A sample of gas initially has a volume of 859 ml at 565 k and 2.20 atm. What pressure will the sample have if the volume changes to 268 ml while the temperature is increased to 815 k?

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.

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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Answer: B

i tried putting explanation but its not working

Answer:

B. [tex]2.8 \times 10^{-5} C[/tex]

Explanation:

As we know that the electric field due to Van de graff generator is same as that of a point charge

so it is given by

[tex]E = \frac{kQ}{r^2}[/tex]

here we know that

[tex]E = 4.5 \times 10^5 N/c[/tex]

also we know that

[tex]r = 0.75 m[/tex]

now from above formula we have

[tex]4.5 \times 10^5 = \frac{(9\times 10^9)(Q)}{(0.75)^2}[/tex]

here we will have

[tex]Q = 2.8 \times 10^{-5} C[/tex]

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.

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.

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:

(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:

The 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)

(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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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.

Answer: d

Explanation:

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.

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

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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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:

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.  

Learn more:

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]

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:

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.

Hello There!

Let's first talk about "What Is Friction"

Friction is a force that pulls when two object touch each-other. Friction happens because the molecules on one surface interlock with the molecules on another surface.

Now, let's get back to our original question "What Effect Does Friction Have On A Roller Coaster"

On a roller coaster, friction is a force that opposes motion and significantly slows the cars as they move on the track.

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:

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

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.

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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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.

A) Into a region of higher potential

Explanation:

Let's remind that:

- Like charges repel each other

- Unlike charges attract each other

Here we have a proton, which is a positive charge, which is brought to rest by an electric field. This means that the electric field has slowed down the proton: so, the force exerted by the electric field on the proton was opposite to the direction of motion of the proton. But the lines of an electric field go from points at higher potential to points at lower potential - this means that the proton was actually moving towards a point at higher potential. (for example, it was moving towards another positive charge source of the field, so the potential increases as the proton approaches the source charge).

B) 3,338 V

The initial kinetic energy of the proton is given by:

[tex]K_i = \frac{1}{2}mv^2[/tex]

where

[tex]m=1.67\cdot 10^{-27} kg[/tex] is the proton mass

[tex]v=800,000 m/s=8\cdot 10^5 m/s[/tex] is the initial speed

Substituting,

[tex]K_i = \frac{1}{2}(1.67\cdot 10^{-27}kg)(8\cdot 10^5 m/s)^2=5.34\cdot 10^{-16}J[/tex]

When the proton is brought to rest, all this energy is converted into electric potential energy, given by

[tex]\Delta U = q \Delta V[/tex]

where

[tex]q=1.6\cdot 10^{-19} C[/tex] is the proton charge

[tex]\Delta V[/tex] is the potential difference

Since [tex]\Delta U = K_i[/tex], we can solve to find the potential difference:

[tex]\Delta V=\frac{K_i}{q}=\frac{5.34\cdot 10^{-16} J}{1.6\cdot 10^{-19} C}=3,338 V[/tex]

C) 3,338 eV

We already found the initial kinetic energy of the proton in part B), and it is given by

[tex]K_i =5.34\cdot 10^{-16}J[/tex]

Now we want to convert it into electron volts; keeping in mind the conversion factor between eV and Joules,

[tex]1 eV = 1.6\cdot 10^{-19}J[/tex]

we find:

[tex]K_i = \frac{5.34 \cdot 10^{-16} J}{1.6\cdot 10^{-19} J}=3,338 eV[/tex]

The proton moved into a region of higher potential. The potential difference that brought it to rest and its initial kinetic energy can be calculated using formulas and given values.

Part A: The proton moved into a region of higher potential. This is because the electric field does work on the proton to bring it to a stop, which indicates that the proton moved against the direction of the electric field and hence into a region of higher potential.

Part B: The potential difference that stopped the proton can be calculated using the formula ΔU = ΔK/e, where ΔK is the change in kinetic energy and e is the charge of the proton. Given that the initial speed of the proton is 800000 m/s (which implies a kinetic energy of ½ mv^2), and knowing that the charge of a proton is 1.6 x 10^-19 C, you can solve this equation to find ΔU.

Part C: The initial kinetic energy of the proton can be calculated using the formula K = ½ mv^2. Converting this to electron volts (eV) involves dividing by the charge of an electron (e), which is also 1.6 x 10^-19 C.

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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

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.

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.

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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Answer:

D. An image that is smaller than the object and is behind the mirror

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

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.

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.

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.

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.

So, the correct answer is option C blueshift.

Learn more about Blueshift here -

https://brainly.com/question/3294257

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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.

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.

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]