in a race a car travels 60 times around a 3.6km track, this takes 2.4 hours.
What is the average speed of the car ?

The resistance values from the givens are having a consistent value ranging from 27 to 33 ohms depending upon the supply of voltage and current.

Ohm's law gives the relation between the voltage, current, and resistance. Ohm's law states that the applied voltage is directly proportional to the current flow in the conductor. Thus, V∝I, V is the voltage and I is the current.

Ohm's law can also be written as, V = IR, where R is the resistance and the unit of resistance is ohm. The resistance is the factor that resists or opposes the current flow in the curcuit. When resistance increases, current flow in the circuit decreases and vice-versa.

From the given values, the resistance values changes when there is a change in voltage and current. The current and voltage have linear relationship, and these calculations make the idealsitic assumptions and can not hold true in practical applicaitons, when woeking with non-ohmic or non-linear components.

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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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The intrepid hero has done 2.332 x Joules of work in pushing the crate.

To ascertain the work done by the traveler, we first need to find the power he applied on the case. As per Newton's subsequent regulation, force is equivalent to mass times speed increase, so the power applied by the traveler on the container is:

Force = mass x speed increase = 220 kg x 2  = 440 N

Then, we really want to work out the distance the case was moved. The pilgrim pushed the box a distance of 5.3 km, or 5,300 m.

At long last, we can compute the work done by the pioneer utilizing the equation:

Work = force x distance = 440 N x 5,300 m = 2.332 x 10^6 Joules

Thusly, the valiant legend has done 2.332 x  Joules of work in pushing the case.

The space pilgrim takes care of business on the case by applying a power that makes it speed up. The work done is equivalent to the power duplicated by the distance over which the power is applied. Involving the recipe for force, F=ma, and the given qualities for mass and speed increase, we can ascertain the power applied. Then, at that point, involving the recipe for work, W=Fd, and the given distance, we can ascertain the work done. The work done by the adventurer is 2.332 x  J.

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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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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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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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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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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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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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If the magnification is -4 the the image is larger and is it inverted in nature.

The magnification of the body is given is -4, here the signs show the nature of the image and the numerical value is the times to which the object has been magnified. Here, it is inverted because there is negative sign and it is enlarged as the magnification is more than 1.

Therefore, a magnification of -4 indicates that the image is both inverted and larger than the object. The resultant picture will be a larger, inverted version of the object if it is seen via a lens or mirror with a magnification of -4.

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The voltage drop across R3 is 34.5 volts.

Voltage, also known as electric potential difference, is a measure of the difference in electric potential energy between two points in an electrical circuit, and it is measured in volts. It is the driving force that moves electric charges through a circuit, from a higher potential to a lower potential.

To determine the voltage drop across R3 in this circuit, we need to first find the equivalent resistance of R2, R3, and R4, since they are connected in parallel. We can then find the total resistance of the circuit by adding the equivalent resistance in series with R1, and finally use Ohm's Law to calculate the voltage drop across R3.

The equivalent resistance of R2, R3, and R4 in parallel can be calculated as:

1/R_parallel = 1/R2 + 1/R3 + 1/R4

1/R_parallel = 1/20 + 1/25 + 1/10

1/R_parallel = 0.15

R_parallel = 1/0.15

R_parallel = 6.67 ohm

The total resistance of the circuit can be found by adding R1 and the equivalent resistance in series:

R_total = R1 + R_parallel

R_total = 15 + 6.67

R_total = 21.67 ohm

Now we can use Ohm's Law to calculate the current flowing through the circuit:

I = ET / R_total

I = 30 / 21.67

I = 1.38 A

Finally, we can use Ohm's Law again to calculate the voltage drop across R3:

V_R3 = I * R3

V_R3 = 1.38 * 25

V_R3 = 34.5 V

Therefore, the voltage drop across R3 is 34.5 volts.

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The charge R is positive and the charge L is negative hence, the correct option is option D.

The electric charges are of two types and they are positive and negative charge. There is an attractive force between the two charges. The electric field is formed around the charges and the electric field is denoted by using field lines around it.

The field lines are emerging from positive charge and end up in a negative charge. The field lines never intersect with each other. Hence, from the given, the field lines emerging from the positive charge(R) and end up in the negative charge(L).

Hence, the ideal solution is option D.

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The order of increasing mass is:

An electron < An atomic nucleus < a DNA molecule < a cell.

     An electron is a fundamental particle with a negative charge and has a very small mass of order [tex]10^{-31}\ kg[/tex]. Which makes it the lightest in the group.

     An atomic nucleus consists of protons and neutrons which both individually have larger mass than an electron. The order of magnitude of the nucleus's mass is [tex]10^{-27}\ kg[/tex] or higher.

     A DNA molecule is a polymer chain of nucleotides and it carries genetic information. It contains many atoms, therefore it makes sense that it should have more mass than an atomic nucleus. Each human cell has about DNA of mass of order [tex]10^{15}\ kg[/tex].

     Lastly, A cell has the most mass in the given list since it encompasses everything else in the list.

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Chemical change involves the rearrangement of atoms or molecules in a substance, resulting in the formation of new substances with different properties than the original substance.

A chemical change in a chemical reaction is a process where one or more substances undergo a chemical transformation, resulting in the formation of new substances with different chemical and physical properties than the original substances. During a chemical change, the atoms or molecules of the reacting substances are rearranged through chemical reactions, leading to the breaking and forming of chemical bonds and the release or absorption of energy.

Chemical changes occur when a material's chemical makeup is changed, creating one or more new compounds with qualities that are distinct from those of the original substance. Chemical reactions that result in the breaking and building of chemical bonds occur when the atoms or molecules of the substances involved are reorganized during a chemical shift. The release or absorption of energy in the form of heat, light, or sound is typically accompanied by this process.

Therefore, Rearranging atoms or molecules in a substance causes chemical change, which creates new compounds with differing properties from the original one.

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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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The magnitude of the resultant vector is obtained by finding the sum of two horizontal vectors and the sum of two vertical vectors. Thus, option C is correct.

The resultant vector is the single vector that has the same effect in the number of vectors collectively produced. The resultant vector in the horizontal and vertical direction is obtained by drawing a diagonal and hence by using the Pythagoras theorem.

To find resultant vector is obtained by finding the sum of two horizontal vectors and vertical vectors and then using the Pythagoras theorem.

Thus, the ideal solution is option C.

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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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Segment a, which is drawn from the center of the polygon perpendicular to one of its sides, is called the apothem.

The typical hexagon has a surface area of around 874.6 square meters.

To find the area of a regular polygon, use the formula:

Area = (1/2) x Perimeter x Apothem

Find the perimeter of the polygon. Since the polygon has n sides, use the formula:

Perimeter = n x s

where s = length of one side.

Since s = 18m, find n by using the formula:

n = 360° / (180° - (360° / n))

where n = number of sides.

Plugging in the values:

n = 360° / (180° - (360° / n))

n = 360° / (180° - (360° / 6))

n = 6

So the polygon has 6 sides, which makes it a hexagon.

Now find the perimeter:

Perimeter = n x s

Perimeter = 6 x 18

Perimeter = 108m

Next, find the apothem, use the formula:

Apothem = s / (2 x tan(π/n))

Plugging in the values:

Apothem = 18 / (2 x tan(π/6))

Apothem = 9√3 m

Now use the formula for the area:

Area = (1/2) x Perimeter x Apothem

Area = (1/2) x 108 x 9√3

Area ≈ 874.6 m²

Therefore, the area of the regular hexagon is approximately 874.6 square meters.

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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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Answer: 3.8

Explanation: speed = wavelength x frequency

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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(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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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?"

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 final temperature of the mixture of cold and hot water is 60°C.

To calculate the final temperature of the mixture of hot and cold water, we can use the principle of conservation of energy. The heat gained by the cold water is equal to the heat lost by the hot water.

Given:

Rate of hot water flow (m₁) = 20 kg/min

The initial temperature of hot water (T₁) = 80°C

Rate of cold water flow (m₂) = 10 kg/min

The initial temperature of cold water (T₂) = 20°C

Duration (t) = 3 minutes

Let's calculate the heat gained by the cold water and the heat lost by the hot water using the formula:

Heat gained = mass * specific heat capacity * temperature change

For the cold water:

Heat gained by cold water = m₂ * specific heat capacity * (final temperature - T₂)

For the hot water:

Heat lost by hot water = m₁ * specific heat capacity * (T₁ - final temperature)

Since the heat gained by the cold water is equal to the heat lost by the hot water, we can set up the equation:

m₂ * specific heat capacity * (final temperature - T₂) = m₁ * specific heat capacity * (T₁ - final temperature)

Let's substitute the given values and solve for the final temperature:

10 kg/min * specific heat capacity * (final temperature - 20°C) = 20 kg/min * specific heat capacity * (80°C - final temperature)

Simplifying the equation:

10 * (final temperature - 20) = 20 * (80 - final temperature)

10 * final temperature - 200 = 1600 - 20 * final temperature

30 * final temperature = 1800

final temperature = 1800 / 30

final temperature = 60°C

Therefore, the final temperature of the mixture of hot and cold water after 3 minutes will be 60°C.

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The formula you mentioned is known as the dot product formula or the scalar product formula. It is used to find the angle between two vectors u and v.

Let's start by defining the vectors u and v. Suppose we have two vectors u and v in a two-dimensional space.

u = (u1, u2)

v = (v1, v2)

The dot product of these vectors is defined as:

u . v = |u| |v| cos(β)

where |u| and |v| are the magnitudes of the vectors u and v respectively, and β is the angle between the vectors u and v.

Now, let's derive this formula. The dot product of two vectors u and v is given by:

u . v = (u1 × v1) + (u2 × v2)

The magnitude of a vector is given by:

|u| = sqrt(u1² + u2²)

|v| = sqrt(v1² + v2²)

We can use the dot product and magnitude equations to obtain:

cos(β) = (u . v) / (|u| × |v|)

Multiplying both sides by |u| × |v| gives us:

|u| × |v| × cos(β) = u . v

Therefore, we have derived the dot product formula:

u . v = |u| × |v| × cos(β)

This formula can be used to find the angle between two vectors u and v in any two-dimensional or three-dimensional space.

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The complete question is-

Write the proof of the formula

u.v=|u||v|cosβ

It takes 2.04 seconds for the object to accelerate from -10.0 m/s to -30.0 m/s.

Velocity is a measure of an object's speed in a specific direction. It is calculated by dividing the distance traveled by the time taken and specifying the direction of the motion.

We can use the following formula to find the time taken for an object to change velocity under constant acceleration:

Δv = a × Δt

where Δv is the change in velocity, a is the acceleration, and Δt is the time taken.

In this case, the initial velocity is -10.0 m/s, and the final velocity is -30.0 m/s. So, the change in velocity is:

Δv = (-30.0 m/s) - (-10.0 m/s) = -20.0 m/s

The acceleration of the falling object is -9.80 m/s² (negative because the object is falling downward).

Now, we can rearrange the formula to solve for Δt:

Δt = Δv / a

Substituting the values we have:

Δt = (-20.0 m/s) / (-9.80 m/s²) = 2.04 seconds

Therefore, The object accelerates from -10.0 m/s to -30.0 m/s in 2.04 seconds.

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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 medium in which electromagnetic waves and mechanical waves travel is one of their primary distinctions. Light and other electromagnetic waves, including radio waves, can move through void space without the aid of a physical medium.

They may move through vacuum, air, or other materials and are made up of oscillating electric and magnetic fields. The propagation of mechanical waves, such as sound or water waves, on the other hand, depends on a physical medium.

To transport energy, they rely on particle interactions and displacements in the medium. Since mechanical waves need a physical medium to carry their energy, they cannot move through a vacuum.

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