a wisconsin boy is playing the traditional winter game of shoot the cheese. he shoots his pellet gun at a piece of cheese that sits on a massive block of ice. on one particular shot, his 1.2 g pellet gets stuck in the cheese, causing the cheese to slide 25 cm before coming to a stop. if the muzzle velocity of the gun is 65 m/s and the cheese has a mass of 120 g, what is the coefficient of friction between the cheese and ice?

Answer:

Fitness Level

Explanation:

The most likely difference between the two joggers who have similar levels of commitment and purpose but one is clearly better at jogging than the other is their fitness level. Fitness level refers to the level of physical fitness and conditioning of an individual, which is usually determined by their exercise habits, lifestyle, and genetics. A person with a higher level of fitness will generally be able to perform physical activities, such as jogging, more easily and with greater proficiency than a person with a lower level of fitness. Therefore, the jogger who is better at jogging is likely to have a higher level of fitness than the other jogger.

While factors such as body composition, motivation, and body positivity can also impact jogging performance, they are less likely to be the most significant difference between the two joggers in this scenario.

Answer:

The most likely difference between the two joggers is

B. body composition.

Body composition refers to the ratio of fat mass to lean mass in the body, and it plays an important role in athletic performance. A jogger with a lower percentage of body fat and a higher percentage of lean muscle mass is likely to have better endurance and speed than a jogger with a higher percentage of body fat and lower muscle mass, even if both have similar levels of commitment and motivation. Fitness goals, motivation, and body positivity can also play a role in jogging performance, but body composition is the most likely factor in this scenario.

The correct option is B: Longitudinal, because the waves travel away from the source, parallel to the movement of the source.

Sound waves are mechanical waves that require a medium to travel through, such as air, water, or solids. These waves are characterized by their frequency, wavelength, amplitude, and speed.

Sound waves are longitudinal waves because the particles of the medium vibrate in the same direction as the wave travels. In other words, the wave compresses and rarefies the medium in the same direction as the wave propagation. This means that the particles of the medium move parallel to the direction of the wave propagation.

Therefore, option B is the correct option as it correctly explains that sound waves are longitudinal and travel away from the source parallel to the movement of the source.

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The rise in temperature of the iron part of the machine is approximately 9.6°C.

Heat transfer is the movement of thermal energy from one object or system to another due to a temperature difference. Thermal energy, which is the energy associated with the motion of particles within a substance, always moves from higher temperature regions to lower temperature regions until thermal equilibrium is achieved.

To calculate the rise in temperature of the iron part of the machine, we can use the following formula:

Q = m × c × ΔT

where Q is the heat absorbed by the iron part of the machine, m is the mass of the iron part, c is the specific heat capacity of iron, and ΔT is the rise in temperature.

First, we need to calculate the heat generated by the machine in 2 minutes, using its input power and efficiency:

P = 2 kW

Efficiency = 80% = 0.8

t = 2 minutes = 120 seconds

Energy generated by the machine = P × t × efficiency

Energy generated by the machine = 2 kW × 120 s × 0.8

Energy generated by the machine = 192 kJ

Next, we can calculate the mass of the iron part of the machine using its density:

Density of iron = 7.87 g/cm³ = 7870 kg/m³

Mass of iron part = 40 kg

Now, we can substitute the values into the formula and solve for ΔT:

Q = m × c × ΔT

ΔT = Q / (m × c)

ΔT = 192000 J / (40 kg × 500 J/kgK)

ΔT = 9.6°C

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The given options to the question must also include 9.6°C.

The representative elements are those with unfilled energy levels in which the "last electron" was added to an s or p orbital. Therefore the correct option is option C.

The "last electron" in an atom refers to the outermost electron that is not part of a filled electron shell. This electron is also called the valence electron, and it is the electron that is most likely to participate in chemical reactions and bond formation with other atoms.

The properties of the valence electron largely determine the chemical behavior and reactivity of an element.

This includes elements in groups 1, 2, and 13-18 of the periodic table. The electrons in these elements occupy the outermost s and p orbitals, which are collectively known as the valence shell.

These valence electrons are responsible for the chemical properties of the elements and their reactivity. Therefore the correct option is option C.

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The equation used to calculate elastic potential energy is:

Elastic potential energy = 1/2 * k * x^2

Elastic potential energy is the energy stored in an object when it is stretched or compressed. It is dependent on the spring constant and the displacement of the object from its equilibrium position. The equation to calculate elastic potential energy is 1/2 * k * x^2, where k is the spring constant and x is the displacement from the equilibrium position. To calculate the elastic potential energy using software, you need to input the values of k and x into the equation, and the software will calculate the value.

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The magnitude of the radial component of acceleration is 1.25 m/s^2.

To solve this problem, we can use the formula for radial acceleration:

a_r = r * (Δω / Δt)

where a_r is the radial acceleration, r is the distance from the axis of rotation, Δω is the change in angular velocity, and Δt is the time interval over which the change occurs.

Plugging in the given values, we get:

a_r = 0.75 m * ((10 rad/s - 15 rad/s) / 3 s)

a_r = 0.75 m * (-1.67 rad/s^2)

a_r = -1.25 m/s^2

Note that the negative sign indicates that the acceleration is directed inward, towards the axis of rotation. So the magnitude of the radial component of acceleration is:

|a_r| = 1.25 m/s^2

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--The complete question is, What is the magnitude of the radial component of acceleration of a point located at a distance of 0.75 meters from the axis of rotation of a centrifuge, just as it begins its deceleration from an angular velocity of 15 rad/s to 10 rad/s over a time period of 3 seconds?--

Answer:

Force of friction = 3.33 N

Explanation:

The distance the bag slides can be calculated using the velocity and time

d = vt

d = 4m/s(3s)

d = 12 m

W = Fd and W = ∆KE=1/2mv^2

[tex]\frac{1}{2} mv^2=Fd\\\\F=\frac{mv^2}{2d}\\ \\F=\frac{(5kg)(4m/s)^2}{2(12m)} \\\\F=3.333 N[/tex]

The average force of friction acting on the 5-kg bag of groceries tossed across the surface of a table at 4 m/s and slides to a stop in 3 s is 16.7 N.

What is friction?

The resistance that a surface or object encounters when it comes into touch with another object or surface that has a different motion is known as friction. Friction happens when two objects slide against one another. Friction is the resistance that opposes motion. For instance, when a car accelerates, the friction between the road and the tires opposes the car's motion, and the car accelerates more slowly.

The following equation is used to compute the force of friction:

F_f = μF_n

Where F_f is the force of friction,

μ is the coefficient of friction,

and F_n is the normal force.

It's worth noting that the force of friction is proportional to the force holding two items together and the type of material on the surfaces in contact. The coefficient of friction is a measure of the force of friction between two objects. The unit of coefficient of friction is N (Newton).

How can you calculate the average force of friction?

We can use the formula, f = m x a to calculate the force of friction, where 'm' is the mass of the object and 'a' is the acceleration due to friction.

The formula can also be written as F_f = μF_n.

Given that the mass of the bag is 5-kg, the initial velocity of the bag is 4 m/s, and the time taken for the bag to come to a stop is 3s.

Then we can calculate the acceleration using the formula,

a = (v - u)/t, where 'v' is the final velocity, 'u' is the initial velocity and 't' is the time taken.

a = (0-4)/3 = -4/3 m/s^2.

We can now calculate the force of friction using the formula,

f = m x a. f = 5 kg x (-4/3 m/s^2) = -20/3 N.

However, the force of friction is negative since it acts in the opposite direction of the motion of the object.

Therefore, the average force of friction acting on the bag is 20/3 N or 6.67 N (rounded off to two decimal places).

The average force of friction acting on the 5-kg bag of groceries tossed across the surface of a table at 4 m/s and slides to a stop in 3 s is 16.7 N.

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Answer: It would take 5 seconds for an object traveling at 12 m/s to go 60 m. Rounded to the nearest whole number, the answer is 5 seconds.

Explanation:

To find the time it would take an object to travel a certain distance at a given speed, we can use the formula:

time = distance / speed

Plugging in the given values, we get:

time = 60 m / 12 m/s

time = 5 seconds

A battery with an emf of 25.10 v provides an appliance with a constant current of 2.50 ma, the work done by the battery in two minutes is 7.53 J.

The power P in watts can be calculated using the formula given below.P = V x Iwhere V is the voltage, and I is the current. In this scenario, V = 25.10 V and I = 2.50 mA = 2.50 x 10⁻³A = 0.0025A. Therefore, P = 25.10 V x 0.0025A = 0.06275 W.

The work W in joules (J) done by the battery in two minutes can be calculated using the formula given below.W = P x t where t is the time in seconds. In two minutes, the time t is 2 x 60 = 120 seconds.Therefore, W = 0.06275 W x 120 s = 7.53 J.

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Answer:

The energy stored in a cloud due to separation of charges that causes a lightning strike can be estimated using the equation:

E = (1/2) * C * V^2

where E is the energy stored, C is the capacitance of the cloud, and V is the potential difference between the cloud and the ground.

Assuming that the capacitance of the cloud is 10 microfarads and the potential difference between the cloud and the ground is 100 million volts, the energy stored in the cloud is:

E = (1/2) * 10^-5 F * (10^8 V)^2 = 5*10^13 J

Now, if 12.5 coulombs of charge is transferred from the cloud to the ground, the energy dissipated can be calculated as:

W = V * Q = V * (12.5 C)

where W is the work done, Q is the charge transferred, and V is the potential difference between the cloud and the ground during the lightning strike.

Assuming that the potential difference remains constant at 100 million volts, the work done or energy dissipated is:

W = (10^8 V) * (12.5 C) = 1.25 * 10^10 J

Therefore, the fraction of stored energy dissipated is:

Fraction = (energy dissipated) / (energy stored)

Fraction = (1.25 * 10^10 J) / (5*10^13 J)

Fraction = 0.00025 or 0.025%

Thus, only a very small fraction of the energy stored in the cloud is dissipated during a lightning strike.

As per the given statement, if 12.5 C of charge is transferred from the cloud to the ground in a lightning strike, the fraction of the stored energy that is dissipated is (25/2 * V1²) * (Q1 - 6.25) / Q1².

We know that the energy stored in a charged capacitor can be calculated using the formula:E = (1/2) * C * V²Where,E is the energy storedC is the capacitance of the capacitorV is the potential difference between the plates of the capacitorLet E1 be the initial energy stored in the cloud before the lightning strike.And E2 be the energy stored in the cloud after the lightning strike.From the law of conservation of energy, we know that the total energy of a closed system remains constant. Therefore,E1 = E2 + EdWhere Ed is the energy dissipated during the lightning strike.Let the capacitance of the cloud be C.So, the initial energy stored in the cloud can be calculated as:E1 = (1/2) * C * V1²Similarly, the final energy stored in the cloud after the lightning strike can be calculated as:E2 = (1/2) * C * V2²And the energy dissipated can be calculated as:Ed = E1 - E2Therefore,Ed = (1/2) * C * (V1² - V2²)But we know that,Charge Q = C * VTherefore,The initial charge stored in the cloud can be calculated as:Q1 = C * V1And the final charge stored in the cloud can be calculated as:Q2 = C * V2Now, let's consider the given statement:"12.5 C of charge is transferred from the cloud to the ground in a lightning strike".So, the final charge stored in the cloud can be written as:Q2 = Q1 - 12.5We need to find the fraction of energy dissipated.Using the above expressions for Ed and Q2, we get:Ed = (1/2) * C * [(Q1/C)² - ((Q1 - 12.5)/C)²]Ed = (1/2C) * [Q1² - (Q1 - 12.5)²]Ed = (1/2C) * [(Q1² - Q1² + 25Q1 - 156.25)]Ed = (25/2C) * (Q1 - 6.25)Ed/Q1 = (25/2C) * (1 - 6.25/Q1)Now, the fraction of energy dissipated can be obtained by using the expression:Ed/E1 = Ed/(1/2 * C * V1²)= (25/2C) * (1 - 6.25/Q1) / (1/2 * C * V1²)= (25/2 * V1²) * (1 - 6.25/Q1) / Q1= (25/2 * V1²) * (Q1 - 6.25) / Q1²Hence, the fraction of energy dissipated is (25/2 * V1²) * (Q1 - 6.25) / Q1².

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Moving a spring toy up and down creates a disturbance that is parallel to the wave motion. This type of disturbance is called a longitudinal wave.

In a longitudinal wave, the particles of the medium oscillate parallel to the direction of the wave motion. When you move a spring toy up and down, you create a series of compressions and rarefactions in the spring, where the coils are compressed together and then spread apart.

This creates a longitudinal wave that travels through the spring. Sound waves are also examples of longitudinal waves, where the compression and rarefaction of air particles create the wave motion.

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Answer:

The period (T) of a satellite in a circular orbit can be determined using the formula:

T = 2πr/v

where r is the radius of the orbit and v is the speed of the satellite.

We can rearrange this formula to solve for v:

v = 2πr/T

Substituting the given values, we get:

v = 2π(3.5 × 10^7 m)/(6.3 h × 3600 s/h) ≈ 5,099 m/s

The centripetal force (F) that keeps the satellite in orbit is given by:

F = mv^2/r

where m is the mass of the satellite.

The gravitational force (Fg) between the satellite and Roton is given by:

Fg = GmM/R^2

where G is the gravitational constant, M is the mass of Roton, and R is the radius of Roton.

Since the satellite is in a circular orbit, the centripetal force is equal to the gravitational force:

mv^2/r = GmM/R^2

Simplifying and solving for M, we get:

M = v^2r/GR^2

Substituting the given values, we get:

M = (5,099 m/s)^2 × 3.5 × 10^7 m/(6.674 × 10^-11 m^3/kg s^2) × (1.75 × 10^7 m)^2 ≈ 8.35 × 10^23 kg

The free-fall acceleration (g) on the surface of Roton is given by:

g = GM/R^2

Substituting the calculated value of M and the given value of R, we get:

g = (6.674 × 10^-11 m^3/kg s^2) × (8.35 × 10^23 kg)/(1.75 × 10^7 m)^2 ≈ 8.73 m/s^2

Therefore, the magnitude of the free-fall acceleration on the surface of Roton is approximately 8.73 m/s^2

1. The ratio of the masses of the two balls is 3(x + L/2) / (x - L/2), and

2. da, the distance from ball a to the system's center of mass, is (2Lma) / (3(ma + mb)).

To solve this problem, we can use the parallel axis theorem, which states that the moment of inertia of a system about an axis parallel to an axis through the center of mass is given by:

I = Icm + Md^2

where Icm is the moment of inertia of the system about an axis through the center of mass, M is the total mass of the system, and d is the distance between the two axes.

To find the ratio of the masses of the two balls, we can set up a system of equations using the parallel axis theorem.

Let ma and mb be the masses of balls a and b, respectively, and let x be the distance from ball a to the center of mass of the system. Then we have:

Ia = Icm + ma(x - L/2)^2

Ib = Icm + mb(x + L/2)^2

We are given that Ia / Ib = 3, so we can substitute Ia = 3Ib into the above equations and simplify:

3Ib = Icm + ma(x - L/2)^2

Ib = Icm + mb(x + L/2)^2

Dividing the first equation by the second equation, we get:

3 = (ma / mb) * ((x - L/2)^2 / (x + L/2)^2)

Simplifying, we get:

3 = (ma / mb) * ((x - L/2) / (x + L/2))^2

Taking the square root of both sides and rearranging, we get:

ma / mb = 3 * (x + L/2) / (x - L/2)

To find da, the distance from ball a to the system's center of mass, we can use the fact that the center of mass is located at a point that balances the torques about both axes.

Let xm be the distance from ball b to the center of mass. Then we have:

ma(x - L/2) = mb(xm + L/2)

ma(x - L/2)^2 = mb(xm + L/2)^2

Solving for xm in terms of x, we get:

xm = (ma / mb)(x - L/2) - L/2

The center of mass is located at a point that balances the torques about both axes, so we also have:

Ia(x - da) = Ib(xm - da)

Substituting xm in terms of x, we get:

Ia(x - da) = Ib[(ma / mb)(x - L/2) - L/2 - da]

Simplifying, we get:

(x - da) / [(ma / mb)(x - L/2) - L/2 - da] = Ib / Ia

Substituting Ia / Ib = 3, we get:

(x - da) / [(ma / mb)(x - L/2) - L/2 - da] = 1/3

Cross-multiplying and simplifying, we get:

da = (2Lma) / (3(ma + mb))

Therefore, the ratio of the masses of the two balls is 3(x + L/2) / (x - L/2), and da, the distance from ball a to the system's center of mass, is (2Lma) / (3(ma + mb)).

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if a constant force is applied to an object, causing the object to accelerate at 5.00 m/s2, the acceleration will remain at 5.00 m/s2 unless there is a change in the force applied or the object's mass.

This is due to Newton's Second Law of Motion, which states that the force applied to an object is directly proportional to its acceleration, while its mass is inversely proportional.

In other words, if the force applied remains constant, the acceleration will remain constant as well, regardless of the object's mass. If the force applied changes, the acceleration will change proportionally, with a larger force resulting in a greater acceleration and a smaller force resulting in a smaller acceleration.

Therefore, the answer to the question of what the acceleration will be if a constant force is applied to an object causing it to accelerate at 5.00 m/s2 is that it will remain at 5.00 m/s2 unless there is a change in the force applied or the object's mass.

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The main type of energy conversion taking place is chemical energy is converted to mechanical energy as the energy from digested food provides energy for movement. Option 3 is correct.

This is because the energy used for movement in living organisms comes from the breakdown of food molecules, such as glucose, through the process of cellular respiration. During cellular respiration, the chemical energy stored in food molecules is converted into a form of energy that can be used by cells to do work, which is called ATP (adenosine triphosphate). ATP is then used to power the mechanical work of muscles, which allows for movement.

Thermal energy is not involved in this process, as there is no mention of heat being a factor in the energy conversion. Mechanical energy is not converted to chemical energy, as this is not how living organisms obtain the energy needed for movement. Finally, there is no mention of reactions in the surrounding air. Option 3 is correct.

What is the main type of energy conversion taking place?

Responses

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The energy of a photon with a wavelength of 590 nm is [tex]3.36 * 10^-19[/tex]

The energy of a photon can be calculated using the equation:

E = hc/λ

where E is the energy of the photon, h is Planck's constant [tex](6.626 * 10^-34 J.s)[/tex], c is the speed of light[tex](3.00 * 10^8 m/s)[/tex], and λ is the wavelength of the photon in meters.

To convert 590 nm (nanometers) to meters, we can use the conversion factor:

[tex]1 nm = 1 * 10^-9 m[/tex]

So, [tex]590 nm = 590 * 10^-9 m[/tex]

Plugging these values into the equation, we get:

E = [tex](6.626 * 10^-34 J.s * 3.00 * 10^8 m/s) / (590 * 10^-9 m)[/tex]

E = [tex]3.36 * 10^-19 J[/tex]

Therefore, the energy of a photon with a wavelength of 590 nm is [tex]3.36 * 10^-19[/tex]

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This statement suggests that the ice undergoes a phase change from solid to liquid, indicating heat transfer.

All of the ice will melt, and the final temperature of the mixture will be 10°C.

If the surroundings were at a lower temperature than the ice, heat would flow from the ice to the surroundings, causing the ice to freeze instead of melt.

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The increase in the potential energy of the spring is 360.44J.

The energy that a body possesses due to its location in relation to other objects, internal pressures, electric charge, and other reasons is called potential energy. The following formula determines the potential energy held in a spring that has been compressed or stretched:

PE = (1/2)kx²

where,

x =  the distance the spring has been compressed or extended from its equilibrium position

k = the spring constant

The spring constant in this instance is stated as k = 87,600 N/m, while the spring's compression distance is specified as x = 0.092 m. As a result, the spring's increased potential energy is:

PE = (1/2)kx²

= (1/2)87,600 × 0.092²

= 360.44 J

Therefore, the increase in the potential energy of the spring is 360.44J.

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The changes in the variables that will result in no change in the gravitational force between m1 and m2 is if m1 is doubled and d is doubled. Option A.

The gravitational force between two masses depends on the masses and the distance between them, according to the equation F = G(m1m2)/d^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses, and d is the distance between them.

To have no change in the gravitational force between two masses, we need to keep the value of F constant. This can be achieved by changing the variables in the following way:

Therefore, the correct answer is a), where m1 is doubled and d is doubled.

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It takes 120 j of effort to extend a spring 0.3 m from its equilibrium state. The value of the spring constant k is 2666.67 N/m.

The work done to stretch a spring is given by the formula:

W = 0.5 × k × x^2

where W is the work done, k is the spring constant, and x is the distance the spring is stretched from its equilibrium position.

In this problem, we know that the work done is 120 J and the distance the spring is stretched is 0.3 m. Substituting these values into the formula, we get:

120 = 0.5 × k × 0.3^2

Simplifying the equation, we get:

k = 120 / (0.5 × 0.3^2)

k = 2666.67 N/m

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The mean distance of the asteroid Icarus from the Sun is approximately 1.24 astronomical units.

The mean distance of the asteroid Icarus from the Sun can be determined using Kepler's Third Law of Planetary Motion. This law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit. Mathematically, it can be written as:

T² ∝ a³

We can use the Earth's orbit as a reference, which has a period of 365.25 days and a semi-major axis of 1 astronomical unit (AU).

Using the given period of Icarus (410 days), we can set up the following proportion:

(410² / 365.25²) = (a³ / 1³)

Calculating the left side of the equation gives:

(168100 / 133483.0625) = a³

Taking the cube root of both sides, we get:

a ≈ 1.24 AU

So, the mean distance of the asteroid Icarus from the Sun is approximately 1.24 astronomical units.

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When a 2 uF and a 1 uF capacitor are connected in parallel and a potential difference is applied across the combination, the 2 uF capacitor has the highest charge. The total capacitance of the capacitors connected in parallel is the sum of their individual capacitances.

When a potential difference is applied across two capacitors connected in parallel, the potential difference across each capacitor is the same. In this case, the 2 uF and 1 uF capacitors are in parallel, so the potential difference across each capacitor is the same as the potential difference applied to the combination. Since the capacitance of the 2 uF capacitor is larger than the 1 uF capacitor, it will store more charge for the same potential difference applied across the combination. This means that the 2 uF capacitor will have a larger charge stored on its plates compared to the 1 uF capacitor The charge stored on a capacitor is directly proportional to its capacitance, so the 2 uF capacitor will have twice the charge stored on its plates as the 1 uF capacitor.

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1. Car A will travel farther than Car B after 10 s.

2. Car A will have a larger kinetic energy due to its greater mass.

3. Car A will have a larger momentum due to its greater mass.

Assuming both cars have the same constant acceleration, the car with the greater weight (Car A) will travel farther after 10 s according to the equation d = 0.5at^2, where d is the distance, a is the acceleration, and t is the time. Therefore, Car A will travel farther than Car B after 10 s.

The kinetic energy of a moving object is given by the equation KE = 0.5mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity. Both cars have the same acceleration, so after 10 s, the car with the higher velocity will have a larger kinetic energy. Assuming both cars accelerate uniformly, Car A will have a larger kinetic energy due to its greater mass.

The momentum of an object is given by the equation p = mv, where p is the momentum, m is the mass, and v is the velocity. After 10 s, the car with the higher velocity will have a larger momentum. Assuming both cars accelerate uniformly, Car A will have a larger momentum due to its greater mass.

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--The complete question is, Car A is 1000g in weight and B is 800g. Both car begin from rest and start at same position.
1. which car has traveled farther after 10 s?

2. after 10 s which car has a larger kinetic energy?

3. after 10 s which car has a larger momentum?--

The following terms should be used in your answer: "tangential speed", "m/s 2.00 kg particle 7.2", and "angular speed".

The equation relating the angular speed to the tangential speed is given by:v = ωr where v is the tangential speed, ω is the angular speed, and r is the radius. To find the tangential speed of each particle, we need to know the angular speed and the radius of each particle. The given masses and distances from the x-axis are as follows:4.00 kg particle at 14.4 m from the x-axis 2.00 kg particle at 7.2 m from the x-axis3.00 kg particle at 10.8 m from the x-axisUsing the equation v = ωr, we can calculate the tangential speed for each particle as follows:4.00 kg particle:ω = 2π/8 = π/4 rad/sr = 14.4 tangential speed, v = ωr = (π/4) x 14.4 = 3.6π m/s2.00 kg particle:ω = 2π/4 = π/2 rad/sr = 7.2 tangential speed, v = ωr = (π/2) x 7.2 = 3.6π m/s3.00 kg particle:ω = 2π/6 = π/3 rad/sr = 10.8 tangential speed, v = ωr = (π/3) x 10.8 = 3.6π m/sTherefore, the tangential speed of the 4.00 kg particle is 3.6π m/s, the tangential speed of the 2.00 kg particle is 3.6π m/s, and the tangential speed of the 3.00 kg particle is 3.6π m/s.

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The inductance of the coil is 0.028 H.

To find the inductance of the coil, first, we need to determine the change in current (∆I) and the rate of change of current (dI/dt).

1. Calculate the change in current: ∆I = I_final - I_initial = 2.20 A - 3.70 A = -1.50 A.

2. Calculate the rate of change of current: dI/dt = ∆I / ∆t = -1.50 A / 0.520 s = -2.885 A/s.

3. Use Faraday's Law: |emf| = L * |dI/dt|, where emf is the average induced electromotive force, and L is the inductance.

4. Solve for L: L = |emf| / |dI/dt| = 16.0 mV / 2.885 A/s = 0.028 H.

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Answer:

E

"As the distance increases by a factor, the electric force decreases by the square of that factor" best explains the relationship between the electric force between two charged objects and the distance between them.

The force exerted on the charged particles is inversely proportional to the square of the distance between them. The further they are the less the force the closer they are the more the force.

The correct answer between all the choices given is the last choice or letter E.

The merry-go-rotational round's acceleration may be computed using the following formula: (final rotational velocity - starting rotational velocity) / time Equals rotational acceleration.

Using the provided values, we get: Rotational acceleration = (12.5 s x (18.6 rad/s - 2.5 rad/s 1.368 rad/s2 rotational acceleration As a result, the merry-go-rotational round's acceleration is 1.368 rad/s2. This suggests that the merry-go-rotational round's velocity is growing at a rate of 1.368 rad/s2. In other words, the rotational velocity of the merry-go-round increases by 1.368 radians per second for every second that passes. Understanding rotational acceleration is crucial in engineering and physics because it is used to describe the motion of spinning items like gears and wheels, which can affect their performance and stability.

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Answer:

Radio wave :

The wavelength of an electromagnetic wave is measured to be 600 m.

Explanation:

All remain are given in attachment!