specific heat of B is greater than that of A.
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case, you know:
For liquid A: Mass= 200 g Initial temperature= 100 °C Final temperature= 50 ºC Specific heat = [tex]c_{A}[/tex] For liquid B: Mass of water = 100 g Initial temperature of water= 0 ºC Final temperature of water= 50 ºC Specific heat of water = [tex]c_{B}[/tex]Replacing in the expression to calculate heat exchanges:
For liquid A: QliquidA= [tex]c_{A}[/tex] × 200 g× (50 C - 100 C)
For liquid B: QliquidB= [tex]c_{B}[/tex] × 100 g× (50 C - 0 C)
If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.
Then, the heat that the liquid A gives up will be equal to the heat that the liquid B receives. Therefore:
- QliquidA = + QliquidB
- [tex]c_{A}[/tex] × 200 g× (50 C - 100 C)= [tex]c_{B}[/tex] × 100 g× (50 C - 0 C)
Solving:
- [tex]c_{A}[/tex] × 200 g× ( - 50 C)= [tex]c_{B}[/tex] × 100 g× (50 C)
[tex]c_{A}[/tex] × 10,000 gC= [tex]c_{B}[/tex] × 5,000 gC
([tex]c_{A}[/tex] × 10,000 gC) ÷ 5,000 gC= [tex]c_{B}[/tex]
[tex]c_{A}[/tex] × 2= [tex]c_{B}[/tex]
Finally, the correct answer is option 5: The specific heat of B is greater than that of A.
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Which term best describes how light waves from a star are affected as the star moves toward earth?
A.Diffraction
B.Redshift
C.Blueshift
D.Cosmological Redshift
Answer:
Answer:C.
Explanation:
Blueshift"If a star is moving towards the earth, its light is shifted to higher frequencies on the color spectrum (towards the green/blue/violet/ultraviolet/x-ray/gamma-ray end of the spectrum). A higher frequency shift is called a "blue shift"
Blueshift is the best term to describe how light waves from a star are affected as the star moves toward earth.
What is Blueshift?Blueshift is the displacement of the spectrum to shorter wavelengths in the light coming from distant celestial objects moving toward the observer.Blueshift is an example of doppler effect.How light waves from a star are affected as the star moves toward earth?Due to blueshift light wave coming from the star shift to longer wavelength to shorter wavelength.As a result we see stars blinking and shining.
So, the correct answer is option C blueshift.
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Suppose that an object travels from one point in space to another. Make a comparison between the magnitude of the displacement and the distance traveled by this object.
Answer:
- Distance is a scalar quantity, defined as the total amount of space covered by an object while moving between the final position and the initial position. Therefore, it depends on the path the object has taken: the distance will be minimum if the object has travelled in a straight line, while it will be larger if the object has taken a non-straight path.
- Displacement is a vector quantity, whose magnitude is equal to the distance (measured in a straight line) between the final position and the initial position of the object. Therefore, the displacement does NOT depend on the path taken, but only on the initial and final point of the motion.
If the object has travelled in a straight path, then the displacement is equal to the distance. In all other cases, the distance is always larger than the displacement.
A particular case is when an object travel in a circular motion. Assuming the object completes one full circle, we have:
- The distance is the circumference of the circle
- The displacement is zero, because the final point corresponds to the initial point
Hydrogen-2 and Hydrogen-3 fuse to form Helium-4 and a neutron. How much energy is released in this nuclear reaction?
Final answer:
In a nuclear fusion reaction, when hydrogen-2 and hydrogen-3 combine to form helium-4 and a neutron, a certain amount of energy is released. The exact amount of energy released can be calculated using the equation E = mc^2, where E is the energy, m is the change in mass, and c is the speed of light.
Explanation:
In a nuclear fusion reaction, when hydrogen-2 (deuterium) and hydrogen-3 (tritium) combine to form helium-4 and a neutron, a certain amount of energy is released. The exact amount of energy released can be calculated using the equation E = mc2, where E is the energy, m is the change in mass, and c is the speed of light.
Based on the given information, we can calculate the change in mass by subtracting the mass of the reactants from the mass of the products. The mass of deuterium (hydrogen-2) is 2 grams, the mass of tritium (hydrogen-3) is 3 grams, the mass of helium-4 is 4 grams, and the mass of a neutron is negligible. Therefore, the change in mass is 2 grams + 3 grams - 4 grams = 1 gram.
Using the equation E = mc2, where c is the speed of light (approximately 3 x 108 m/s), we can calculate the energy released:
E = (1 gram) x (3 x 108 m/s)2 = 9 x 1016 joules
What can one say about the image produced by a thin lens that produces a positive magnification?
Answer:
The image is virtual and upright
Explanation:
The magnification of a lens can be written as follows:
[tex]M=\frac{y'}{y}=-\frac{q}{p}[/tex]
where
y' is the size of the image
y is the size of the object
q is the location of the image with respect to the lens
p is the location of the object with respect to the lens
In this situation, the magnification is positive. This means that:
- y' (the image) has same sign as y (the object) --> the image is upright (same orientation as the object)
- q has opposite sign to p --> this means that the image is located on the same side as the object, so it is a virtual image
In 1909 Robert Millikan was the first to find the charge of an electron in his now-famous oil drop experiment. In the experiment tiny oil drops are sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops are observed with a magnifying eyepiece, and the electric field is adjusted so that the upward force q E on some negatively charged oil drops is just sufficient to balance the downward force m g of gravity. Millikan accurately measured the charges on many oil drops and found the values to be whole-number multiples of 1.6 × 10−19 C — the charge of the electron. For this he won the Nobel Prize. If a drop of mass 1.51837 × 10−12 kg remains stationary in an electric field of 1 × 106 N/C, what is the charge on this drop? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of C.
Answer:
[tex]1.49\cdot 10^{-17}C[/tex]
Explanation:
The oil drop remains stationary when the electric force on it and the gravitational force are balanced, so we have:
[tex]F_E = F_G\\qE = mg[/tex]
where
q is the charge of the oil drop
E is the electric field strength
m is the mass of the drop
g is the acceleration due to gravity
here we have
[tex]E=1\cdot 10^6 N/C[/tex]
[tex]m=1.51837\cdot 10^{-12} kg[/tex]
[tex]g=9.8 m/s^2[/tex]
So the charge of the drop is
[tex]q=\frac{mg}{E}=\frac{(1.51837\cdot 10^{-12} kg)(9.8 m/s^2)}{1\cdot 10^6 N/C}=1.49\cdot 10^{-17}C[/tex]