What is the correct formula for power? A. Power = work / time B. Power = work * time C. Power = force * distance D. Power = work * distance

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:

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:

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].

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.

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.

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.

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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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.

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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Answer:inner core?

Explanation:

Answer: your answer would be inner core hope it helps

Explanation:

tell me i am wrong?

Answer:

The answer is ionic bond

Explanation:

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]

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.

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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Physical sciences study nonliving matter, including fields like geology, astronomy, physics, and chemistry, with a focus on matter and energy interactions.

The term physical sciences pertains to the study of matter and energy. The field of physical science includes subjects like geology, astronomy, physics, and chemistry, which all explore various aspects of nonliving matter. Physics, being the most fundamental of these sciences, deals with concepts of energy, matter, space and time, and their interactions. It is essential for understanding the general truths of nature that are expressed through scientific laws and theories, which describe the rules that all natural processes appear to follow.

See the attached picture:

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.

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

(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]

it is the vertical distance between the waterline and the bottom of hull and determines the minimum depth of water

If we approximate the orbit of the planets around the Sun to circular orbits with a uniform circular motion, where the velocity [tex]\vec{V}[/tex] is a vector, whose direction is perpendicular to the radius [tex]r[/tex] of the trajectory; the acceleration [tex]\vec{a}[/tex] is directed towards the center of the circumference (that's why it's called centripetal acceleration).  

Now, according to Newton's 2nd law, the force [tex]\vec{F}[/tex] is directly proportional and in the same direction as the acceleration:  

[tex]\vec{F}=m.\vec{a}[/tex]  

Therefore the net force resulting from the movement of a planet orbiting the Sun points towards the center of the circle, this is called Centripetal Force which is a central force that in this case is equal to the gravity force.

Answer:

Force increases nine times its initial value

Explanation:

For charges, charged bodies very small compared to the distance [tex]r[/tex] that separates them, Coulomb discovered that the electric force is proportional to [tex]\frac{1}{r^{2}}[/tex]

So, if the distance is made a third of what it was, the force will be increased nine times its initial value

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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The total rotational kinetic energy of this system is : ( C ) none of these

Ker =  Iw²

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

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;

The 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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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%

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

A) The forward force is equal to the backward force

In this problem:

- the forward force is the force that Megan applies to the pedal to go forward

- the backward force is due to the air resistance and the friction between the wheels and the track

In this case, Megan is travelling at constant speed. This means that her acceleration is zero:

a = 0

According to Newton's second law, the resultant of the forces acting on Megan is equal to the product between mass (m) and acceleration (a):

[tex]\sum F = ma[/tex]

However, a = 0, so the resultant of the forces is also zero:

[tex]\sum F =0[/tex]

and this implies that the forward force and the backward force are equal in magnitude and opposite in direction.

B) The forward force is larger than the backward force

In this case, Megan crouched down in order to make herself more streamlined. As a result, the air resistance acting on Megan will decrease: so, the backward force will decrease, and therefore the forward force (which has remained the same) will be larger than the backward force.

So, the resultant force

[tex]\sum F[/tex]

will be no longer zero, and therefore the acceleration will be different from zero, which means that Megan will increase her speed.

Explanation:

that don't look bright

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.

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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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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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:

For [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]  

For [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]  

For [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.

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.

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.

A 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.

Learn more about white dwarfs

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