Derive the following equations of motion
1. v = at
2. s = ut + at²
3. v² = u² + 2as​

a negative charge Q1 = -5.50 is located at a point x1=-6.00 abd. positive charge Q2= 6.50 uc is located at point x2= 4.00 m then the magnitude and direction of the electric force between the charges is 3.56 mN and from left to right.

Coulomb's Law, which says that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of their distance, may be used to determine the size and direction of the electric force between the provided charges.

Given:

Q1 = -5.50 μC (negative charge)

Q2 = 6.50 μC (positive charge)

x1 = -6.00 m (position of Q1)

x2 = 4.00 m (position of Q2)

distance between the charges is

Distance (d) = |x2 - x1| = |4.00 m - (-6.00 m)| = 10.00 m

electric force (F) by using Coulomb's Law:

F = (k * |Q1| * |Q2|) / (d^2)

Where k is the electrostatic constant, which is equal to 9 × 10^9 N·m^2/C^2.

Substituting the given values:

F = (9 × 10⁹ N·m²/C² * |(-5.50 μC)| * |6.50 μC|) / (10.00 m)^2

F ≈ 3.2 mN

The magnitude of the electric force between the charges is approximately 3.56 mN.

Since Q1 is negative and Q2 is positive, the electric force acts in the direction from Q1 towards Q2. Therefore, the direction of the electric force is from left to right.

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Answer: [tex]f=3.125 Hz[/tex]

Explanation:

frequency = 1 / period

[tex]f=\frac{1}{T}[/tex]

[tex]f=\frac{1}{0.32}=3.125Hz[/tex]

Therefore, the frequency of the wave is 3.125 Hz.

Thermal expansion is defined as the property of metals to expand when they are heated.

The metal expands when heated because the atoms move apart more.

An excellent method to demonstrate the concept of thermal expansion is using the bar and gauge experiment.

Both the diameter and the length are precisely comparable when both are at room temperature.

The bar will not fit within the gauge once it has been heated and cooled. Similar to the last instance; the bar will no longer fit flush inside the gauge if the gauge is heated and the bar is cooled.

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In this configuration, the bar-cylinder system is in static equilibrium, which means that the net force and the net torque acting on the system are both zero. The system remains motionless and does not rotate around the peg in hole C because the weight of the cylinder hanging from hole B and the weight of the bar itself create equal and opposite torque around the peg.

To understand why the system is in equilibrium, we can consider the forces and torques acting on the system. Let's assume that the weight of the bar is W, and the weight of the cylinder is also W. When the system is hung in this configuration, the weight of the cylinder hanging from hole B creates a clockwise torque around the peg, while the weight of the bar creates a counter-clockwise torque around the same peg. The two torques cancel each other out, resulting in zero net torque.

The location of the center of mass of the bar at the center of hole D ensures that the weight of the bar acts vertically downward through the center of hole D. Since the peg passes through hole C, the weight of the bar does not act at a distance from the peg, and so it does not create any torque around the peg.

Therefore, the bar-cylinder system remains in equilibrium and does not rotate around the peg in hole C because the weight of the cylinder and the weight of the bar create equal and opposite torques around the peg, resulting in zero net torque. The location of the center of mass of thebar at the center of hole D ensures that the weight of the bar itself does not create any torque around the peg. Hence, the system remains motionless and balanced in this configuration.

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If two buses are moving in the same direction with the same velocity, a passenger in one bus would perceive the other bus as being at rest relative to their own frame of reference. This phenomenon is known as relative motion or the principle of relativity.

The reason behind this perception is that both buses are moving with the same velocity. From the perspective of the passenger in one bus, they are also moving at the same speed and in the same direction as their own bus. Therefore, the relative velocity between the two buses is zero, and the passenger perceives the other bus as being stationary.

This perception is similar to how passengers inside a moving train may perceive nearby objects, such as other stationary trains on parallel tracks, as being at rest relative to their own moving train. It's important to note that this perception of relative rest is valid only within the frame of reference of the passenger in one bus. Observers outside the buses, or in a different frame of reference, would see both buses moving with the same velocity.

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Answer:

Charged

Explanation:

When an atoms loses or gains electrons they become charged(ion)

When an atom loses electrons it becomes positively charged and it is called a Cation

When an atom gains electrons it becomes negatively charged and it is called an Anion

(a) The  force pulling the apple is the weight of the apple and the magnitude is determined as 0.98 N.

(b) The gravity needed is 0.98 N.

(c) The energy transfer that takes place is 5.88 J.

The gravity needed is calculated as follows;

Fg = mg

where;

Fg = 0.1 x 9.8

Fg = 0.98 N

The  force pulling the apple is equal in magnitude to force of gravity and the magnitude is determined as 0.98 N.

So the pulling force on the apple is the weight of the apple.

The energy transfer that takes place is calculated as follows;

E = P.E = mgh

E = 0.98 N x 6 m

E = 5.88 J

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The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

This observation that electromagnetic radiation behaves like particles is known as the photoelectric effect.

It provides evidence that electromagnetic radiation exhibits particle-like properties. When EMR with sufficient energy (above a certain threshold) interacts with a metal surface, it can cause the ejection of electrons from the metal.

This behavior indicates that EMR behaves as discrete packets of energy called photons, which transfer their energy to the electrons and cause their release. The photoelectric effect supports the particle nature of EMR and is a fundamental concept in the field of quantum mechanics.

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The total electric field produced by the charges at the midpoint is

8 x 10³ N/C.

Amount of charge on the first particle, q₁ = 40 x 10⁻⁹C

Amount of charge on the second particle, q₂ = 60 x 10⁻⁹C

Distance between the charges, r = 30 cm = 0.3 m

The expression for the electric field due to the charges is given by,

E = 1/4πε₀ x q/r²

At the midpoint between the two charges, d = 0.15 m

So, the electric field produced by the charge q₁ at the midpoint is,

E₁ = 1/4πε₀ x q₁/d²

E₁ = 9 x 10⁹ x 40 x 10⁻⁹/(0.15)²

E₁ = 1.6 x 10⁴ N/C

So, the electric field produced by the charge q₂ at the midpoint is,

E₂ = 1/4πε₀ x q₂/d²

E₂ = 9 x 10⁹ x 60 x 10⁻⁹C/(0.15)²

E₂ = 2.4 x 10⁴ N/C

Therefore, the total electric field produced by the charges at the midpoint is,

E = E₂ - E₁

E = 2.4 x 10⁴- 1.6 x 10⁴

E = 8 x 10³ N/C

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a. Savannah would debit retained earnings and credit dividends payable for the declaration of a cash dividend, and debit dividends payable and credit cash for the payment, b. To declare and pay a stock dividend, Savannah would debit retained earnings and credit common stock dividend distributable for declaration, then debit common stock dividend distributable and credit common stock for payment, and c. Dividends payable to shareholders like Elena and Josh can be computed by multiplying the number of shares owned by the dividend per share.

Dividends Payable is a liability account used to record the amount of dividends that a company owes to its shareholders but has not yet paid out. It represents the obligation of the company to pay out the declared dividend amount to its shareholders, and the balance is typically cleared out when the dividend payment is made. Companies usually set a record date and payment date to declare and pay dividends respectively.

a. To journalize the declaration and payment of the cash dividend, Savannah would make the following entries:

Declaration:

Debit: Retained Earnings

Credit: Dividends Payable

Payment:

Debit: Dividends Payable

Credit: Cash

b. To journalize the declaration and payment of the stock dividend, Savannah would make the following entries:

Declaration:

Debit: Retained Earnings

Credit: Common Stock Dividend Distributable (70,000 x 5% = 3,500 shares)

Payment:

Debit: Common Stock Dividend Distributable

Credit: Common Stock ($7.25 x 3,500 shares)

c. To compute the dividends payable to Elena and Josh, we would use the following formula:

Dividends Payable = Number of Shares x Dividend per Share

For Elena:

Dividends Payable = 300 shares x $0.10 per share = $30.00

For Josh:

Dividends Payable = 75 shares x $0.10 per share = $7.50

Therefore, To declare and pay a cash dividend, Savannah would debit retained earnings and credit dividends payable for declaration, and debit dividends payable and credit cash for payment. To declare and pay a stock dividend, Savannah would debit retained earnings and credit common stock dividend distributable for declaration, then debit common stock dividend distributable and credit common stock for payment. Dividends payable to shareholders like Elena and Josh can be computed by multiplying the number of shares owned by the dividend per share.

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A physical standard is a physical item used as a standard measurement

A physical standard is an object that is carefully designed, manufactured and maintained to serve as a reference or benchmark for measurements. It represents a specific quantity or characteristics with a well defined value, making it a point of comparison for other similar measurements. Physical standards are used in various fields, including science, engineering, metrology , and quality control.

Physical standards can take different forms depending on the quantity being measured. For a example a standard kilogram represent the unit of mass and a standard meter represent unit of length.

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A parallel circuit can be modelled by connecting the components parallel.

A parallel circuit is a form of electrical circuit in which the components are linked in parallel to one another, each having a separate channel for current flow and being directly connected to the power source. The circuit must first be schematically represented, with the power supply and each component linked in parallel. Next, the total resistance the total current in the circuit by using Ohm's Law must be calculated.

Additionally, it is necessary to confirm that the total current entering the circuit equals the total current leaving the circuit. In a parallel circuit, the total current passing through all of the components equals the current entering the circuit. The voltage drop between each component must then be once more computed using Ohm's Law.

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Move numbers to the boxes to show the factor pairs of 14:

Blank space 1: 1

Blank space 2: 1

Blank space 3: 2

Blank space 4: 7

A factor pair of a number is a pair of whole numbers that can be multiplied together to give the original number. For the number 14, the factor pairs are (1,14) and (2,7). So, we can put 1 in the first blank, 14 in the second blank, 2 in the third blank, and 7 in the fourth blank to show the factor pairs of 14.

To show the factor pairs of 14 in the given response area with 4 blank spaces, we need to find the two numbers that can be multiplied together to give 14. These two numbers are called factor pairs of 14.

To begin, we can start listing the factors of 14. The factors of 14 are 1, 2, 7, and 14. We can then use these factors to form factor pairs by multiplying them together. The factor pairs of 14 are (1, 14) and (2, 7).

To show these factor pairs in the given response area, we can put the first factor of each pair in the first and third blank spaces, and the second factor of each pair in the second and fourth blank spaces. Therefore, we can put 1 in the first blank, 14 in the second blank, 2 in the third blank, and 7 in the fourth blank to show the factor pairs of 14.

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According to the information, the steel ring should be heated to a maximum temperature of approximately 50.67°C to enable fitting; and the stresses in the materials on cooling will be approximately 17.04 MPa in the steel ring and 2.11 MPa in the brass wheel.

To enable fitting, the steel ring needs to expand in size so that its inner diameter matches the outer diameter of the brass wheel. We can use the formula for linear thermal expansion:

Where ΔL is the change in length, α is the coefficient of linear expansion, L0 is the initial length, and ΔT is the change in temperature. By substituting the given values and solving for ΔT, we find that the steel ring needs to be heated to a maximum temperature of approximately 50.67°C.

The stresses in the materials can be calculated using the formula:

Since the dimensions of the cross-sections are given, we can calculate the areas of the steel ring and brass wheel. The force can be calculated using the equation:

By substituting the given values for Young's modulus, change in length, and initial length, we can calculate the forces in both materials. Finally, by dividing the forces by their respective areas, we can determine the stresses in the steel ring and brass wheel, which are approximately 17.04 MPa and 2.11 MPa, respectively.

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The average speed of the campers down the river was 5 miles per hour.

Total distance traveled by the campers = 25 miles

Total time taken by the campers = 5 hours

Speed is defined as the rate of change in distance or altitude reached. It is a time-based quantity. In the given question, a camper's average speed can be calculated by dividing the total distance travelled by the time taken.

Calculating the average speed of the campers -

Average speed = Total distance / Time taken

Substituting the values -

= 25 miles / 5 hours

= 5

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A and B are the correct statements. A. The ‘Moon’ should be 240 steps away from the ‘Earth’. The ‘Earth’ should be 93 steps away from the ‘Sun’.

To create a model of the distances between the Earth, the Moon, and the Sun, we represent one million miles with one step. Based on this model:

A. The ‘Moon’ should be 240 steps away from the ‘Earth’. This is a correct statement, as the distance between the Earth and the Moon is about 240 thousand miles. In the model, we represent each million miles with one step, so the distance between the Earth and the Moon in the model would be 240 steps.

B. The ‘Earth’ should be 93 steps away from the ‘Sun’. This is also a correct statement, as the distance between the Earth and the Sun is about 93 million miles. In the model, we represent each million miles with one step, so the distance between the Earth and the Sun in the model would be 93 steps.

C. The ‘Moon’ should be less than one step away from the ‘Earth’. This statement is incorrect, as the actual distance between the Earth and the Moon is much greater than one million miles. In the model, the distance between the Earth and the Moon would be represented by 240 steps.

D. The ‘Earth’ should be 5 steps away from the ‘Sun’. This statement is incorrect, as the actual distance between the Earth and the Sun is much greater than five million miles. In the model, the distance between the Earth and the Sun would be represented by 93 steps.

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The activation energy for the diffusion process of Al in Al2O3 is approximately 225,744 J/mol.

To calculate the activation energy for the diffusion process of Al in Al2O3, we can use the Arrhenius equation:

D = D0 * exp(-Q/RT)

where D is the diffusion coefficient, D0 is a constant, Q is the activation energy, R is the gas constant, and T is the absolute temperature.

Taking the natural logarithm of both sides, we get:

ln(D/D0) = -Q/RT

We can then use the given diffusion coefficients and temperatures to obtain two equations:

ln(D1/D0) = -Q/RT1

ln(D2/D0) = -Q/RT2

where D1 and T1 are the diffusion coefficient and temperature at 1000C, and D2 and T2 are the diffusion coefficient and temperature at 1500C.

Taking the ratio of the two equations, we get:

ln(D2/D1) = Q/R * (1/T1 - 1/T2)

Solving for Q, we get:

Q = -R * ln(D2/D1) / (1/T1 - 1/T2)

Plugging in the given values, we get:

Q = -8.314 J/mol-K * ln(2.48 x 10^-14 m^2/s / 7.48 x 10^-23 m^2/s) / (1/1273 K - 1/1773 K)

Q ≈ 225,744 J/mol

Therefore, the activation energy for the diffusion process of Al in Al2O3 is approximately 225,744 J/mol.

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The Falcon 9 rocket is so important in today's space exploration world because It's a reusable rocket and has saved money by being reused over 60 times already.

option D.

The Falcon 9 rocket, is developed by SpaceX, and it holds  a significant importance in today's space exploration world due to several key features and achievements, and some of the importance include the following;

From the given options, we can conclude that the Falcon 9 rocket is so important in today's space exploration world because It's a reusable rocket and has saved money by being reused over 60 times already.

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The circuit given is a series resistance circuit.

1) The given circuit is a series resistance circuit.

2) Resistance, R₁ = 20 Ω

Resistance, R₂ = 5 Ω

The effective resistance of the circuit is given by,

Reff = R₁ + R₂

Reff = 20 + 5

Reff = 25 Ω

3) Total resistance of the circuit, R = 25 Ω

Current flowing through the circuit, I = 50 A

Therefore, the voltage across the circuit is,

V = IR

V = 50 x 25

V = 1250 V

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If the coefficient of kinetic friction between Josh's sled and the snow is 0.08, he slides 6.97 meter from the base of the hill.

To find how far from the base of the hill Josh's sled ends up, we need to first find the speed of the sled at the bottom of the hill using the conservation of energy principle,

mgh = (1/2)mv², plugging in the values given in the problem, we get,

m(9.81 m/s²)(3.5 m) = (1/2)mv²

Simplifying and solving for v, we get,

v = √(2gh)

v = √(2(9.81 m/s²)(3.5 m))

v = 8.29 m/s

Now we can use the kinematic equation,

d = vt - (1/2)at, to find how far the sled slides on the horizontal patch of snow before coming to a stop, where d is the distance traveled, v is the initial velocity (8.29 m/s), a is the acceleration due to friction (-μg), and t is the time it takes to come to a stop (which we can find by setting v = 0 and solving for t),

0 = 8.29 m/s - μg*t

t = 8.29 m/s / μg

Substituting this value of t back into the kinematic equation, we get,

d = (8.29)(8.29/μg) - (1/2)μg(8.29/μg)²

d = 6.97 m

Therefore, Josh's sled ends up 6.97 meters from the base of the hill.

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Answer: 3.8

Explanation: speed = wavelength x frequency

To calculate the average speed of the car, we can use the formula:

Average speed = Total distance / Total time

Given information:

Number of laps = 60

Track length per lap = 3.6 km

Total time = 2.4 hours

First, let's calculate the total distance traveled by the car:

Total distance = Number of laps * Track length per lap

Total distance = 60 * 3.6 km

Total distance = 216 km

Now, we can calculate the average speed:

Average speed = Total distance / Total time

Average speed = 216 km / 2.4 hours

Average speed ≈ 90 km/h

So, the average speed of the car is approximately 90 km/h.

The magnitude of the frictional force is 5 N.

The force exerted by the boy on the box is 40.0 N. According to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration:

F(net) = m*a

where F(net) is the net force, m is the mass of the object, and a is its acceleration.

In this case, the net force acting on the box is the force applied by the boy minus the frictional force:

F(net) = F(applied) - F(friction)

By substituting the values:

F(applied) = 40.0 N

m = 5 kg

a = 7 m/s^2

Now, by using frictional force:

F(friction) = F(applied) - F(net)

F(friction) = F(applied) - ma

                = 40.0 N - (5 kg)(7 m/s^2)

                = 5 N

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Explanation: ΔL = τ(average) * Δt
Change in angular momentum = average torque * change in time

solve for average torque for each objects
τ(average) = ΔL / Δt

Object y average torque
τy = ΔLy / Δt = 20 / 5 = 4
τy = 4

Object x average torque
τx = ΔLx / Δt = 10 / 5 = 2
τx = 2

Relates τy and τx
2τx = τy

Two objects, X and Y, experience external net torques that vary over a period of 5 seconds. Object X has a moment of inertia I0, and Object Y has a moment of inertia 2I0. The average value of the magnitude of the external net torque exerted on Object X from time t=0 to t=5s is torquex. Similarly, the average value for ObjectY is torquey.

The magnitudes of the angular momenta L of Objects X and Y versus t are shown in the graph. The precise relation between torquey and torquex is torquey = 2 * torquex.

To relate torquey to torquex, we are able to use the concept of angular momentum and torque. Angular momentum is described because the manufactured from the moment of inertia and angular velocity:

L = I * ω

Differentiating this equation with an appreciation of time, we get:

dL/dt = d(I * ω)/dt

Using the product rule of differentiation, we've got:

dL/dt = I * dω/dt + ω * dI/dt

Now, we realize that torque (τ) is described because of the charge of the exchange of angular momentum:

τ = dL/dt

Substituting the expression for dL/dt in terms of angular velocity and second of inertia:

τ = I * dω/dt + ω * dI/dt

Let's denote the common price of torque for item X as torquex. Since object X has a moment of inertia I0, we can write:

torquex = I0 * dω/dt + ω * dI0/dt

Now, let's consider item Y. It has a moment of inertia 2I0. Using the identical expression, we will write:

torquey = (2I0) * dω/dt + ω * d(2I0)/dt

torquey = 2I0 * dω/dt + ω * (2 * dI0/dt)

torquey = 2I0 * dω/dt + 2ω * dI0/dt

Comparing the expressions for torquex and torquey, we will see that:

torquey = 2 * torquex

Therefore, the precise relation between torquey and torquex is;

torquey = 2 * torquex.

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The correct question is;

"Two objects, X and Y, experience external net torques that vary over a period of 5 seconds. Object X has a moment of inertia I0, and Object Y has a moment of inertia 2I0. The average value of the magnitude of the external net torque exerted on Object X from time t=0 to t=5s is torquex. Similarly, the average value for ObjectY is torquey. The magnitudes of the angular momenta L of Objects X and Y versus t are shown in the graph. Which of the following expressions correctly relates torquey to torquex?"

the physics of a domino chain shows how momentum and kinetic energy can combine to produce an amazing display of dominoes falling.

First of all, momentum is a vector that denotes an object's mass and speed. When a domino chain is pushed over, the initial domino that was knocked over transfers momentum to the next domino, which then transfers momentum to the next, and so on. Each domino does not have to travel at the same pace as the others as long as they are in time with one another since the group of dominoes offers a coordinated transfer of momentum.

The second definition of kinetic energy is the energy that an item has as a result of motion. Kinetic energy increases each domino's speed and the power necessary to topple the subsequent domino in a chain of dominoes as momentum is transferred from one domino to the next. The largest domino, which frequently topples onto a designated catch or platform that intensifies the effect for spectators, is where the chain's greatest amount of kinetic energy is stored when it reaches its conclusion.

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The magnitude of the magnetic field's torque on the loop is 0.034 Nm.

When a current-carrying conductor is put in a magnetic field, it experiences a force known as the Lorentz force, which is proportional to the magnetic field's strength and the current running through the conductor. As a result of this force, a torque is imparted to the conductor, which tends to turn it around an axis perpendicular to the conductor's plane.

The following formula can be used to calculate the torque operating on the conductor:

= NIABsin

where is the torque, N is the number of turns, I is the current, A is the area of the loop, B is the strength of the magnetic field, and is the angle between the magnetic field and the plane of the loop.

If the magnetic field is parallel to the plane of the loop, then θ is equal to 0°, and sinθ is equal to 0.

As a result, the torque formula can be shortened to: = NIAB.

When the numbers for N, I, A, and B are entered into this formula, the magnetic field's torque on the loop is found to be:

= (1)(0.017 A)(π(1 m)²)(0.80 T)τ = 0.034 Nm.

As a result, the magnitude of the magnetic field's torque on the loop is 0.034 Nm.

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The required acceleration of the car, when the driver hits the brake, is [tex]-5 m/s^2[/tex], the negative sign indicates that the car is slowing down.

From the question we can see that the:

initial velocity(u) is + 100 m/s (to the right)

Final velocity (v) is 0 since the car comes to stop

Time (t) is 20 s

We can use the formula

a= (v-u)/t to find the acceleration(a) of the car.

Now Substituting the values in the formula given above we get:

a=(0-100)/20

a= [tex]-5m/s^2[/tex]

Therefore, the acceleration of the car is [tex]-5 m/s^2[/tex]. The negative sign in the answer indicates that the acceleration is in the opposite direction to the initial velocity, which means that the car is slowing down or decelerating.

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The metal with the highest heat capacity between metals A.25 B.35 C.10 and D.15 is metal A.

Heat capacity is the amount of heat required to raise the temperature of a substance by one degree Celsius. Metal A has a heat capacity of 400 J/kg°C, which means that it takes 400 joules of heat to raise the temperature of one kilogram of metal A by one degree Celsius.

Metal B has a heat capacity of 285.7 J/kg°C, metal C has a heat capacity of 1000 J/kg°C, and metal D has a heat capacity of 666.7 J/kg°C. Therefore, metal A has the highest heat capacity of the four metals.

Metal A's high heat capacity means that it can absorb a lot of heat without its temperature changing very much. This makes metal A a good material for things like heat sinks and thermal insulation.

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