If you were to examine the profile of a typical river, you would probably find the fastest flowing water closer to the __.

Answer:

Explanation:

for vertical displacement h = 1.6 , let time taken be t , initial vertical velocity = 0

h = ut + 1/2 g t²

= 1/2 g t²

1.6 = .5 x 9.8 x t²

t² = .3265

t = .57 s

b )

velocity in horizontal direction = 3.2 m /s

horizontal displacement

= 3.2 x .57 = 1.82 m

c )

for vertical movement

v = u + gt

= 0 + 9.8 x .57

= 5.58 m /s

The mass of the object that the star is orbiting is approximately 5.57 x 10^39 kg.

Kepler's third law states that the square of the orbital period of a planet is proportional to the cube of its semi-major axis (or in the case of a circular orbit, the radius of the orbit):

T^2 ∝ r^3

where T is the orbital period in years and r is the radius of the orbit in astronomical units (AU).

We can rearrange this equation to solve for T:

T = √ (r^3 / GM)

where G is the gravitational constant and M is the mass of the object being orbited.

The orbital speed of the star can be related to the radius and period of the orbit using the centripetal force equation:

F = mv^2 / r = GMm / r^2

where m is the mass of the star.

We can rearrange this equation to solve for M:

M = v^2 r / G

Substituting in the given values, we get:

M = (1200 km/s) ^2 * (22 light-days * c) / G

where c is the speed of light and G is the gravitational constant. We need to convert the units to a consistent system, so let's convert the radius to kilometers and the time to seconds:

22 light-days = 22 * 24 * 60 * 60 seconds = 1,900,800 seconds

1 light-day = 24 hours * 3600 seconds/hour * c = 86400 seconds * 299792458 m/s ≈ 2.5902 x 10^13 km

Therefore,

M = (1200 km/s) ^2 * (1.9 x 10^6 s) * (2.59 x 10^13 km) / (6.674 x 10^-11 N m^2/kg^2)

M ≈ 5.57 x 10^39 kg

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The flux per unit area of the toxic gas towards the residential room is 0.351 mg/(m^2 * minute).

To calculate the flux per unit area of the toxic gas towards the residential room, we can use Fick's law of diffusion, which states that the flux (J) is proportional to the concentration gradient (ΔC) and the diffusion coefficient (D), and inversely proportional to the distance (Δx):

J = -D * (ΔC / Δx)

In this case, we want to calculate the flux per unit area, so we need to divide the flux by the area of the tunnel.

Length of the tunnel (Δx) = 5 m

Diameter of the tunnel = 1.5 m (radius = 0.75 m)

Concentration in the chamber (C1) = 30 mg/m^3

Concentration in the residential room (C2) = 3 mg/m^3

Diffusion coefficient (D) = 0.065 m^2/minute

First, let's calculate the concentration gradient:

ΔC = C2 - C1 = 3 mg/m^3 - 30 mg/m^3 = -27 mg/m^3

Next, let's calculate the area of the tunnel:

Area = π * (radius)^2 = π * (0.75 m)^2 = 1.767 m^2

Now, we can calculate the flux per unit area:

J = -D * (ΔC / Δx) = -0.065 m^2/minute * (-27 mg/m^3 / 5 m) = 0.351 mg/(m^2 * minute)

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Answer:

false

Explanation:

Answer:

Find the given attachments

The force that pushed the cow back into the barn produced work using the formula W = -831.45 J.

The total vector of all applied forces to an object is known as the net force. The result of the fact that a force is a vector and that two forces of equal magnitude and opposite direction cancel each other out is the net force.

W = ∫ F(x) dx

In this instance,

Fx = -[20.0 N + 3.0 N/m]x gives the force.

The displacement of the cow is:

Δx = 6.9 m

Integrating this expression gives:

W = ∫ F(x) dx

= -∫ (3.0 N/m) x dx - ∫ 20.0 N dx

= -1.5 N/m * x² - 20.0 N * x + C

where C is the constant of integration.

W(x=0) = -1.5 N/m * 0² - 20.0 N * 0 + C = 0

Therefore, C = 0, and the work done by the force is:

W = -1.5 N/m * x² - 20.0 N * x

W = -1.5 N/m * (6.9 m)² - 20.0 N * 6.9 m

W = -831.45 J

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Question:

A balky cow is leaving the barn as you try harder and harderto push her back in. In coordinates with the origin at the barndoor, the cow walks from x = 0 tox = 6.9{rm m}as you apply a force withx-componentF_x= -[20.0{rm N}+ (3.0{rm N/m})x].

How much work does the force you apply do on the cow duringthis displacement?

Answer:

5.88 m/s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 7 m/s

Acceleration (a) = –0.6 m/s²

Distance (s) = 12 m

Final velocity (v) =.?

We, can determine how fast the sprinter is running when he crosses the goal line as follow:

v² = u² + 2as

v² = 7² + (2 × –0.6 × 12)

v² = 49 + (–14.4)

v² = 49 – 14.4

v² = 34.6

Take the square root of both side

v = √34.6

v = 5.88 m/s

Therefore, the sprinter is running at a speed of 5.88 m/s when he crosses the goal line.

The distance from the equilibrium position at which the kinetic and potential energies are equal can be calculated using the amplitude A of the oscillator and the equation for the potential energy of a harmonic oscillator, which is given by: U = (1/2)kx2.

To answer your question, let's first understand the terms involved:

1. Harmonic: refers to the motion of the oscillator, which is sinusoidal in nature.
2. Potential: The stored energy of the oscillator, related to its position in equilibrium.
3. Amplitude: The maximum displacement of the oscillator from its equilibrium position.

Now, let's find the position when the kinetic and potential energies are equal for a simple harmonic oscillator with amplitude A.
At maximum displacement, the potential energy is equal to the maximum value of kinetic energy, which is given by:

K = (1/2) mv2 = (1/2) kA2

where m is the mass of the oscillator and v is its velocity. Equating U and K and solving for x, we get:

(1/2)kx^2 = (1/2)kA^2
x^2 = A^2
x = A
Step 1: Write the expressions for potential energy (PE) and kinetic energy (KE).
PE = (1/2)kx^2, where k is the spring constant and x is the displacement from equilibrium.
KE = (1/2)mv^2, where m is the mass and v is the velocity of the oscillator.

Step 2: Equate the potential and kinetic energies.
(1/2)kx^2 = (1/2)mv^2

Step 3: Use the relationships for a simple harmonic motion to substitute v2.
v2 = (k/m)(A^2 - x^2), as the maximum potential energy is (1/2)kA2.
So, kx^2 = m(k/m)(A^2 - x^2)

Step 4: Simplify the equation.
x^2 = A^2 - x^2

Step 5: Solve for x.
2x^2 = A^2
x^2 = (1/2)A^2
x = A / sqrt(2)

So, when the kinetic and potential energies are equal, the simple harmonic oscillator is A/sqrt(2) far from its equilibrium position.

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The center of mass of the shaded region will be located at a distance of (3/4)a from the center of the larger circle, along the line connecting the centers of both circles.

The center of mass of an object can be thought of as the point where the object would balance perfectly. In this case, we have a circular region with a circular hole cut out of it, resulting in a shaded region.

When the shape has geometric symmetry, such as in this case, the center of mass will lie along the line connecting the centers of symmetry. Since the larger circular region has a radius of a, the center of this circle is at the origin (0, 0).

By symmetry, we know that the center of mass of the shaded region will also lie on this line. The distance from the origin to the center of mass of the shaded region can be determined by considering the ratio of the areas of the shaded region and the remaining circle.

The area of the shaded region is (3/4)π\(a^2\), and the remaining circle has an area of (1/4)π\(a^2\). Using the ratio of the areas, we find that the center of mass of the shaded region is located at a distance of (3/4)a from the origin along the line connecting the centers of both circles.

Therefore, the center of mass of the shaded region left over after cutting out a circular hole of radius a/2 from a circular region of radius a is located at a distance of (3/4)a from the center of the larger circle, along the line connecting the centers of both circles.

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Answer:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."

Explanation:

The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.

D sin θ = m λ

if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."

This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.

Hope this helps!

Answer:

the motion of an object at a constant speed along a circular path is known as Uniform circular motion

The substances that existed as solid flakes within the inner 0.3 au of the solar system before planets began to form are called dust grains.

Dust grains are the substances that exist as solid flakes within the inner 0.3 au of the solar system before planets began to form. Dust is composed of small solid particles that are less than 1mm in diameter. When dust particles collide and stick together, they form larger objects known as planetesimals. Over time, planetesimals collide and combine to form planets.

The inner 0.3 au of the solar system is a dust-rich environment. It is where the temperatures are high enough to vaporize water and other volatile materials, and the radiation pressure from the sun is strong enough to blow gas and dust away from the region. Therefore, the dust is concentrated in this region, providing the raw materials for planet formation.

The dust grains in the solar nebula also played a crucial role in the formation of the solar system. They served as the building blocks for the planets, and their composition gives us clues about the conditions in the early solar system. Scientists study dust grains found in meteorites to learn about the composition of the solar nebula and the conditions under which the planets formed.

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The mass of the gold is determined as 37,952.4 g.

The mass of the gold is calculated by applying the following formula as follows;

m = density x volume

The volume of the gold is calculated as follows;

Volume = L x w x h

where;

The volume = 25.4 cm x 10.16 cm x 7.62 cm = 1,966.45 cm³

The mass of the gold is calculated as;

m = 19.3 g/cm³ x 1,966.45 cm³

m = 37,952.4 g

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If we know the stall torque and no load speed of a pm dc motor, the maximum power motor can deliver is P max = 1/2* Tstall * 1/2* ω(no load).

A mechanical device that produces stall torque has an output rotational speed of zero. Electric motors are capable of developing torque when stalled.

However, as the current flowing is at its maximum while an electric motor is halted, they are vulnerable to overheating and potential damage.

The no load speed is the speed when running at nominal voltage and zero load.

So, the maximum power motor can deliver is P max = 1/2* Tstall * 1/2* ω(no load).

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Answer:

Option (a) is correct

Explanation:

The acceleration of an object is defined as the rate of change of velocity. Mathematically, it can be written as :

\(a=\dfrac{v-u}{t}\)

Where

v and u are final and initial velocity

It is clear that if there is some change in velocity, it means the object is experiencing either acceleration or deceleration. Hence, the correct option is (a).

Answer:

a

Explanation:

Kinetic energy rises along with the skater's pace. The kinetic energy rises as the speed falls.

A skateboard will have greater kinetic energy as it glides more quickly up or down a slope. Some of this kinetic energy, which was transformed into motion through friction, will be lost as heat when skaters reach the bottom of the ramp and begin travelling again in a horizontal direction (between surfaces).

As the skateboarder changes positions along the track and changes velocities, her potential energy is transformed into kinetic energy (KE), or the energy of motion. The system's total potential energy determines how much kinetic energy the skateboarder can have at any given time.

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Answer:

I feel like it is C

Explanation:

I think it is C because it involves plant, which is a form of life and it involves the amount of water being given to the plants.

Answer:

true

Explanation:

The reason galaxies that are distant from our galaxy move away from our galaxy more rapidly is more space expands between us and distant galaxies.

A galaxy is a group of millions of stars and their systems that are grouped due to gravitational forces.

According to the Big Bang theory, galaxies are expanding and separate among them.

In conclusion, the reason galaxies that are distant from our galaxy move away from our galaxy more rapidly is more space expands between us and distant galaxies.

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Answer:

the galaxies are moving as the universe keep expanding .

Explanation:

The value of x, y, and z in the equation for F is x = 1, y = 1/2, and z = 2.

To use the method of dimensions, we need to first identify the fundamental dimensions involved in the problem. The fundamental dimensions in this problem are:

Length (L)

Mass (M)

Time (T)

Now let's consider each term in the equation for F:

K is non-dimensional, so it doesn't have any fundamental dimensions.

a has dimensions of length (L).

p has dimensions of mass per unit volume, or density, which is mass (M) divided by length cubed (L³).

v has dimensions of length per unit time, or velocity, which is length (L) divided by time (T).

Using these fundamental dimensions, we can write the dimensional formula for each term in the equation for F:

[F] = M L T⁻² (force)

[K] = 1 (dimensionless)

[a] = L

[p] = M L⁻³

[v] = L T⁻¹

Substituting these dimensional formulas into the equation for F, we get:

M L T⁻² = \((KL)^x (ML^{-3})^{y} (LT^{-1}})^{z}\)

Simplifying, we can rewrite this as:

M L T⁻² = K \(M^y L^{x-3y+z} T^{-z}\)

Equating the dimensions of both sides, we get the following system of equations:

\(M = K M^{y}\)

Solving for x, y, and z, we get:

x = 1

y = 1/2

z = 2

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The length of the vibrating section of the violin string with a linear density of 0.710 g/m and tension of 160 N would be 9.89 cm.

The speed of a wave on a string is given by the equation:

v = √(T/μ)

where v is the wave speed, T is the tension in the string, and μ is the linear density of the string.

The wavelength of the fundamental mode of vibration of a string fixed at both ends is twice the length of the vibrating section. Therefore, the length of the vibrating section can be calculated as:

L = λ/2

where L is the length of the vibrating section, and λ is the wavelength of the fundamental mode of vibration.

Given T = 160 N and μ = 0.710 g/m, we have:

v = √(T/μ) = √(160 N / (0.710 g/m)) = 75.6 m/s

The fundamental frequency of a string fixed at both ends is given by:

f = v/2L

where f is the fundamental frequency.

At the fundamental frequency, the wavelength is twice the length of the vibrating section, so we have:

λ = 2L

Substituting λ = v/f and simplifying, we get:

L = v/(2f)

To find the length of the vibrating section, we need to know the fundamental frequency of the string. This can be calculated from the tension and linear density using the formula:

f = (1/2L) √(T/μ)

Substituting the given values, we get:

f = (1/(2L)) √(160 N / (0.710 g/m)) = (1/(2L)) √(160000 N/m^2 / 0.710 g/m) = (1/(2L)) √(2.25 × 10^5 Hz)

Simplifying, we get:

f = 0.0496/L Hz

Substituting this expression for f into the equation for L, we get:

L = v/(2f) = (75.6 m/s)/(2 × 0.0496/L) = 764 L

Solving for L, we get:

L = 0.0989 m = 9.89 cm

Therefore, the length of the vibrating section of the violin string is approximately 9.89 cm.

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This code prompts the user to enter the voltage and current values, creates a Circuit object with those values, increases the voltage using the IncreaseVoltage() member function .

```cpp

#include <iostream>

#include <iomanip>

using namespace std;

class Circuit {

public:

   Circuit(double voltageValue, double currentValue);

   void IncreaseVoltage();

   void Print();

private:

   double voltage;

   double current;

};

Circuit::Circuit(double voltageValue, double currentValue) {

   voltage = voltageValue;

   current = currentValue;

}

void Circuit::IncreaseVoltage() {

   voltage = voltage * 8.0;

   cout << "Circuit's voltage is increased." << endl;

}

void Circuit::Print() {

   cout << "Circuit's voltage: " << fixed << setprecision(1) << voltage << endl;

   cout << "Circuit's current: " << fixed << setprecision(1) << current << endl;

}

int main() {

   double voltageInput, currentInput;

   cout << "Enter the voltage: ";

   cin >> voltageInput;

   cout << "Enter the current: ";

   cin >> currentInput;

   Circuit* myCircuit = new Circuit(voltageInput, currentInput);

   myCircuit->IncreaseVoltage();

   myCircuit->Print();

   delete myCircuit;

   return 0;

}

```

In the modified code, the main function prompts the user to enter the voltage and current values. Then, a new Circuit object is created using the entered values, and the IncreaseVoltage() member function is called on that object.

Finally, the Print() member function is called to display the updated voltage and current values. The dynamically allocated memory for myCircuit is released using the delete operator at the end.

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The given statement is true. Because since the net electric field at point P is twice the electric field due to charge Q₂, this means that the electric field due to charge Q₁ must be three times greater than the field by charge Q₂. Therefore, R₁ = R₂.

The electric field, E is a vector quantity that exists in the vicinity of electric charges. The electric field, E exists in the region of space where an electric charge or a collection of charges is present. Electric charges have an effect on each other, which is accomplished via the electric field.

The electric field is a force field that surrounds the electric charges. If there are two charges Q₁ and Q₂ placed at a distance R from a point P, the electric field at P due to each charge is given by,

E₁ = (1/4πε₀) × (Q1/R²) and E2 = (1/4πε₀) × (Q2/R²)

The electric field, E, at point P due to both charges is given by,

E = E₁ + E₂

If the magnitude of the net electric field at P is twice the electric field due to charge Q₂, this means that the electric field due to charge Q₁ is three times greater than the field due to charge Q₂ and that charges Q₁ and Q₂ are of the same sign.

Therefore, R₁ = R₂ is true.

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The magnitude of the magnetic field inside the solenoid is 9.42 × 10^-4 T.

To find the magnitude of the magnetic field inside the solenoid, we can use the formula:

B = μ₀ * N * I / L

Where:
B is the magnetic field
μ₀ is the permeability of free space (constant)
N is the number of turns
I is the current
L is the length of the solenoid

Given:
Length of solenoid (L) = 1.00 cm = 0.01 m
Radius of solenoid (r) = 0.950 cm = 0.0095 m
Number of turns (N) = 21
Current (I) = 1.75 A

First, we need to calculate the effective length of the solenoid (Leff) using the formula:

Leff = L * (1 + r^2 / (2L^2))

Plugging in the values, we get:

Leff = 0.01 * (1 + (0.0095^2) / (2 * 0.01^2))

Leff = 0.01 * (1 + 0.000045 / 0.0002)

Leff = 0.01 * (1 + 0.225)

Leff = 0.01 * 1.225

Leff = 0.01225 m

Next, we can calculate the magnitude of the magnetic field using the formula mentioned earlier:

B = μ₀ * N * I / L

Plugging in the values, we get:

B = (4π × 10^-7 T·m/A) * 21 * 1.75 A / 0.01225 m

B = (4π × 10^-7 T·m/A) * 36.75 A / 0.01225 m

B = 3.14 × 10^-7 T·m/A * 36.75 A / 0.01225 m

B = 3.14 × 10^-7 T·m/A * 3000 T·m/A

B = 9.42 × 10^-4 T

Therefore, the magnitude of the magnetic field inside the solenoid is 9.42 × 10^-4 T.

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Answer:

New Hampshire

New York.

Virginia.

Minnesota.

Connecticut.

Vermont

New Jersey.

Massachusetts.

Answer:

Minnesota

Explanation:

It is FALSE to state that the distance from the point of no return to the intersection is the same no matter what speed you are going.

If you are 100 feet or fewer from the junction, you have passed "the point of no return" and cannot safely halt before the intersection. As a result, it is preferable to proceed through the junction at your present, legal speed, but with extreme caution.

The point of no return is the moment at which you can no longer halt without entering that space, which is two seconds away. Time yourself in this scenario. Red, green, and flashing lights Yellow arrows: Each traffic signal turn is a dangerous 4-second danger zone. The riskiest is a left turn that requires you to stop and surrender.

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According to the Huygens-Fresnel principle, when a wave with wavelength λ travels through a slit with a width d, no quiet spots will be created outside the room due to diffraction of waves when:

Therefore, since the wavelength of the sound waves is 50.0cm, the widest we can leave the door open so that everyne on the other side can hear it, is:

The length of the day on planet x would be 2.65 * 10⁴. On the other hand, the orbital speed of the satellite would be: 8.9 * 10³.

To find the length of the day on planet X we must take into account the information provided in the question and carry out the following mathematical procedure:

The distance between North Pole and equator over the surface that is equal to the one-fourth of the circumference.

\(d = \frac{2\pi R}{4} \\\\R= \frac{2d}{\pi} \\= \frac{(18850 km)}{\pi } \\\\= 12000 km\)

Therefore,

\(8.837 m/s^{2} = 9.509 m/s^{2} - (12000 * 10^{3} m ) 2 cos (rad)\\\\\\\sqrt{\frac{9.509 m/s^{2} - 8.8357 m/s^{2} }{12000 * 10^{3 m} } } \\\\= 2.2368 * 10^{-4} rad/s\)

Therefore, the time period of the planet is

\(T = \frac{2\pi }{2.368 * 10^{-4} rad/s} \\= 2.65 * 10^{4} seg\)

This is the time of one day in the planet X.

To calculate the orbital speed of the satellite we must take into account the data provided by the question and carry out the following procedure:

The mass of the planet X is,

M = gx-northpole R 2/G

\(M = \frac{gx - north pole R2}{ G}\\= \frac{(9.509 m/s^{2})(12000 * 10^{3})}{6.67 * 10 x^{-11} N * m^{2}/kg^{2} } \\= 2.053 * 10^{25} kg\\\)

The orbital speed of the satellite is,

\(Vo = \sqrt[2]{x} \frac{GM}{r} \\= \sqrt[n]{\frac{(6..67*10^{-11})(2.053 * 1025kg}{12000*10^{3} + 2000 * 10^{3}) m } } \\= 9.88 * 10 x^{3} m/s\)

The orbital speed of the satellite is,

\(T = \frac{2\pi r}{Vo} \\= \frac{2\pi (12000*10^{3} m + 2000 * 10^{3} m }{9.88 * 10^{3} m/s } \\=8.9 * 10^{3} s\)

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Answer:

\(a_1 = 1.446m/s^2\)

Explanation:

Given

\(m_1 = 70.0kg\) -- Mass of the boy

\(m_2 = 45.0kg\) -- Mass of the girl

\(a_2 = 2.25m/s^2\) -- Acceleration of the girl towards the boy

Required

Determine the acceleration of the boy towards the girl (\(a_1\))

From the question, we understand that the surface is frictionless. This implies that the system is internal and the relationship between the given and required parameters is:

\(m_1 * a_1 = m_2 * a_2\)

Substitute values for \(m_1, m_2\) and \(a_2\)

\(70.0 * a_1 = 45.0 * 2.25\)

Make \(a_1\) the subject

\(\frac{70.0 * a_1}{70.0} = \frac{45.0 * 2.25}{70.0}\)

\(a_1 = \frac{45.0 * 2.25}{70.0}\)

\(a_1 = \frac{101.25}{70.0}\)

\(a_1 = 1.446m/s^2\)

Hence, the acceleration of the boy towards the girl is 1.446m/s^2

Block 4 shows the greatest rise in temperature. The rise in the temperature is inversely proportional to the initial length. The concept of thermal expansion is used in the given problem.

Thermal expansion is the tendency of matter to alter its form, area, volume, and density in response to a change in temperature.

The formula for the thermal expansion is found as;

\(\rm \triangle L = L \alpha \triangle T \\\\\ \triangle T = \frac{\triangle L}{L\alpha}\)

The rise in the temperature is inversely proportional to the initial length;

Block 4 has the smallest initial length.

Hence, the block 4 has the greatest rise in temperature.

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