as explained by kinetic molecular theory, why does a gas take the shape and volume of its container?

a) Megan's average speed is 12.5 meters per second.

b) The forward force on the bike becomes greater than the backward force on the bike.

As according to Newton's Third Law of Motion, every action must have an equal and opposite reaction. This means that when one object exerts a force on a second object, The second object enforces an equal magnitude and opposite direction force on the first object.

a) Megan's average speed can be calculated by dividing the total distance by the time taken:

Average speed = Total distance / Time taken

Here, the total distance covered by Megan is 400 m and the time taken is 32 seconds.

Average speed = 400 m / 32 sec = 12.5 m/s

When Megan is travelling at a constant speed, the forward force on the bike must be equal and opposite to the backward force on the bike, as per Newton's Third Law of Motion. Therefore, the forward force and backward force on the bike are equal.

b) When Megan crouches down on the handles to be more streamlined, the air resistance on the bike reduces. This leads to a reduction in the backward force acting on the bike, while the forward force remains the same. Therefore, the forward force on the bike becomes greater than the backward force on the bike.

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The common rock that can be dissolved by water and weak acids is limestone. Hence, the correct option is (C) limestone.

Limestone is a sedimentary rock that is made up primarily of calcium carbonate (CaCO3) mineral. It is composed of calcite mineral and often contains fossils of marine organisms like shells, coral, and mollusks, making it an important rock type in the construction of buildings.Limestone is easily dissolved by water and weak acids. Because of its calcite content, it is very susceptible to acid rain's erosive effects. When dissolved in water or acid, limestone can be transformed into the chemical compound calcium bicarbonate. Calcium bicarbonate is a soluble material that can be transported by water.

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A planet whose mass is half of the mass of Earth and radius equal to the radius of the Earth, the acceleration due to gravity on the surface of the planet would be  19.6m/s².

Large, circular celestial objects that are neither stars nor their remnants are known as planets. The nebular hypothesis, which states that an interstellar cloud collapses out of a nebula to create a young protostar orbited by a protoplanetary disc, is currently the best theory for planet formation. The formula for the acceleration caused by gravity on earth's surface is,

g = GM/R

Now, Mp = M/2

Rp = R/2

The planet's gravitational acceleration is,

gp =  GMp/Rp

gp = (GM/2)/(R²/4) = 2g

gp = 2 × 9.8 = 19.6m/s²

Therefore, the acceleration due to gravity on the surface of the planet will be 19.6m/s².

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The output voltage of a 3.0000 V lithium cell in a digital wristwatch that draws 0.280 mA, with the cell's internal resistance of 2.15 ohms can be calculated using Ohm's Law.

Step 1: Convert the current to Amperes (A)
0.280 mA = 0.000280 A

Step 2: Calculate the voltage drop across the internal resistance
Voltage drop = Current (I) × Resistance (R)
Voltage drop = 0.000280 A × 2.15 ohms ≈ 0.000602 V

Step 3: Calculate the output voltage
Output voltage = Cell voltage - Voltage drop
Output voltage = 3.0000 V - 0.000602 V ≈ 2.9994 V

The output voltage of the lithium cell in the digital wristwatch is approximately 2.9994 V.

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Object A will have a negative charge when it is touched by a negatively charged, conducting sphere.

We need to first understand the properties of conductors and charges.

A conductor is a material that allows electric charges to flow freely through it. In other words, it has low resistance to electric current. Most metals, such as copper and aluminum, are good conductors.

An electric charge is a fundamental property of matter. It is an electrical property of the atomic particles (such as electrons and protons) that make up matter. Charges can be either positive or negative.

When a charged object touches a conductor, the charge spreads out evenly over the surface of the conductor. This is called electrostatic induction. If the conductor is neutral, it will become charged. If the conductor is already charged, the charge will distribute itself evenly over the conductor's surface.

This happens because charges of the same type repel each other and charges of the opposite type attract each other. When the charged object is pulled away, the charge on the conductor remains.

In this scenario, object A is a neutral conductor. When it is touched by a negatively charged sphere, the negative charge spreads out evenly over the surface of the conductor. This means that object A becomes negatively charged. When the sphere is pulled away, the negative charge on object A remains.

Therefore, the charge on object A is negative.

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The speed of the joined cars after the collision is most likely less than the speed of car b before the collision.

This is due to the principle of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act on it. In this case, the system is the two toy cars, and the collision is a closed system because no external forces are acting on the cars.

Before the collision, car b has a certain momentum, and car a has zero momentum since it is at rest. After the collision, the two cars stick together and move in the same direction as car b before the collision. The total momentum of the system remains constant, but now it is shared between the two cars. Since the total mass of the system has increased, the speed of the joined cars after the collision will be less than the speed of car b before the collision.

To calculate the exact speed of the joined cars after the collision, we would need to know the masses and initial velocities of the two cars. However, we can say with certainty that the speed will be less than the speed of car b before the collision due to the conservation of momentum.

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The angular acceleration is -1.33 revolutions per second.

The angular acceleration can be calculated using the following formula:
Angular acceleration (α) = (Final angular velocity - Initial angular velocity) / Time


Initial angular velocity = 1200 revolutions per minute (rpm) = 20 revolutions per second (rps)
Final angular velocity = 0 revolutions per second (rps)
Time = 15 seconds
Angular acceleration (α) = (0 - 20) / 15 = -1.33 revolutions per second squared (rps2)

Moreover, the temporal rate at which angular velocity changes is known as angular acceleration. The standard unit of measurement is radians per second per second. Therefore, = d d t. Rotational acceleration is another name for angular acceleration. A rigid body's points all share the rotating velocity and acceleration. Here, The rotation is in the clockwise direction and the angular acceleration is negative.

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The primary rationale for the technician's precautions about radiation exposure is to protect their health. Ionizing radiation can cause damage to the body's cells and the risk increases with increased exposure.

The technician wants to minimize their exposure to radiation to reduce their risk of developing cancer or any other health issues that may arise from over exposure to radiation. Even though the levels of radiation used in contemporary radiologic imaging are relatively low, it is important for the technician to take precautions to protect their health.

This includes standing behind protective shields, wearing protective clothing and avoiding long-term exposure to the radiation source.

Additionally, the technician should ensure that any equipment they use is regularly tested to ensure it is working properly and is providing the lowest possible dose of radiation. By following these safety precautions, the technician can ensure their safety and reduce their long-term risk of developing any health issues from radiation exposure.

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The sample size required to be 90% confident that the estimated mean has an error less than 0.02 is 5534.

The formula used to calculate the required sample size is given by:

n = ((zα/2 × σ) / E)²

where:

n = sample size

zα/2 = z-value for the level of confidence (α/2

)σ = population standard deviation

E = maximum error

Population standard deviation, σ = 6.8

Maximum error, E = 0.02

Confidence level, α = 0.9

Therefore, α/2 = 0.45 (since the confidence interval is symmetric)

The z-value for 0.45 level of confidence is 1.645.

Thus:

n = ((1.645 × 6.8) / 0.02)²

n = (11.066 / 0.02)²

n = 5533.0256

Rounding up, the required sample size is n = 5534.

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The average force exerted by the oxygen molecule on one of the walls of the container is 8.5 x 10^-20 N. This value is calculated using the formula   [tex]F = (2mv^2)/(Δt * d) where Δt = 5.0 x 10^-4 s[/tex] and d = 0.10 m.

When an oxygen molecule bounces back and forth between opposite sides of a cubical vessel, it exerts a force on each of the walls it strikes. To calculate the average force exerted by the molecule on one of the walls, we use the formula [tex]F = (2mv^2)/(Δt * d)[/tex], where m is the mass of the molecule, v is its velocity, Δt is the time it takes for the molecule to travel from one wall to the opposite wall and back, and d is the distance between the two opposite walls.

Substituting the given values, we get [tex]F = 8.5 x 10^-20 N[/tex]. This is a very small force, which is expected for a single oxygen molecule in a container. However, for a large number of molecules, the total force exerted on the walls of the container can be significant, leading to the pressure inside the container. This principle is used in many industrial and everyday applications, such as gas storage, refrigeration, and air conditioning.

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The car would have been moving too fast for a safe stop and would have been moving at a speed even higher if it could have produced a skid mark of 65 mph with a friction coefficient of 0.25.

A skid mark is measured from the beginning of one, which may start out light if two tyres lock and gradually darken as more tyres lock.

Just dividing the vertical load by the horizontal force yields the coefficient of friction. For each section, the high, low, and average SN values are presented. The test speed has an impact on SN results; the slower the speed, the higher the SN.

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The force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This is known as the inverse-square law of gravitational attraction. Mathematically, this can be expressed as F = G(m1m2)/d^2, where F is the force of attraction between the two objects, G is the gravitational constant, m1, and m2 are the masses of the two objects, and d is the distance between their centers. Therefore, the greater the masses of the objects, the stronger the force of attraction between them. Similarly, the farther apart the objects are, the weaker the force of attraction between them. This fundamental relationship is what governs the behavior of celestial bodies in space and is crucial for understanding many natural phenomena in our universe.

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In the wave, the particles of the medium are closer together in B. compressions.

Particles in a compressional wave compress and expand in the direction of wave propagation. In a compressional wave, when a force is exerted on one end of the chain of particles, it is transferred from one particle to the next, resulting in a disturbance in the medium that spreads outward. As a result, compressions and rarefactions travel through the medium, causing the particles to be either close together or far apart.

In a compressional wave, the particles in the medium vibrate parallel to the direction of wave propagation. This is due to the fact that the direction of particle movement in a compressional wave is in the same direction as the wave's motion. The particles in the medium are closer together in compressions, whereas they are further apart in rarefactions. The reason for this is that the particles in a medium are pushed together by the wave's pressure and then pushed apart as the wave continues. Therefore, the correct option is B.

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When a sound gets louder, the following factors change:

a. Intensity

d. Decibel level

f. Amplitude

When a pitch gets higher, the following factors change:

c. Frequency

e. Wavelength

Sound is a type of energy that is transmitted through the vibration of the particles in a medium. When sound is produced, it has characteristics such as frequency, intensity, and amplitude. Changes in these characteristics can result in a different perception of sound by our ears.

When a sound gets louder, the intensity, decibel level, and amplitude of the sound waves increase. Intensity is the amount of energy per unit area that a sound wave carries while the amplitude is the maximum displacement of particles from their equilibrium position. The decibel level also increases with an increase in sound intensity.

When the pitch of a sound gets higher, the frequency of the sound waves increases. Frequency is the number of complete cycles of a wave that occur per second. The wavelength of the sound waves also decreases with an increase in frequency. The speed of sound waves does not change when the pitch of a sound gets higher.

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A softball weighs 0.18427 kg. Its average rebound height on concrete, grass, and wood is 80.8 cm, 75.3 cm, or 87.6 cm, respectively. Kinetic energy 81 multiplied by a speed of 30 m/s and mass equals 2430.

The amount of matter inside an item is referred to as mass in mathematics. The most common way to determine mass is to weigh something. Anything will weigh more the more matter it contains. For instance, a mouse will have a higher mass than an ant since it contains more stuff.

The quantity of substance contained within an item is expressed in terms of mass. Typically, mass is expressed in kilogrammes (kg) or grammes (g) (kg). No matter where in the cosmos it is or how much gravitational force is exerted on it, mass is a measure of how much matter there is.

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Answer:

The coefficient of cubical expansion of a solid is defined as the increase in volume of a solid per unit volume per degree Celsius rise in temperature. It is denoted by the symbol α.

Mathematically, we can express the coefficient of cubical expansion as:

α = (1/V) x (dV/dT)

where V is the volume of the solid and dV/dT is the rate of change of volume with respect to temperature.

The variation of density with temperature can be expressed using the coefficient of cubical expansion as follows:

ρ = m/V, where ρ is the density of the solid and m is its mass.

Differentiating this expression with respect to temperature, we get:

dρ/dT = (1/V) x (dm/dT) - (m/V^2) x (dV/dT)

Using the relationship between the coefficient of cubical expansion and the rate of change of volume with respect to temperature, we can substitute (dV/dT) = αV into the above expression to obtain:

dρ/dT = (1/V) x (dm/dT) - αρ

This equation shows that the variation of density with temperature is directly proportional to the coefficient of cubical expansion of the solid.

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The outer Jovian planets are larger and mostly gaseous because they formed beyond the frost line, where water and methane condensed into large icy planetesimals that attracted the abundant light gases.

The outer planets are composed mainly of hydrogen and helium, which are the most abundant elements in the universe. These elements were able to condense into solid form beyond the frost line, where the temperature was low enough for them to freeze. The solid cores of the outer planets were able to attract and hold onto the abundant light gases, which formed the thick atmospheres of the outer planets.

In contrast, the inner terrestrial planets are smaller and denser because they formed inside the frost line, where the temperature was too high for hydrogen and helium to condense into solid form. Instead, the inner planets are composed mainly of rock and metal, which are denser than hydrogen and helium and were able to condense into solid form inside the frost line.

The Jovian planets are the outer planets of our solar system: Jupiter, Saturn, Uranus, and Neptune. These planets are also called “gas giants” or “giant planets” because they are much larger than the inner, rocky planets and are composed mainly of hydrogen and helium, which are the most abundant elements in the universe. The Jovian planets are named after their mythological namesake Jupiter, which is also known as Jove.

The Jovian planets have several characteristics in common, including their large size, low density, and thick atmospheres. They also have many moons and are surrounded by rings. The Jovian planets are located beyond the asteroid belt, which is the region of the solar system between Mars and Jupiter.

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Wind speed and Air temperature considerations are used to calculate a windchill factor. Option a and option b are correct.

The considerations used to calculate a windchill factor are,

Wind speed: The faster the wind blows, the faster the body loses heat, and therefore, the colder the air feels.

Air temperature: The lower the air temperature, the colder the air feels, and the greater the effect of the wind on the body.

Therefore, the windchill factor takes into account the air temperature and wind speed to calculate how cold the air feels on exposed skin. Other factors such as humidity and sun angle can also affect the windchill factor. However, air pressure, wind direction, and atmospheric heating are not directly used in the calculation of the windchill factor. Hence, option a and b are correct.

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Answer:
The bullet accuracy is influenced by a combination of factors, including bullet spin, shape, design, muzzle velocity, air resistance, and the shooter's skill and technique.

Explanation:
Several physics factors contribute to the accuracy of a fired bullet, but one of the most important factors is the bullet's stability in flight. This stability is primarily determined by the following aspects:

Bullet spin: When a bullet is fired, it is spun by the rifling (spiral grooves) inside the barrel of the gun. This spin imparts a gyroscopic stability to the bullet, helping it maintain a stable and consistent trajectory in flight.

Bullet shape and design: The shape and design of the bullet also play a significant role in its accuracy. Aerodynamic bullets with a streamlined shape are more stable in flight, as they reduce air resistance and minimize the effects of crosswinds.

Muzzle velocity: The speed at which the bullet leaves the barrel, known as muzzle velocity, can also impact accuracy. Higher muzzle velocities generally lead to flatter trajectories, which can make it easier to hit targets at longer distances. However, too high a muzzle velocity may cause instability and reduced accuracy.

Air resistance and drag: As the bullet travels through the air, it experiences air resistance and drag, which can cause it to slow down, lose stability, and deviate from its intended path. Factors such as altitude, humidity, and wind can influence air resistance and, consequently, bullet accuracy.

Shooter's skill and technique: The skill and technique of the person firing the bullet also play a crucial role in the accuracy of a fired bullet. Proper trigger control, sight alignment, and breath control can significantly impact the accuracy of the shot.

The physics factor that contributes to the accuracy of a fired bullet is its stability in flight, which depends on several factors including its velocity, spin, and shape.

Velocity and spin are two different concepts that are related to the motion of an object.

Velocity refers to the rate at which an object changes its position in a particular direction over time. In other words, velocity is a vector quantity that has both magnitude (speed) and direction, and it is often expressed in units of meters per second (m/s) or kilometers per hour (km/h). Velocity can be positive or negative, depending on the direction of the object's motion.

Spin, on the other hand, refers to the rotation of an object around its axis. When an object spins, it rotates around a fixed point or axis, and it can have a variety of different spin rates or angular velocities. Spin is often measured in units of revolutions per minute (RPM) or radians per second (rad/s).

In some cases, velocity and spin can be related. For example, in sports like baseball, the spin of a ball can affect its trajectory and the way it moves through the air. A baseball pitcher can throw a pitch with backspin or topspin, which will cause the ball to curve or drop in a particular way as it travels towards the batter. In this case, the spin of the ball can be used to control its velocity and direction, and to deceive the opposing team.

Here in the Question,

The velocity of the bullet, or its speed, is important for accuracy because it determines the distance that the bullet will travel in a given amount of time. A faster bullet will cover more distance in a shorter amount of time, which can make it more difficult to aim accurately. On the other hand, a bullet that is too slow may not have enough momentum to travel a long distance or penetrate a target effectively.

The spin of the bullet is also crucial for stability in flight. Most bullets are designed with a spiral groove pattern on the inside of the barrel, which causes the bullet to spin as it travels through the air. This spin stabilizes the bullet by causing it to rotate around its longitudinal axis, which helps to counteract any irregularities or disturbances in the air. A bullet that is not spinning or is spinning too slowly can be more prone to wobbling or tumbling in flight, which can reduce accuracy.

Finally, the shape of the bullet can also affect its stability in flight. Bullets that are streamlined and have a high ballistic coefficient, or resistance to air resistance, will maintain their velocity and trajectory better than bullets with a more irregular shape. Additionally, a bullet that is too long or too short for its caliber can be unstable in flight and can affect accuracy.

Therefore, the accuracy of a fired bullet depends on a combination of factors, including its velocity, spin, and shape, which must be carefully optimized to ensure consistent and reliable performance.

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Each rocket will need to provide a steady force of 6.98 N to accelerate the satellite to 25 rpm in 6.5 minutes.

We can use the following equation to solve this problem,

ω = (1/2) α t^2

where,

ω = final angular velocity (25 rpm)

α = angular acceleration

t = time taken to reach final velocity (6.5 minutes = 390 seconds)

Since the satellite starts from rest, the initial angular velocity is 0. Thus, rearranging the equation,

α = (2ω) / t^2

α = (2 x 25 rpm x 2π rad/rpm) / (390 s)^2

α = 0.0008727 rad/s^2

Each rocket will need to provide a torque to generate this angular acceleration. The required torque is given by,

τ = Iα

where,

τ = torque required

I = moment of inertia of the satellite

The moment of inertia of a solid sphere is given by,

I = (2/5) mr^2

where, m = mass of the satellite and r = radius of the satellite.

Let's assume the satellite has a mass of 1000 kg and a radius of 2 meters.

I = (2/5) x 1000 kg x (2 m)^2

I = 8000 kg m^2

Therefore, the torque required by each rocket is,

τ = Iα

τ = 8000 kg m^2 x 0.0008727 rad/s^2

τ = 6.98 Nm

Since there are two rockets, each rocket will need to provide half of this torque, or,

F = τ / r

where, F = force required by each rocket and r = distance from the center of the satellite to the rocket (let's assume this is 1 meter)

F = 6.98 Nm / 1 m

F = 6.98 N

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My diagnosis for the left knee is that the patient likely has a torn or injured MCL. The sonogram showed thickening and irregularity of the MCL, indicating an injury.

This data means that there is damage to the MCL, which could lead to instability of the knee joint and ongoing pain and swelling.

When performing an ultrasound examination of the knee, the sonographer places the transducer on the skin over the area of interest, in this case, the medial aspect of the knee joint. The transducer sends out high-frequency sound waves that travel through the skin and soft tissues of the knee. When these sound waves encounter a boundary between tissues of different densities, some of the waves are reflected back to the transducer, while others are transmitted further into the body.

The reflected sound waves are detected by the transducer, which converts them into electrical signals that are processed by a computer to create an image of the knee. The image shows the different structures within the knee joint, including the MCL.

To diagnose an injury to the MCL, the sonographer will look for changes in the appearance of the ligament on the sonogram. The MCL appears as a hyperechoic band of tissue that extends from the femur to the tibia. If the MCL is torn or injured, it may appear thicker or thinner than normal or have an irregular shape.

In conclusion, my diagnosis for the left knee is that the patient likely has a torn or injured MCL. The sonogram showed thickening and irregularity of the MCL, indicating an injury.

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The magnetic force experienced by the particle when it enters a magnetic field of 2.2 T is given by:

F = qVBsinθ

where q is the charge of the particle, V is its velocity, B is the magnetic field, and θ is the angle between the velocity vector and the magnetic field vector. Since the angle is not given, we can assume it to be 90 degrees. So, sinθ = 1. Hence, the formula reduces to:

F = qVB

Now, we need to find the radius of the circular path that the particle traverses. The magnetic force provides the necessary centripetal force for circular motion.

Hence, we can equate the two forces as follows:

F = mv^2/r

where m is the mass of the particle and v is its velocity. Since the velocity of the particle is not given, we can assume it to be constant.

Hence, we can combine the two formulas as follows:qVB = mv^2/rSolving for r, we get r = mv/qBSubstituting the given values, we get:

r = (1.6 × 10^-19 C)(5 × 10^7 m/s)/(1.6 × 10^-31 kg)(2.2 T) = 0.665 m

Therefore, the radius of the circular path that the particle traverses are 0.665 meters.

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The work done by gravity on the block is 346.4 Joules.

The work done by gravity on an object is given by the formula,

W = Fd cos(theta)

where, W = work done, F = force applied, d = distance moved in the direction of the force, theta = angle between the force and the direction of motion. In this case, the force applied by gravity is the weight of the block, which is given by,

F = mg

where, m = mass of the block and g = acceleration due to gravity (9.8 m/s^2).

Substituting the given values,

F = 80 N

m = F/g = 80 N / 9.8 m/s^2 = 8.16 kg

The angle of the incline is 30 degrees, which means the angle between the force of gravity and the direction of motion is also 30 degrees. The distance moved by the block in the direction of the force is the length of the incline, which is given as 5m.

Therefore, the work done by gravity on the block is,

W = Fd cos(theta) = (80 N) x (5 m) x cos(30) = 346.4 J

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The initial similarity between the surface of Mercury and the Moon is primarily due to the existence of numerous impact craters caused by meteoroid and asteroid impacts.

An initial glance at the surface of Mercury indicates that it is somewhat similar to the surface of the Moon. This initial similarity is primarily due to the existence of impact craters, which are formed when a meteoroid or asteroid collides with the surface. Both the Moon and Mercury lack significant atmospheres and have solid, rocky surfaces that are geologically inactive, which means that craters are not erased by erosion or tectonic processes. The craters on both surfaces are also similar in appearance, with raised rims, flat floors, and central peaks. Additionally, both surfaces have extensive plains that were formed by volcanic activity, although the plains on Mercury are thought to be younger and more extensive than those on the Moon. Overall, the similarity between the surfaces of Mercury and the Moon is primarily due to their shared lack of geological activity and the prevalence of impact craters.

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Answer:

a) A photon with energy greater than or equal to the work function of the metal will cause photoemission.

From the given data, the work function of the metal is 1.7 eV. Therefore, photons with energies of 2.0 eV and 4.0 eV will cause photoemission, but the photon with an energy of 1.0 eV will not.

b) The maximum kinetic energy (KEmax) of a liberated electron is given by:

KEmax = E(photon) - work function

where E(photon) is the energy of the incident photon.

For the photon with an energy of 2.0 eV:

KEmax = 2.0 eV - 1.7 eV

KEmax = 0.3 eV

In joules:

KEmax = (0.3 eV)(1.602 x 10^-19 J/eV)

KEmax = 4.81 x 10^-20 J

For the photon with an energy of 4.0 eV:

KEmax = 4.0 eV - 1.7 eV

KEmax = 2.3 eV

In joules:

KEmax = (2.3 eV)(1.602 x 10^-19 J/eV)

KEmax = 3.69 x 10^-19 J

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Kinetic energy just before it hits the ground will be equal to mgh by law of conservation of energy.

An apple hanging from a tree contains potential energy due to its elevated position. whenever an apple falls from a tree. The fruit contains kinetic energy, which is why it is falling. As a result, when the apple starts falling, the potential energy that was previously held in it at its raised position is converted into kinetic energy.

The height is always the vertical distance between the starting point and the lowest point of fall, not necessarily the entire distance the body may go. Almost all mechanical processes include the exchange of work and kinetic and potential energy.

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The specific heat capacity of the object is 1.145 J/(g°C)

The specific heat capacity of the object can be calculated using the formula:

specific heat capacity = (energy required)/(mass x change in temperature)

In this case, the energy required to increase the temperature of the object by 10.0 degrees Celsius is 1145 J, the mass of the object is 100.0 g, and the change in temperature is 10.0 degrees Celsius.

specific heat capacity = (1145 J)/(100.0 g x 10.0 °C)

specific heat capacity = 1.145 J/(g°C)

Therefore, the specific heat capacity of the object is 1.145 J/(g°C). This means that it takes 1.145 joules of energy to raise the temperature of 1 gram of the object by 1 degree Celsius.

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