mark needs to learn the planets of the solar system listed in the order of their proximity to the sun: mercury, venus, earth, mars, jupiter, saturn, uranus, and neptune. if we consider the serial learning curve, which planets will probably be the most difficult for mark to learn?

The conductivity of a material is determined by the number of free electrons. Insulators have few or no free electrons.

The insulators are composed of a few electrons and only a tiny amount of electrical current can pass through them. The wires, however, make for superior electrical conductors. In contrast to metal, which is a very excellent conductor, rubber is a poor insulator.

A few electrons make up the insulator, which allows very little electrical current to pass through it. While the wires are excellent electrical conductors. Rubber is a poor insulator while metal is a very excellent conductor.

As a result, the conductor has been used for wiring and can be efficiently converted to use energy. Because the insulator is a weak absorber, energy cannot be transferred.

Therefore, the correct answer is insulators.

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This is true because the process of dry ice sublimating and the puddle freezing both involve chemical bonds breaking.

In the case of dry ice (solid CO2), the sublimation process occurs when the CO2 molecules absorb enough energy to break the bonds holding them together in a solid state. This energy is supplied by the surrounding air, causing the CO2 to transition directly from a solid to a gas.

In the case of the puddle freezing, the water molecules lose energy and slow down, allowing hydrogen bonds to form between the molecules. As more and more bonds form, the liquid transitions to a solid (ice). The breaking of hydrogen bonds in the liquid and the formation of new bonds in the solid are chemical processes that represent a breaking and forming of chemical bonds.

"

Complete and correct question

I dry ice sublimating and becoming a ga and a puddle freezing into ice on a clod night both involve chemical bond breaking

True or false

"

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Part A: The work done on the particle as it moves from x=0 to x=L is W=0.
Part B: The work done on the particle as it moves from x=0 to x=2L is W=0.


Part A: The velocity of the particle is given by vx = v0 sin(πx/2L). To find the work done, we need to find the force acting on the particle, which can be found by taking the time derivative of the momentum (F = dp/dt = mdv/dt). However, since the velocity function is symmetric about x=L/2, the net change in velocity is zero as the particle moves from x=0 to x=L, meaning there's no net force acting on the particle, and thus, no work is done (W=0).

Part B: Similarly, the velocity function is periodic with a period of 2L, which means the net change in velocity is also zero as the particle moves from x=0 to x=2L. Therefore, the work done is also zero (W=0).

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sis i love the eren season 1-2-3-4 but the eren season 5?....... i just :')

Answer:

Explanation:

Aaron yogurt supremacy

Spiral galaxies are classified based on spiral arm tightness, bulge size, and amount of gas and dust present. This allows astronomers to categorize them into subtypes such as Sa, Sb, and Sc.

The classification of spiral galaxies is based on three properties:

1. Spiral arm tightness: This refers to how tightly wound the spiral arms are around the galaxy's center. Galaxies with more tightly wound arms are classified as "Sa," while those with more loosely wound arms are classified as "Sc."

2. Bulge size: The central bulge of a spiral galaxy can vary in size. Larger bulges are typically found in early-type spiral galaxies (such as Sa), while smaller bulges are found in late-type spiral galaxies (like Sc).

3. Amount of gas and dust: The presence and distribution of gas and dust within a spiral galaxy also play a role in its classification. Early-type spiral galaxies generally have less gas and dust compared to late-type spiral galaxies.

By considering these three properties, astronomers can classify spiral galaxies into various subtypes (such as Sa, Sb, and Sc) within the broader spiral galaxy category.

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The slope of the plot of d vs \(t^2\)should be equal to one-half of the acceleration of the object.

If a graph plotting the distance travelled (d) of an object against time squared (t2) is well-fitted by a straight line, then its slope can be used to calculate the acceleration of said object.

We can observe this relationship by rearranging the kinematic equation for distance travelled (d) with a constant acceleration (a):\(a = 2 * d / t^2.\)

In order to calculate the constant acceleration of an object, we need to draw a straight line with the slope equal to 2a on a graph of d vs \(t^2\). By rearranging the equation we get

:\(a = 2d / t^2.\)

Now, by dividing both sides of this equation by 2, we can obtain the acceleration, which is simply given by

:\(a = d / t^2.\)

\(a/2 = d / t^2\)

This tells us that the slope of the plot of d vs \(t^2\)should be equal to one-half of the acceleration of the object.

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According to the standard atmosphere, the air temperature at an altitude of 30,000 feet is 93.15 K, the air pressure is 1394.6 Pa, and the air density is 52.18 kg/m^3.

The standard atmosphere is a model of the Earth's atmosphere that describes how the temperature, pressure, and density of air change with altitude. The values for air temperature, pressure, and density at an altitude of 30,000 feet can be found in the standard atmosphere table.

The air temperature at 30,000 feet is 93.15 K, which is about -130 degrees Celsius. The air pressure at this altitude is 1394.6 Pa, which is about 1.4 psi. The air density at 30,000 feet is 52.18 kg/m^3, which is about one-tenth the density of air at sea level.

The decrease in air temperature, pressure, and density with altitude is due to the fact that there are fewer air molecules at higher altitudes. As the altitude increases, the weight of the air above decreases, and the air molecules spread out more. This results in a lower air pressure and density.

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So, the distance of the wall from Nahi is (C). 498 m.

Hi ! I'm here to help you solve this problem. This matter states that Nahi shouted and from her scream echoed, so that he could hear his own screams at intervals of time. Echoes are caused by the reflection of sound on objects that are too hard, so the sound will experience twice movement (from sound to the reflected plane and from the reflected plane in all directions, including the sound source). Based on the image I gave, the relationship between the distance of the sound source to the reflected sound field can be expressed in the following equation:

\( \boxed{\sf{\bold{s = \frac{v \times t}{2}}}} \)

With the following condition :

We know that :

What was asked :

Step by step :

\( \sf{s = \frac{v \times t}{2}} \)

\( \sf{s = \frac{\cancel{332} \:_{166} \times 3}{\cancel{2} \:_1}} \)

\( \boxed{\sf{s = 498 \: m}} \)

So, the distance of the wall from Nahi is (C). 498 m

The pilot's apparent weight during the plane's turn is 3665.3 N.

To determine the apparent weight of the pilot during the plane's turn, we need to consider the centripetal force acting on the pilot due to the turn. The apparent weight is the sum of the actual weight and the centripetal force.

Calculate the centripetal force:

The centripetal force (Fc) can be calculated using the equation\(Fc = (m * v^2) / r\), where m is the mass of the pilot, v is the velocity of the plane, and r is the radius of curvature.

Fc = \((61 kg) * (450 m/s)^2 / 4000 m\)

Fc = 3067.5 N

Calculate the apparent weight:

The apparent weight (Wa) is the sum of the actual weight (W) and the centripetal force (Fc).

Wa = W + Fc

Wa = 597.8 N + 3067.5 N

Wa = 3665.3 N

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Answer:

1º all

2º 1

3º 2

4º the same

Explanation:

Answer:

2

Explanation:

To find force it's force = mass times acceleration so to find mass you would divide force by acceleration

Answer:

Potential difference, V = -0.000224 Volts

Explanation:

Given the following data;

Quantity of charge = 6.93*10-4 C

Energy = -3.09 J

To find the electric potential V at that point?

Mathematically, the energy transferred per unit of charge is given by the formula;

E = QV

Where:

E is the energy

Q is the quantity of charge

V is the potential difference.

Substituting into the formula, we have;

\( -3.09 = 6.93*10^{-4} * V \)

\( V = \frac {6.93*10^{-4}}{-3.09} \)

Potential difference, V = -0.000224 Volts

Answer:

The correct answer is: Energy is conserved, and it cannot be created or destroyed. This is known as the law of conservation of energy, which states that in a closed system, the total amount of energy remains constant and cannot be created or destroyed, only transformed from one form to another. This means that energy can be converted from one form to another, such as from potential energy to kinetic energy, but the total amount of energy in the system remains the same.

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

=  B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

An object appears white because it reflects all wavelengths of visible light . Option A is correct.

Ultraviolet rays are a kind of electromagnetic radiation that extends from the visible light spectrum to the X-ray area.

A white item reflects white light in all hues equally. We see the hue that an item does not absorb if it absorbs all except one. Red, orange, green, blue, indigo, and violet light are absorbed by the yellow strip in the diagram.

Harmful rays have a long-term effect on humans, causing skin and eye problems. It is also a source of vitamin C.

An object appears white because it reflects all wavelengths of visible light .

Hence, the correct answer for the blanks is reflects all wavelengths of visible light .

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The friction which acts on the object that is not moving is static friction, so, option A is correct.

Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.

Static friction is 0 for an object at rest on a level surface. Static friction generates an equal and opposite force that holds the book at rest if you push horizontally with a small force.

The static friction force grows to meet the force as you push harder. The book eventually moves when the maximum static friction force is reached.

Thus, the friction is static.

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the velocity of the wave is 400m/s

The formula for the velocity of the wave is, V = w/k

where ,  w is the coefficient of t and k is the coefficient of x

now putting values we get, v = 200/0.5 = 400

Hence the velocity of the wave is 400 m/s

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Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Answer:

Ok, sabemos que la velocidad inicial de la pelota es 15m/s.

Desconocemos la posición inicial a la que es lanzada la pelota, pero vamos a suponer que es a una altura igual a cero, es decir, la pelota es lanzada al ras del suelo.

Una vez lanzada, la única fuerza actuando en la pelota es la gravitatoria, entonces la aceleración de la pelota es:

a = -g = -9.8m/s^2

El signo negativo es por que esta aceleración apunta hacia abajo.

Ahora, para la velocidad, necesitamos integrar sobre el tiempo.

v(t) = (-9.8m/s^2)*t + v0

donde v0 = 15m/s

v(t) = (-9.8m/s^2)*t + 15m/s.

De aca podemos obtener el tiempo en el que la pelota llega a la altura máxima, que es el punto donde la velocidad es igual a cero.

0 = (-9.8m/s^2)*t + 15m/s.

t = (15/9.8)s = 1.53 s

Ahora, para la ecuación de la posición integramos la ecuación de la velocidad sobre el tiempo:

p(t) = (1/2)(-9.8m/s^2)*t^2 + 15m/s*t + p0

donde p0 es la pocision inicial, pero arriba dijimos que era igual a cero, entonces la ecuación queda:

p(t) = (-4.5m/s^2)*t^2 + 15m/s*t

ahora reemplazamos t por el tiempo que encontramos antes, y descubrimos que:

p(1.53s) =  (-4.5m/s^2)*(1.53s)^2 + 15m/s*1.53s = 12.41m

La máxima altura que alcanza la pelota es 12.41 metros arriba del punto desde el que se la lanzo.

Ahora, el tiempo total que esta en el aire puede ser calculado de tal forma que la posición vuelva a ser cero, es decir, la pelota llega a la misma altura desde la que fue lanzada inicialmente (y es agarrada por la persona, podemos suponer)

Entonces:

p(t) = 0 =  (-4.5m/s^2)*t^2 + 15m/s*t

Ahora resolvemos la eq cuadrática, usando la eq. de Bhaskara:

\(t = \frac{-15 +- \sqrt{15^2 - 4*(-4.5)*0} }{-2*4.5} = \frac{-15 +-15}{-9.8}\)

Entonces las soluciones son:

t = (-15 + 15)/-9.8 = 0s

t = (-15 - 15)/-9.8 = 3.06s

Tomamos la segunda solución, ya que la primera corresponde al tiempo inicial.

Entonces concluimos con que la pelota estuvo 3.06 segundos en el aire.

The velocity is 6.75

The velocity in the equation stated above can be calculated as follows

m= 2,000
p= 2.25

y= 6000

velocity= 2.25 × 6000/ 2000
= 13500/2000
= 6.75

Hence the velocity is 6.75

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Answer:

See below ~

Explanation:

Part (a) :

We can say a body is in uniform acceleration if the acceleration of the object remains constant with respect to time throughout its motion.

Part (b) :

We can say a body is non-uniform acceleration if the acceleration of the body varies with respect to time throughout its motion.

Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

Explanation:

Answer:

Physics is a natural science that involves the study of matter and its motion through space and time, along with related concepts such as energy and force.

An aptitude test is a systematic means of testing a job candidate's abilities to perform specific tasks and react to a range of different situations. The tests each have a standardised method of administration and scoring. The results are quantified and compared with all other test takers.

Answer:

Hi There!

Explanation:

An aptitude test is a systematic means of testing a job candidate’s abilities to perform specific tasks and react to a range of different situations. The tests each have a standardised method of administration and scoring, with the results quantified and compared with all other test takers.

~Your Welcome!

Answer:

multiply the number of moles of Na by the conversion factor 6.02214179×1023 atoms Na/ 1 mol Na, with 6.02214179×1023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.

Answer:

the answer is

6.022×10²³ na...

One of the most significant observations that has provided the most accurate values of the age, dark matter, dark energy content, and the average density of the universe is the cosmic microwave background (CMB) radiation.

The CMB radiation is the afterglow of the Big Bang and is a remnant of the hot, dense early universe. The CMB radiation provides a snapshot of the universe when it was only 380,000 years old, and its properties can be analyzed to infer the universe's current state.

By analyzing the CMB radiation, cosmologists have determined that the universe is approximately 13.8 billion years old. Furthermore, they have found that dark matter constitutes around 27% of the universe's total energy density, and dark energy constitutes around 68%.

The CMB radiation has also provided insight into the universe's average density. By measuring tiny fluctuations in the CMB, scientists have determined that the average density of the universe is very close to the critical density required for a flat universe. This result is consistent with the inflationary Big Bang model and the concept of a flat universe.

Therefore, the observation of the cosmic microwave background radiation has been crucial in providing some of the most accurate values of the age, dark matter, and dark energy content, and the average density of the universe.

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Answer:

S=29.298m/s

Explanation:

Use S=d/t

So, S=120m/4.1s= 29.298m/s