Answer:jyjyjjhjh
Explanation:
hjjhjhjhjjh
What can engineers do to prevent structures from collapsing
Explanation:
that don't look bright
The electrical force between two objects 36N. If the distance between the two objects is doubled the force between them will be
Answer:
The electrical force is 9N
Explanation:
For point loads, charged bodies very small compared to the distance r that separates them, Coulomb discovered that the electric force is proportional to [tex]\frac{1}{r^{2}}[/tex]. So, if the distance is doubled, the force will decrease a [tex]\frac{1}{4}[/tex] of its initial value.
[tex]F=\frac{36N}{4}=9N[/tex]
When the distance between two charged objects is doubled, the electrostatic force between them becomes one-fourth of the original force. Thus, if the original force is 36 N, the new force would be 9 N.
Explanation:The force between two charges is governed by Coulomb's Law, which states that the electrostatic force (F) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them. The law is mathematically expressed as F = k * (|q1*q2|) / r², where k is Coulomb's constant. If the distance between two charges is doubled, since the force is inversely proportional to the square of the distance, the new force will be one-fourth of the original force.
Thus, if the initial force is 36 N and the distance is doubled, the new force is calculated as:
Initial force: 36 N
New distance: 2r
New force (F') = F / (22) = 36 N / 4 = 9 N
Therefore, the new electrostatic force between the two objects when the distance is doubled would be 9 N.
The most massive planet in the solar system is
Answer:
Jupiter
Largest Planet: Jupiter. The largest planet in our solar system by far is Jupiter, which beats out all the other planets in both mass and volume. Jupiter's mass is more than 300 times that of Earth, and its diameter, at 140,000 km, is about 11 times Earth's diameter.
Explanation:
A metal sphere is neutral because it has an equal number of protons and electrons. Draw how the charges in the sphere are redistributed when a negatively charged rod is brought near it.
Answer:
The protons will shift towards the negatively charged rod and the electrons will shift away
Explanation:
When negatively charged rod is brought near it , in sphere protons ( positively charge ) get attracted on the surface and electron get away (negatively charge) due to induction
What is meant by charging by induction ?
Charging by induction is a process by which a neutral body can be charged electrostatically in the presence of a negatively or positively charged body.
Whenever a charged body is placed over a neutral conducting material that conducting material will induce an opposite charge to the charged body because of induction . Example : if charged body have positive charge than conducting material will induce a negative charge on it
hence , when negatively charged rod is brought near it , in sphere protons ( positively charge ) get attracted on the surface and electron get away (negatively charge) due to induction
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A container holds a pure sample of a radioactive substance with a half-life of 2 hours.Part AWhich of the following statements are true?a) After 1 hour, less than 50\% of the original atoms in the container will have decayed.b) After 1 hour, more than 50\% of the original atoms in the container will have decayed.c) After 2 hours, 50\% of the original atoms in the container will have decayed.d) After 4 hours, 25\% of the original atoms will have decayed.e) After 4 hours, the total number of atoms in the container will be reduced by 75\%.
Answer: Statements A, C, and E are true
Explanation: Half life is the time required for one half of a radioactive substance to decay into its daughter material. Since the half life of the radioactive material is 2 hours, it means that in 2 hours the material will have decayed by 50% and after 4 hours, the material will have decayed by 75%
Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but with opposite angular velocities ~ω and −~ω. what is the total rotational kinetic energy kr of this system of two disks? 1. 1 2 i ω 2 2. zero 3. none of these
The total rotational kinetic energy of this system is : ( C ) none of these
Ker = Iw²
Determine the Total rotational kinetic energy
Given that the two disks have the same rotational inertia and the same angular speed but opposite angular velocities
w and -w
Total rotational kinetic energy ( Kr )
K.Er = K₁ + K₂
= [tex]\frac{1}{2} * Iw^2 + \frac{1}{2} * I (-w)^2[/tex]
= [tex]Iw^2[/tex]
Hence the total rotational kinetic energy of the system is : Iw²
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The total rotational kinetic energy of the system of two disks spinning with the same angular speed but in opposite directions is Kr = Iω², since the kinetic energy for both disks will be the same positive value when squared.
Given that both disks have the same rotational inertia (I) and angular speed (ω), we can calculate the total kinetic energy using the formula for rotational kinetic energy K = ½Iω² for each disk individually and then combine the results.
For the first disk with angular velocity ω:
K1 = ½Iω²
For the second disk with angular velocity -ω:
K2 = ½I(-ω)²
Since squaring a negative number yields a positive result, the kinetic energy for both disks will be positive and the same value.
Therefore, the total rotational kinetic energy is:
Kr = K1 + K2 = ½Iω² + ½Iω² = Iω²
The table represent the thickness, top density, and bottom density of the different layers of the Earth. In most of the layers, the deeper the layer, the thicker and denser the layer becomes. Thickness (km) Density (g/cm3) Types of Rock Found Top Bottom Crust 30 2.2 2.9 Silicic rocks Upper mantle 720 3.4 4.4 Peridotite Lower mantle 2,171 4.4 5.6 Magnesium and silicon oxides Outer core 2,259 9.9 12.2 Iron+oxygen Inner core 1,221 12.8 13.1 Iron+oxygen At which location in Earth’s interior exhibits a change in the trend? inner core outer core lower mantle upper mantle
Answer:inner core?
Explanation:
Answer: your answer would be inner core hope it helps
Explanation:
tell me i am wrong?
. In any energy transformation, energy is _____. destroyed created conserved
Answer: i believe it is conserved or stays the same.
Explanation: energy cant be destroyed no matter what and no energy is being created
I hope this helps a thank and a brainlist would be greatly appreciated
Answer: Nor
Explanation: In any energy transformation, energy is nor created destroyed or conserved
Positively-charged particles consisting of two protons and two neutrons emitted by radioactive materials are
A) beta particles.
Eliminate
B) gamma particles.
C) alpha particles.
D) positron particles
Answer:
The answer to your question is Alpha particles.
Explanation: An electron released by a radioactive nucleus that causes a neutron to change into a proton is called a beta particle.
Final answer:
The question refers to alpha particles, which consist of two protons and two neutrons and are symbolized by He or the Greek letter α. Alpha particles carry a positive charge and result in the atomic number decreasing by two and the mass number by four following emission.
Explanation:
The positively-charged particles emitted by radioactive materials that consist of two protons and two neutrons are known as alpha particles. These particles are the equivalent of a helium nucleus and carry a positive charge due to the protons. The atomic symbol for an alpha particle is either He or the Greek letter α, and this type of radioactive emission results in the reduction of the atomic number by two and the mass number by four. For example, when uranium-238 undergoes alpha decay, it emits an alpha particle and transforms into thorium-234.
In contrast, beta particles are electrons with a 1- charge and are represented as e or β. The emission of a beta particle results in the conversion of a neutron to a proton within the nucleus, increasing the atomic number by one without changing the mass number. Gamma rays, on the other hand, are high-energy electromagnetic radiation with no mass and hence are not particles. Lastly, positron particles are positively charged electrons (anti-electrons) and have negligible mass.
An object is placed 50 cm from a concave mirror of radius 70 cm. (a)What is the distance of the image from the mirror? (b) What is the lateral magnification of the image? Show your work and provide units of measure for each answer.
See the attached picture:
The K-shell ionization energy of iron is 8500 eV, and it's L-shell ionization energy is 2125 eV. What is the wavelength of Kalpha X-rays emmited by iron?
Answer:
[tex]1.95\cdot 10^{-10}m[/tex]
Explanation:
First of all, we need to calculate the energy of the x-ray photon emitted during the transition from K-shell to L-shell, and this energy is equal to the difference in energy between the two levels:
[tex]E=E_K-E_L=8500 eV-2125 eV=6375 eV[/tex]
Converting into Joules,
[tex]E=6375 eV \cdot (1.6\cdot 10^{-19} J/eV)=1.02\cdot 10^{-15} J[/tex]
Now we know that the energy of the photon is related to its wavelength by:
[tex]E=\frac{hc}{\lambda}[/tex]
where
h is the Planck constant
c is the speed of light
[tex]\lambda[/tex] is the wavelength
Re-arranging the equation for [tex]\lambda[/tex], we find
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.02\cdot 10^{-15} J}=1.95\cdot 10^{-10}m[/tex]
A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be:A) 2TB) T square root(2)C) T/4D) T/2E) 4
Applying Kepler's third law and considering the decrease in Planet X's mass, the orbital period of the satellite in the same radius orbit would double, from T to 2T.
Explanation:The question involves the satellite's orbital period in the context of Kepler's third law, which states that the square of a planet's orbital period is directly proportional to the cube of the semi-major axis of its orbit. In the special case of a circular orbit, the semi-major axis is equal to the orbit's radius R. To find the impact of the reduced mass of Planet X, we need to understand the mass influences on the satellite's orbital period.
When the mass of the planet decreases by a factor of 4, the orbital velocity also decreases by the same factor as vorbit is proportional to the square root of Planet X’s mass. Consequently, with a quarter of the orbital velocity, the period must be twice as long. Hence, if the original period was T, the new period would be 2T. So, the correct choice is: A) 2T.
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An organ pipe is 248 cm long.(a) Determine the fundamental overtone if the pipe is closed at one end.(b) Determine the first audible overtone if the pipe is closed at one end.(c) Determine the second audible overtone if the pipe is closed at one end.(d) Determine the third audible overtone if the pipe is closed at one end.(e) Determine the fundamental overtone if the pipe is open at both ends.(f) Determine the first audible overtone if the pipe is open at both ends.(g) Determine the second audible overtone if the pipe is open at both ends.(h) Determine the third audible overtone if the pipe is open at both end.
(a) 34.6 Hz
The fundamental frequency of a pipe closed at one end is given by
[tex]f_1 = \frac{v}{4 L}[/tex]
where
v = 343 m/s is the speed of the sound in air
L is the length of the pipe
In this problem,
L = 248 cm = 2.48 m
So, the fundamental frequency is
[tex]f_1 = \frac{343 m/s}{4 (2.48 m)}=34.6 Hz[/tex]
(b) 103.8 Hz
In a open-closed pipe, only odd harmonics are produced; therefore, the frequency of the first overtone (second harmonic) is given by:
[tex]f_2 = 3 f_1[/tex]
where
[tex]f_1 = 34.6 Hz[/tex] is the fundamental frequency.
Substituting into the equation,
[tex]f_2 = 3 (34.6 Hz)=103.8 Hz[/tex]
(c) 173 Hz
The frequency of the second overtone (third harmonic) is given by:
[tex]f_3 = 5 f_1[/tex]
where
[tex]f_1 = 34.6 Hz[/tex] is the fundamental frequency.
Substituting into the equation,
[tex]f_3 = 5 (34.6 Hz)=173 Hz[/tex]
(d) 242.2 Hz
The frequency of the third overtone (fourth harmonic) is given by:
[tex]f_4 = 7 f_1[/tex]
where
[tex]f_1 = 34.6 Hz[/tex] is the fundamental frequency.
Substituting into the equation,
[tex]f_4 = 7 (34.6 Hz)=242.2 Hz[/tex]
(e) 69.2 Hz
The fundamental frequency of a pipe open at both ends is given by
[tex]f_1 = \frac{v}{2 L}[/tex]
where
v = 343 m/s is the speed of the sound in air
L is the length of the pipe
In this problem,
L = 248 cm = 2.48 m
So, the fundamental frequency is
[tex]f_1 = \frac{343 m/s}{2 (2.48 m)}=69.2 Hz[/tex]
(f) 138.4 Hz
In a open-open pipe, both odd and even harmonics are produced; therefore, the frequency of the first overtone (second harmonic) is given by:
[tex]f_2 = 2 f_1[/tex]
where
[tex]f_1 = 69.2 Hz[/tex] is the fundamental frequency.
Substituting into the equation,
[tex]f_2 = 2 (69.2 Hz)=138.4 Hz[/tex]
(g) 207.6 Hz
The frequency of the second overtone (third harmonic) in an open-open pipe is given by:
[tex]f_3 = 3 f_1[/tex]
where
[tex]f_1 = 69.2 Hz[/tex] is the fundamental frequency.
Substituting into the equation,
[tex]f_3 = 3 (69.2 Hz)=207.6 Hz[/tex]
(h) 276.8 Hz
The frequency of the third overtone (fourth harmonic) in an open-open pipe is given by:
[tex]f_4 = 4 f_1[/tex]
where
[tex]f_1 = 69.2 Hz[/tex] is the fundamental frequency.
Substituting into the equation,
[tex]f_4 = 4 (69.2 Hz)=276.8 Hz[/tex]
What type of galaxy is M82 based on its appearance in the visible-light view?
A) elliptical
B) spiral
C) irregular Part
According to Hubble galaxies are classified into elliptical, spiral and irregular.
It should be noted this classification is based only on the visual appearance of the galaxy, and does not take into account other aspects, such as the rate of star formation or the activity of the galactic nucleus.
The classification is as follows:
1. Elliptical galaxies: Their main characteristic is that the concentration of stars decreases from the nucleus, which is small and very bright, towards its edges. In addition, they contain a large population of old stars, usually little gas and dust, and some newly formed stars.
2. Spiral galaxies: They have the shape of flattened disks containing some old stars and also a large population of young stars, enough gas and dust, and molecular clouds that are the birthplace of the stars.
3. Irregular Galaxies: Galaxies that do not have well-defined structure and symmetry.
In this context, galaxy M82 does not match with the first two types of galaxies, because it has not a defined shape.
Therefore, M82 is an irregular galaxy.
Study the image.
At which point is subduction occurring?
1
2
3
4
Answer:
At point 4
Explanation:
Subduction is a geologic process which occurs at convergent plate margins.
Subduction occurs when a denser lithospheric plate goes beneath the lighter one. Generally, the lithosphere is made up of the crust and part of the mantle. The two moves slowly on the weak and plastic asthenosphere.
In a convergent plate margin, two plates comes together. When two plates comes together, it is either we have collision or subduction. When subduction occurs, denser plates goes beneath the less dense ones.
The average density of the continental crust is 2.9g/cm³ while that of the oceanic crust is 3.3g/cm³. The oceanic crust geos beneath the continental crust because it is denser. This is subduction.
Answer:
the answer is 4
Explanation:
Calculate the energy of a photon having a wavelength in thefollowing ranges.
(a) microwave, with λ = 50.00 cm
eV
(b) visible, with λ = 500 nm
eV
(c) x-ray, with λ = 0.50 nm
eV
The energy [tex]E[/tex] of a photon is given by the following formula:
[tex]E=h.f[/tex] (1)
Where:
[tex]h=4.136(10)^{-15}eV.s[/tex] is the Planck constant
[tex]f[/tex] is the frequency
Now, the frequency has an inverse relation with the wavelength [tex]\lambda[/tex]:
[tex]f=\frac{c}{\lambda}[/tex] (2)
Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum
Substituting (2) in (1):
[tex]E=\frac{hc}{\lambda}[/tex] (3)
Knowing this, let's begin with the answers:
(a) Microwave: 50.00 cmFor [tex]\lambda=50cm=0.5m[/tex]
[tex]E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{0.5m}[/tex]
[tex]E=\frac{1.24(10)^{-6}eV.m }{0.5m}[/tex]
[tex]E=2.48(10)^{-6}eV[/tex]
(b) Visible: 500 nmFor [tex]\lambda=500nm=500(10)^{-9}m[/tex]
[tex]E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{500(10)^{-9}m}[/tex]
[tex]E=\frac{1.24(10)^{-6}eV.m }{500(10)^{-9}m}[/tex]
[tex]E=2.48 eV[/tex]
(c) X-ray: 0.5 nmFor [tex]\lambda=0.5nm=0.5(10)^{-9}m[/tex]
[tex]E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{0.5(10)^{-9}m}[/tex]
[tex]E=\frac{1.24(10)^{-6}eV.m }{0.5(10)^{-9}m}[/tex]
[tex]E=2480 eV[/tex]
As we can see, as the wavelength decreases, the energy increases.
Which of the following methods has led to the most discoveries of massive planets orbiting near their parent stars?A) detecting the starlight reflected off the planetB) detecting the infrared light emitted by the planetC) detecting the gravitational effect of an orbiting planet by looking for the Doppler shifts in the star's spectrumD) detecting the shift of the star's position against the sky due to the planet's gravitational pullE) detecting a planet ejected from a binary star system
Answer:
C
Explanation: when an object gets in front of a light, it temporarily breaks the beaks the beam from the light source to the user. the same occurs when a planet gets in front of a star. i don't feel like explaining the doppler effect but this is basically it.
hope this helped
(please help me)
If a wave has a wavelength of 13 meters and a period of 0.005, what's the velocity of the wave?
A. 260 m/s
B. 1,300 m/s
C. 2,600 m/s
D. 1,560 m/s
(please dont type any wrong answer on porpuse)
Answer:2,600 m/s
Explanation:13/ 0.005=2,600.
ANSWER:
The velocity of the wavelength is [tex]2600 \mathrm{m} / \mathrm{s}[/tex]
Explanation:
Given:
The wavelength of the wave= 13 meters
Time period of the wave=0.005seconds
To find:
velocity of the wave=?
Solution:
The velocity of the wave is defined as the product of frequency and wavelength.
Mathematically,
[tex]v=f \lambda[/tex]
Where f is the frequency and λis the wavelength of the wave.
Finding the frequency using time period,
[tex]f=\frac{1}{T}[/tex]
Substituting the value of time period we have,
[tex]f=\frac{1}{0.005}[/tex]
[tex]f=200 \mathrm{Hz}[/tex]
Now,
[tex]v=f \lambda[/tex]
[tex]v=200 \times 13[/tex]
[tex]v=2600 \mathrm{m} / \mathrm{s}[/tex]
Result:
The velocity of the wave with wavelength 13 meters and time period 0.005seconds is [tex]2600 \mathrm{m} / \mathrm{s}[/tex].
) a 45-mh ideal inductor is connected in series with a 60-ω resistor through an ideal 15-v dc power supply and an open switch. if the switch is closed at time t = 0 s, what is the current 7.0 ms later?
a.250 ma
b.850 ma
c.550 ma
d.280 ma
e.650 ma
Initial current = 0
Final current = (15 V) / (60 ohms) = 0.25 Ampere
Current along the way = 0.25 · (1 - e^- time / time-constant)
"time-constant" = L/R = (0.045 / 60) = 7.5 x 10⁻⁴ second
Current = 0.25 · (1 - e^-10,000t/7.5)
When t = 7 ms,
Current = 0.25 · ( 1 - e^-70/7.5)
Current = 0.25 · (1 - e^-9.33)
Current = 0.25 · (1 - 8.84 x 10⁻⁵)
Current = 0.25 · (0.9999)
Current = so close to 250 mA that you can't tell the difference.
The reason is that 7.0 mS is 9.3 time-constants, and during EVERY time-constant, the current grows by 37% of the distance it still has left to go. So after 9.3 of these, it's practically AT the target.
I have a feeling that the time in the question is SUPPOSED TO BE 7 microseconds. If that's true, then
Current = 0.25 · (1 - e^-[ 7 x 10⁻⁶ / 7.5 x 10⁻⁴ ]
Current = 0.25 · (1 - e^-0.00933)
Current = 0.25 · (1 - 0.9907)
Current = 0.25 · (0.0093)
Current = 2.32 mA ?
No, that can't be it either.
Well ! Now, I'm going to determine the true and correct final answer in the only cheap and sleazy way I have left ... by looking at all the choices offered, and eliminating the absurd ones.
The effect of an inductor in the circuit is to resist any change in current. The final current in this circuit is when it's not trying to change any more. So the final current is just the battery with a resistor across it ... (12 V) / (60 ohms). That's 0.25 Ampere, or 250 mA. The current starts at zero when the switch closes, and it builds up and builds up to 250 mA. It's never more than 250 mA.
So look at the choices ! The only one that not more than 250 mA is choice-A .
THAT has to be it. 7.0 mS is a no-brainer. It's 9.3 time-constants after the switch closes, the current has built up to 99.99% of its final value by then, it's not really trying to change much any more, the inductor is just about finished having any effect on the current, and the current is essentially at its final value of 250 mA. The action is all over.
Now, I fully realize that Mister "Rishwait" is a bot and all, and nobody really needs the answer to this question. But every cloud has a silver lining. It's a numskull question, but it earned me 10 points, and it's been a truly fascinating trip down Memory Lane.
the current approximately 7.0 ms after closing the switch is about 250 mA, which is option (a).
To find the current through the circuit 7.0 ms after the switch is closed, we can use the concept of an RL circuit. The current in an RL circuit follows an exponential growth equation, given by:
I(t) = (V/R)(1 - e^(-t/τ))
Where:
I(t) is the current at time t.
V is the voltage from the power supply (15 V in this case).
R is the resistance (60 Ω).
τ (tau) is the time constant of the circuit, given by L/R, where L is the inductance (45 mH = 0.045 H).
First, calculate the time constant τ:
τ = L/R = 0.045 H / 60 Ω = 0.00075 s.
Now, plug in the values into the equation to find I(7.0 ms):
t = 7.0 ms = 0.007 s.
I(0.007 s) = (15 V / 60 Ω) * (1 - e^(-0.007 s / 0.00075 s))
I(0.007 s) = (0.25 A) * (1 - e^(-9.333...))
Now, calculate the current:
I(0.007 s) ≈ (0.25 A) * (1 - e^(-9.333...))
I(0.007 s) ≈ (0.25 A) * (1 - 0.0000962) [Using e^(-9.333...) ≈ 0.0000962]
I(0.007 s) ≈ (0.25 A) * (0.9999038)
I(0.007 s) ≈ 0.24998 A
I(0.007 s) ≈ 250 mA
So, the current approximately 7.0 ms after closing the switch is about 250 mA, which is option (a).
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What nuclear reaction is the energy source of a white dwarf?
Answer:
Multiple choice answer would be "None"
Explanation:
White dwarfs are radiating stored heat from earlier reactions.
Technically, it would be the last fusion stage the star went through
BEFORE it became a white dwarf, but that's nit-picking.
The energy source of a white dwarf is not a nuclear reaction in the traditional sense, but rather it is supported by a process called electron degeneracy pressure.
What is a white dwarfA white dwarf is the remnant of a low to medium-mass star (up to about 1.4 times the mass of the Sun) after it has exhausted its nuclear fuel. The core of the star collapses under gravity, and the electrons in the core become packed extremely closely together due to the Pauli exclusion principle, which states that no two electrons can occupy the same quantum state simultaneously.
This electron degeneracy pressure provides the counterforce to gravity, preventing further collapse. No nuclear reactions are occurring in a white dwarf as it no longer has the high temperatures and pressures required for nuclear fusion. Instead, it is a stellar remnant that is gradually cooling over time.
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Technician a says in a parallel circuit, the more branches that are added, the more current flow increases. technician b says a series-parallel circuit is made of parallel branches only. who is correct?
Technician-A is correct. His statement: "In a parallel circuit, the more branches that are added, the more current flow increases." is technically true.
Technician-B is incorrect. His statement: "A series-parallel circuit is made of parallel branches only." is technically false.
The more branches a parallel circuit has the more current flow in the circuit therefore ; Technician A is correct while Technician B is wrong
In a parallel circuit the increase in branches will lead a corresponding increase in the amount of current flow through the circuit because the Total amount of current flowing through a parallel circuit is a summation of the individual currents flowing through the branches
i.e. [tex]I_{T} = I_{1} + I_{2} + I_{3}[/tex]
But A series-parallel circuit is made up of both parallel and series branches as the name implies therefore Technician B is wrong
Hence we can conclude that the more branches a parallel circuit has the more current flow in the circuit hence Technician A is correct.
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Suppose that the sun shrank in size but that its mass remained the same. What would happen to the orbit of the earth?
Answer:
Nothing
Explanation:
The radius of the orbit of the Earth does not depend on the radius of the sun.
In fact, the gravitational attraction between the Earth and the Sun provides the centripetal force that keeps the Earth in orbit:
[tex]G\frac{Mm}{r^2} = m\frac{v^2}{r}[/tex]
where
G is the gravitational constant
M is the mass of the sun
m is the mass of the Earth
r is the radius of the orbit of the Earth
v is the orbital speed of the earth
Re-arranging the equation for r:
[tex]r=\frac{GM}{v^2}[/tex]
Also,
[tex]v=\omega r[/tex]
where [tex]\omega[/tex] is the angular velocity of the Earth's orbit. So we can rewrite the equation as
[tex]r=\frac{GM}{\omega^2 r^2}\\r^3 = \frac{GM}{\omega^2}[/tex]
As we see, the radius of the orbit of the Earth, r, does not depend on the mass of the Sun, so if the sun shrank in size, the orbit remains the same.
Final answer:
The Earth's orbit would remain unchanged if the Sun's size decreased but its mass remained the same since gravitational force depends on mass, not size. The orbital period would also be unchanged if the Sun turned into a black hole with the same mass. However, if the Sun shrank to a certain point, it would become a black hole.
Explanation:
If the Sun shrank in size while its mass remained the same, the Earth's orbit would not change. This is because the gravitational force between two objects depends on their masses and the distance between them, not their sizes. According to Newton's law of universal gravitation, the force is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers. Therefore, as long as the mass of the Sun and the distance between the Earth and the Sun remains constant, the gravitational force and thus the orbit would remain unaffected.
If the Sun were to collapse into a black hole of the same mass, the Earth's orbital period would remain the same. That's because the Earth's orbit depends on the mass of the Sun, as described by Kepler's third law, which relates the period of an orbit to the distance from the focus (in this case, the Sun or black hole) and the mass of the object being orbited.
However, if the Sun collapsed beyond a particular point, general relativity tells us that the curvature of spacetime would get larger. If it shrank to a diameter of about 6 kilometers, it would become a black hole, and only light beams sent out perpendicular to the surface would escape. Even a slight further shrinkage would trap all light, rendering the Sun a black hole.
The attraction between two oppositely charged Atoms or groups of Atoms is which type of bond
Answer:
The answer is ionic bond
Explanation:
The circular but relatively flat portion of the galaxy is the
The circular but relatively flat portion of the galaxy is the Disk
A galaxy that resembles a circle is known as a ring galaxy. Art Hoag's 1950 discovery of Hoag's Object is an illustration of a ring galaxy. Many big, relatively young blue stars that are quite brilliant can be found in the ring.
What circular but relatively flat portion of the galaxy?Galactic disks are thin, essentially circular collections of stars, gas, and dust; this matter revolves around a common core in almost circular orbits. As a result of this rotation, many disks have lovely spiral patterns, and some have distinct bars crossing their centres.
Nearly all of our galaxy's gas, dust, hot young stars, and star-forming regions are present there. When viewed from above, the disk reveals spiral arms that contain the majority of the ISM's cool, dense regions.
Therefore, Our galaxy's disk is incredibly narrow, only around 100 times wider than its own height.
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A proton accelerates from rest in a uniform electric field of 635 N/C. At some time later, its speed is 1.30 × 106 m/s. What is the magnitude of its acceleration? The mass of a proton is 1.6726 × 10−27 kg and the fundamental charge is 1.602 × 10−19 C . Answer in units of m/s 2 . 031 (part 2 of 4) 10.0 points How long does it take the proton to reach this speed? Answer in units of s. 032 (part 3 of 4) 10.0 points How far has it moved in this time interval? Answer in units of m. 033 (part 4 of 4) 10.0 points What is its kinetic energy
1) [tex]6.11\cdot 10^{10} m/s^2[/tex]
The force experienced by the proton is
[tex]F=qE[/tex]
where
[tex]q=1.6\cdot 10^{-19}C[/tex] is the proton charge
[tex]E=635 N/C[/tex] is the strength of the electric field
Substituting into the equation,
[tex]F=(1.6\cdot 10^{-19} C)(635 N/C)=1.02\cdot 10^{-16}N[/tex]
The acceleration of the proton is given by Newton's second law:
[tex]a=\frac{F}{m}[/tex]
where
[tex]F=1.02\cdot 10^{-16}N[/tex] is the force exerted on the proton
[tex]m=1.67\cdot 10^{-27} kg[/tex] is the proton's mass
Substituting,
[tex]a=\frac{1.02\cdot 10^{-16}N}{1.67\cdot 10^{-27}kg}=6.11\cdot 10^{10} m/s^2[/tex]
2) [tex]2.13\cdot 10^{-5} s[/tex]
We can use the following equation:
[tex]a=\frac{v-u}{t}[/tex]
where
[tex]a=6.11\cdot 10^{10} m/s^2[/tex] is the acceleration of the proton
[tex]v=1.30\cdot 10^6 m/s[/tex] is the final velocity
u = 0 is the initial velocity
t is the time
Solving the equation for t, we find
[tex]t=\frac{v-u}{a}=\frac{1.30\cdot 10^6 m/s -0}{6.11\cdot 10^{10} m/s^2}=2.13\cdot 10^{-5} s[/tex]
3) 13.86 m
The distance travelled by the proton is given by the equation
[tex]d=ut + \frac{1}{2}at^2[/tex]
where
u = 0 is the initial velocity
[tex]t=2.13\cdot 10^{-5} s[/tex] s the time
[tex]a=6.11\cdot 10^{10} m/s^2[/tex] is the acceleration of the proton
Substituting,
[tex]d=0 + \frac{1}{2}(6.11\cdot 10^{10}m/s^2)(2.13\cdot 10^{-5} s)^2=13.86 m[/tex]
4) [tex]1.41\cdot 10^{-15} J[/tex]
The final kinetic energy of the proton is given by
[tex]K=\frac{1}{2}mv^2[/tex]
where we have
[tex]m=1.67\cdot 10^{-27} kg[/tex] is the proton's mass
[tex]v=1.30\cdot 10^6 m/s[/tex] is the final velocity
Substituting into the formula,
[tex]K=\frac{1}{2}(1.67\cdot 10^{-27}kg)(1.30\cdot 10^6 m/s)^2=1.41\cdot 10^{-15} J[/tex]
The magnitude of the proton's acceleration can be found using Newton's second law. The time it takes for the proton to reach a given speed can be calculated using the equation for linear motion. The distance the proton moves can be determined using the equation for distance traveled. The kinetic energy of the proton can be calculated using the equation for kinetic energy.
Explanation:To find the magnitude of the acceleration of the proton, we can use the equation F = ma, where F is the force exerted on the proton and m is its mass. In this case, the force is given by F = qE, where q is the charge of the proton and E is the electric field. Since the charge of the proton is known, we can calculate the force and then use Newton's second law to find the acceleration.
Once we have the acceleration, we can use the equation v = u + at to find the time it takes for the proton to reach the given speed, where v is the final velocity, u is the initial velocity (0 in this case), a is the acceleration, and t is the time.
To find the distance the proton has moved, we can use the equation s = ut + (1/2)at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.
To calculate the kinetic energy of the proton, we can use the equation KE = (1/2)mv², where KE is the kinetic energy, m is the mass of the proton, and v is its velocity.
The Kinetic Molecular Theory of Gases Learning Goal To understand some aspects of molecular motion in the gas phase The kinetic molecular theory of gases explains how gas molecules behave in terms of motion, speed, and energy One important aspect of this theory deals with the relationship between temperature and the average speed of the gas molecules. Increasing the temperature of a gas sample increases the average kinetic energy of the molecules. The kinetic energy of a molecule determines its speed It is important to realize that not all molecules in a sample will have the same kinetic energy, which is why we refer to the average kinetic energy and the average speed. The speed of a particle with average kinetic energy is called the root mean square (rms) speed, vrms The rms speed may be expressed by the following equation: vrms=√3RT/MWhere R is the ideal gas constant, T is the absolute temperature, and M is the molar mass of the substance in kilograms per mole The constant motion of gas molecules causes diffusion and effusion. Diffusion is the gradual mixing of two substances resulting from the movement of their particles. Effusion is the gradual escape of gas molecules through microscopic holes in their container. Part A Which of the following state ments are true? Check all that apply The average kinetic energy of gas molecules increases with increasing temperature There are gas molecules that move faster than the average The temperature of a gas sample is independent of the average kinetic energy The average speed of gas molecules decreases with decreasing temperature All the gas molecules in a sample cannot have the same kinetic energy
Answer:
The average kinetic energy of gas molecules increases with increasing temperature
There are gas molecules that move faster than the average
The average speed of gas molecules decreases with decreasing temperature
All the gas molecules in a sample cannot have the same kinetic energy
Explanation:
The average kinetic energy of the particles in an ideal monoatomic gas is given by:
[tex]E_k = \frac{3}{2}kT[/tex] (1)
where
k is the Boltzmann constant
T is the absolute temperature of the gas
While the rms speed of the particles in a gas is given by
[tex]v_{rms}= \sqrt{\frac{3RT}{M}}[/tex] (2)
where
R is the gas constant
T is the absolute temperature
M is the molar mass
Let's now analyze each statement:
- The average kinetic energy of gas molecules increases with increasing temperature --> TRUE. If we look at eq.(1), we see that the average kinetic energy is directly proportional to the temperature.
- There are gas molecules that move faster than the average --> TRUE. The distribution of the speed of the particles in a gas is spread around the rms speed, but of course not all the particles are moving at that speed: some particles are moving faster, while some are moving slower.
- The temperature of a gas sample is independent of the average kinetic energy --> FALSE. As we see from eq.(1), the two quantities are related to each other.
- The average speed of gas molecules decreases with decreasing temperature --> TRUE. As we see from eq.(2), the average speed is proportional to the square root of the temperature: so, when the temperature decreases, the average speed decreases as well.
- All the gas molecules in a sample cannot have the same kinetic energy --> TRUE. In fact, each particle will have a different kinetic energy, depending on its speed (different speed means also different kinetic energy).
The Kinetic Molecular Theory denotes that with an increase in temperature, the average kinetic energy and speed of gas molecules also increase. Gas molecules can move faster or slower than the average speed, hence all molecules will not have the same kinetic energy. The temperature of a gas is not independent of the average kinetic energy.
Explanation:The Kinetic Molecular Theory of gases explains some key aspects of molecular motion within gases. The theory denotes that molecules are constantly in motion and the average speed of these molecules is determined by their absolute temperatures. As the temperature increases, so too does the average kinetic energy of the molecules, which in turn increases their speed.
Not all molecules in a gas sample will have the same kinetic energy; some will move faster than the average speed and others slower, lending to what we refer to as the average kinetic energy and average speed. The typical or root mean square (rms) speed of a particle with average kinetic energy can be expressed using the equation: vrms=√3RT/M, where R is the ideal gas constant, T is the absolute temperature, and M is the molar mass.
Moving on towards the state comparisons, the following statements are true: The average kinetic energy of gas molecules increases with increasing temperature; gas molecules can indeed move faster than the average speed; with a decrease in temperature, the average speed of the gas molecules decreases; all the gas molecules in a sample do not possess the same kinetic energy. The statement that classifies temperature of a gas sample as independent of the average kinetic energy is false.
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This is a map that shows hot spots around the world.
Which phrases describe the formation of the Hawaiian Islands? Check all that apply.
formed due to subduction
formed in the middle of a plate
formed at a divergent boundary
formed at a convergent boundary.
formed where mantle erupts through crust
Answer:
formed in the middle of a plate
formed where mantle erupts through crust
Explanation:
The Hawaiian Islands are Volcanoes that formed right in the middle of the Pacific plate which is moving North-westward.
Lithospheric plates lies on the weak and plastic asthenosphere. Such is the Pacific plate too. The weak asthenosphere can erupt on the surface if it gets access through faulting or other geologic conduits. When these mantle magma reaches the surface, they form hotpots on the crust.
The Hawaiian island is a series of these hotspot as it forms when mantle materials upwells to the surface. The hotspot from which the magma is sourced is relatively fixed. The moving plate is what leads to the eruption of the magma at several other parts in the crust.
Calculate the drops per minute “gtts/min” for the following questions. Be sure to write your answer in gtts /min. 1. 250 ml/hour using macro tubing 20 gtts/ml _____________________ 2. 167 ml/hour using macro tubing 10 gtts/ml _____________________ 3. 125 ml/hour using macro tubing 15 gtts/ml _____________________ 4. 75 ml/hour using micro tubing 60 gtts/ml _____________________ 5. 250 ml/hour using macro tubing 20 gtts/ml _____________________ 6. 167 ml/hour using macro tubing 15 gtts/ml _____________________ 7. 125 ml/hour using macro tubing 10 gtts/ml _____________________ 8. 75 ml/hour using macro tubing 10 gtts/ml ______________________ 9. 1000 ml/hour using macro tubing 10 gtts/ml _____________________ 10. 500 ml/hour using macro tubing 10 gtts/ml _____________________
Answer:
Explanation:
There are 12 gtts in 1 mL, and 60 minutes in 1 hr.
250 mL/hr * (12 gtts / mL) * (1 hr / 60 min) = 50 gtts/min
167 mL/hr * (12 gtts / mL) * (1 hr / 60 min) = 33.4 gtts/min
125 mL/hr * (12 gtts / mL) * (1 hr / 60 min) = 25 gtts/min
75 mL/hr * (12 gtts / mL) * (1 hr / 60 min) = 15 gtts/min
1000 mL/hr * (12 gtts / mL) * (1 hr / 60 min) = 200 gtts/min
500 mL/hr * (12 gtts / mL) * (1 hr / 60 min) = 100 gtts/min
The best method for separating an oil and water mixture would be _____.
screening
magnetism
density
filtering
density..................your welcomr
The complete statement is "The best method for separating an oil and water mixture would be density". This is further explained below.
What is separation technique?Generally, Separation techniques are simply defined as methods for separating two distinct matter states.
In conclusion, The best method for separating an oil and water mixture would be through their density
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A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average
speed of 6.0 m/s.
What is her average speed for the total distance of 600m?
A 2.0 m/s
B 4.0 m/s
C 4.5 m/s
D 8.0 m/s
Answer:
B. 4 m/s
Explanation:
v=d/t
Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.
The average speed of the runner for the entire journey is 4 m/s.
The given parameters;
distance covered by the runner, s = 300 minitial speed of the runner, v₁ = 3 m/sfinal speed of the runner, v₂ = 6 m/stotal distance covered by the runner, = 600 mThe average speed of the runner is obtained by diving the total distance covered by the runner by the total time of motion as shown below;
[tex]average \ speed = \frac{total \ distance }{total \ time \ of \ motion} \\\\[/tex]
The time of motion during the first distance covered;
[tex]t_1 = \frac{300}{3} = 100 \ s[/tex]
The time of motion during the second distance covered;
[tex]t_2 = \frac{300}{6} = 50 \ s[/tex]
The average speed is calculated as;
[tex]average \ speed = \frac{600}{100 + 50} \\\\average \ speed = 4 \ m/s[/tex]
Thus, the average speed of the runner for the entire journey is 4 m/s.
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