With what speed would a rock of mass 115g
have to be thrown if it were to have a
wavelength of 2.03 × 10−34 m?
Planck’s constant is 6.63 × 10−34 J · s.
Answer in units of m/s.

The Third Law of Thermodynamics states that as the temperature of a system approaches absolute zero, the entropy of the system approaches a minimum value.

Entropy is a thermodynamic quantity that measures a system's degree of disorder or unpredictability. The higher the entropy of a system, the more disorganised or unpredictable it is. According to the Third Law of Thermodynamics, the entropy of a system at absolute zero is the lowest value that can be achieved.

The Third Law of Thermodynamics and entropy are intimately related since the Third Law limits a system's entropy. The Third Law states that as a system approaches absolute zero, its entropy approaches a minimum value, which is specified by the Third Law.

The Third Law also suggests that reaching absolute zero temperature is unachievable, and so the minimal entropy is an unattainable goal.

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The length of the guitar string required to produce a fundamental frequency of 256 Hz is approximately 0.79 meters.

To determine the length of the guitar string required to produce a fundamental frequency of 256 Hz, we need to use the formula for the speed of waves in a string:

v = fλ

Where v is the speed of waves in the string, f is the frequency, and λ is the wavelength.

We know that the speed of waves in the particular guitar string is 405 m/s and we want to produce a fundamental frequency of 256 Hz. To find the wavelength, we rearrange the formula as:

λ = v/f

λ = 405/256

λ = 1.58 m

Now that we know the wavelength, we can find the length of the string required to produce this frequency using the formula:

L = n(λ/2)

Where L is the length of the string, n is the number of half-wavelengths that fit in the string, and λ is the wavelength.

Since we want to produce the fundamental frequency (1st harmonic), n = 1. Therefore:

L = (1)(1.58/2)

L = 0.79 m

So, the length of the guitar string required is approximately 0.79 meters.

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The book on the top shelf will have the greatest kinetic energy. Option C

When Noah's books books fall from their locations, they turn gravitational potential energy into kinetic energy.

The more gravitational potential energy a book has, the higher it is above the earth. As a result, the book on the highest shelf has the most gravitational potential energy. As the book falls, it has the more kinetic energy immediately before it strikes the ground.

Kinetic energy can be described as a type of energy an object has, given to movement or motion.

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Assuming that both swings have the same period and amplitude, the friend on the longer swing will be slightly behind you in the swing cycle after you complete one complete swing back and forth. This is because the longer swing has a greater distance to cover, so it will take a slightly longer time for it to complete one swing cycle.

As a result, your friend will be slightly lower than you but moving upward toward you when you are at your highest point, and slightly higher than you but moving downward away from you when you are at your lowest point. Therefore, the correct answer is: she will be slightly lower than you but moving upward toward you.

When two swings have slightly different lengths, they have slightly different periods. The period of a swing is the time it takes to complete one full swing cycle, which is the time it takes for the swing to go back and forth once. The period of a swing depends on its length, with longer swings having longer periods than shorter swings.

When two people start swinging at the same time, the person on the longer swing will take slightly longer to complete one full swing cycle. As a result, after one complete swing back and forth, the person on the longer swing will be slightly behind the person on the shorter swing in the swing cycle. This means that the person on the longer swing will be slightly lower than the person on the shorter swing when the shorter swing is at its highest point, and slightly higher than the person on the shorter swing when the shorter swing is at its lowest point.

Overall, the difference in height between the two people on the swings will be very small, but the person on the longer swing will be slightly lower and higher than the person on the shorter swing at different points in the swing cycle.

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The half-life of the radioactive material is approximately 4.96 years.

To find the half-life (h) of the radioactive material, we can use the given formula: [tex]y = y0 (0.5)^(t/h)[/tex]. We have y0 = 133 grams, y = 69 grams, and t = 7 years.

Plugging in the values, we get:
69 = 133 (0.5)^(7/h)

Now, we need to solve for h. First, divide both sides by 133:
69/133 = (0.5)^(7/h)

Next, take the logarithm base 0.5 of both sides:
[tex]log_0.5(69/133) = 7/h[/tex]
Now, solve for h:
[tex]h = 7 / log_0.5(69/133)[/tex]

Using a calculator, we find that:
h ≈ 4.96 years

So, the half-life of the radioactive material is approximately 4.96 years.

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In our finite-element solution the error in temperature is less than the error in the heat flux.
C) The error in temperature is less than the error in the heat flux.

In a finite-element solution, the temperature error is usually of a lower order compared to the heat flux error.

This is because the heat flux is determined by the gradient of the temperature field, and taking the gradient amplifies the error in the solution.

If the finite-element solution is based on the Galerkin approach, then the statement "the error in temperature is less than the error in the heat flux" is generally true.

This is because the Galerkin approach typically leads to more accurate solutions for the temperature field compared to the heat flux field.

Therefore, option C would be the correct answer in this case.

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To determine the magnitude of the angular momentum of the asteroid immediately after the collision, we need to consider the conservation of angular momentum. Assuming there are no external torques acting on the system, the initial angular momentum of the asteroid and the projectile before the collision should be equal to the final angular momentum of the combined object after the collision.

The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. Since the asteroid is assumed to be a point mass, we can use the formula I = mr^2, where m is the mass of the asteroid and r is the distance between the asteroid and the center of the sun.

Assuming that the asteroid and the projectile have the same mass and are traveling at the same speed, we can use the formula for the moment of inertia of a point mass to find the initial angular momentum:

Linitial = Iω = mr^2ω

After the collision, the two objects will stick together and form a single object with a new moment of inertia. We can use the formula for the moment of inertia of two point masses to find the final moment of inertia:

If = (m + m)r^2 = 2mr^2

Since the two objects will be traveling at the same speed after the collision, we can use the formula for the angular velocity of a rotating object to find the final angular momentum:

Lfinal = Ifωf = 2mr^2(ω/2) = mr^2ω

Therefore, the magnitude of the angular momentum of the asteroid immediately after the collision, as measured about the center of the sun, will be the same as the initial angular momentum:

|Lfinal| = |Linitial| = |mr^2ω|

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Since the image is inverted, the negative sign indicates that it is actually 8.25 cm below the principal axis, so the actual height of the image is 8.25 cm.

To determine the location and size of the image formed by a concave mirror, we can use the following ray diagram:

Draw a ray parallel to the principal axis that passes through the focal point F and then reflects back through the mirror along the same path.

Draw a ray that passes through the focal point F and then reflects back parallel to the principal axis.

Draw a ray that passes through the center of curvature C of the mirror and then reflects back along the same path.

The point where these three rays intersect is the location of the image.

Using this method, we can find that the image of the object is located at a distance of 0.50 m behind the mirror, and it is inverted and reduced in size.

To determine the size and location of the image using the mirror and magnification equations, we can use the following formulas:

1/f = 1/p + 1/q, where f is the focal length of the mirror, p is the distance of the object from the mirror, and q is the distance of the image from the mirror.

m = -q/p, where m is the magnification of the image.

Substituting the given values, we get:

1/0.75 = 1/2.0 + 1/q

Solving for q, we get q = 0.50 m.

m = -q/p

Substituting q = 0.50 m and p = 2.0 m, we get m = -0.25.

This means that the image is located 0.50 m behind the mirror, it is inverted, and its height is 0.25 times the height of the object. Therefore, the height of the image is:

h' = m * h

= -0.25 * 33 cm

= -8.25 cm.

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The axis is now perpendicular to the one in part A, the formula simplifies to:
[tex]I₂ = (1/12) * M * b^2[/tex]

For the first part of the question, we are asked to find the moment of inertia of the thin, rectangular sheet of metal with mass M, sides of length a and b, about an axis that lies in the plane of the plate, passes through the center of the plate, and is parallel to side b.

The moment of inertia for this case can be calculated using the formula:

[tex]I₁ = (1/12) * M * (a^2 + b^2)[/tex]

Since the axis is parallel to side b, the formula simplifies to:

[tex]I₁ = (1/12) * M * a^2[/tex]

For the second part of the question, we are asked to find the moment of inertia for an axis that lies in the plane of the plate, passes through the center of the plate, and is perpendicular to the axis in part A.

The moment of inertia for this case can be calculated using the same formula as before:
[tex]I₂ = (1/12) * M * (a^2 + b^2)[/tex]

Since the axis is now perpendicular to the one in part A, the formula simplifies to:

[tex]I₂ = (1/12) * M * b^2[/tex]

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The tension in string A (TA) is 154,335 N, and the tension in string B (TB) is 37,200 N.

Στ = 0
The torques involved are due to the tension in string A (TA), the weight of the bar (m1 * g), and the weight of the object (m2 * g).
τ_A = TA * d
τ_m1 = (m1 * g) * (L/2)
τ_m2 = (m2 * g) * x
Now, we set up the equation:
TA * d - (m1 * g) * (L/2) - (m2 * g) * x = 0
Solve for TA:
TA = ((m1 * g) * (L/2) + (m2 * g) * x) / d
TA = ((80 * 9.807) * (5.1/2) + (3500 * 9.807) * 4.9) / 1.2
TA = 154335.46 N (rounded to four significant figures)
Now, to find TB, we need to use the fact that the net vertical force is also zero since the bar is in equilibrium.
ΣFy = 0
TB - TA - m1 * g - m2 * g = 0
Solve for TB:
TB = TA + m1 * g + m2 * g
TB = 154335.46 + (80 * 9.807) + (3500 * 9.807)
TB = 37199.60 N (rounded to four significant figures)

Hence, the tension in string A (TA) is 154,335 N, and the tension in string B (TB) is 37,200 N.

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If two sources of radiation have waves leaving them that have the same wavelength, same frequency, and bear the same phase relationship to each other, these are called coherent sources.

A coherent source is one that generates light waves with the same frequency, wavelength, and phase or with a stable phase difference. When the maxima and minima positions are fixed and the waves superimpose, a coherent source produces long-lasting interference patterns.

Waves from coherent sources have the same frequency and amplitude and a consistent phase difference. A laser emits coherent, parallel, monochromatic light with continuous wave chains. The image below illustrates the coherent wave's pattern.

Specification-matched prisms, lenses, and mirrors are used to create the coherent source. Fresnel's biprism, Young's double-slit experiment, and Lloyd's mirror arrangement are a few of the methods that aid in the creation of coherent sources.

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The stem loop structure directly involved in transcriptional termination of the trp operon is:  formation of the 1-4 stem loop.(C)

The trpL attenuator region has several stem loop structures, but the 1-4 stem loop is specifically involved in transcriptional termination. When tryptophan levels are high, the ribosome stalls at the leader peptide coding region, allowing the formation of the 1-2 and 3-4 stem loops.

The 3-4 stem loop facilitates the formation of the 1-4 stem loop, which is the transcription terminator structure. This leads to the dissociation of the RNA polymerase, thus terminating transcription of the trp operon and preventing the synthesis of unnecessary tryptophan synthesis enzymes.(C)

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Rounding to two decimal places, the magnification of the image after calculations is -0.57

Using the formula for magnification, we have:

magnification = - image distance / object distance

where the negative sign indicates that the image is inverted.

We can use the thin lens equation to find the image distance:

1/f = 1/do + 1/di

where f is the focal length, do is the object distance, and di is the image distance.

Substituting the given values, we get:

1/75.2 = 1/74 + 1/di

Solving for di, we get:

di = 41.95 cm

Now we can substitute the values for di and do into the magnification formula:

magnification = - di / do

magnification = -41.95 cm / 74 cm

magnification = -0.57

Rounding to two decimal places, the magnification of the image is -0.57.

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To prevent the car from falling off the track at the top of the circular loop, it must reach a minimum height of 4.76 meters, where the force of gravity is balanced by the centripetal force.

To find the minimum height h so that the car will not fall off the track at the top of the circular part of the loop, we can use the centripetal force and gravitational force acting on the car. At the top of the loop, the gravitational force acting on the car will be equal and opposite to the centripetal force, otherwise, the car will either leave the track or fall off.

Let's assume the mass of the car is m, and the velocity of the car at the top of the loop is v. Then the gravitational force acting on the car is given by Fg = mg, where g is the acceleration due to gravity (approximately 9.8 m/s²). The centripetal force acting on the car is given by Fc = mv²/r, where r is the radius of the loop.

Since the car is at the top of the loop, the net force acting on it will be the difference between the gravitational force and the centripetal force, which is given by:

[tex]F_n_e_t[/tex] = Fg - Fc

= mg - mv²/r

For the car to remain on the track, the net force should be greater than or equal to zero. Thus:

[tex]F_n_e_t[/tex] >= 0

mg - mv²/r >= 0

mg >= mv²/r

g >= v²/r

Now, solving for the minimum height h, we can equate the gravitational potential energy of the car at the minimum height with the kinetic energy of the car at the top of the loop:

mgh = (1/2)mv²

Solving for h:

h = v²/(2g) + r

Substituting the given values of r = 20.0 m and g = 9.8 m/s², and assuming that the car will not lose any speed at the top of the loop, since there is no friction to cause energy losses, we get:

h = (3.5 m/s)²/(2 x 9.8 m/s²) + 20.0 m

h = 4.76 m

Therefore, the minimum height h so that the car will not fall off the track at the top of the circular part of the loop is 4.76 meters.

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The magnetic field amplitude of the electromagnetic wave with an electric field amplitude of 0.465 V/m is approximately 1.55 x 10^-9 T.

The magnetic field amplitude (B0) of an electromagnetic wave can be found using the relationship between the electric field amplitude (E0) and the speed of light (c) in a vacuum. The formula is:

B0 = E0 / c

Given that the electric field amplitude E0 is 0.465 V/m, and the speed of light c is approximately 3 x 10^8 m/s, the magnetic field amplitude B0 can be calculated as follows:

B0 = 0.465 V/m / (3 x 10^8 m/s)

B0 ≈ 1.55 x 10^-9 T (tesla)

So, the magnetic field amplitude of the electromagnetic wave with an electric field amplitude of 0.465 V/m is approximately 1.55 x 10^-9 T.

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The yield strength of a material is significantly less than the "ideal" yield strength due to the presence of defects, dislocations, and imperfections in the material's microstructure.

In an ideal, perfect crystal lattice structure, the atoms are perfectly aligned, and the yield strength would be at its maximum value.

However, in reality, materials have dislocations, defects, and imperfections within their microstructure. These imperfections act as stress concentrators, making it easier for the material to deform when subjected to an external force.
When a material is subjected to stress, the dislocations within the material begin to move, causing plastic deformation. This movement of dislocations is hindered by the presence of defects and imperfections, which prevent the material from reaching its ideal yield strength.

Therefore, the actual yield strength of a material is lower than the ideal value due to the presence and interaction of these microstructural imperfections.
The yield strength of a material is significantly less than the "ideal" yield strength because the presence of defects, dislocations, and imperfections in the material's microstructure negatively affects the material's ability to withstand stress, leading to a lower yield strength value in practice.

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The closest distance of approach, d, the alpha particle will come to the nucleus is given by d = k * q₁ * q₂ / K.E. = k * 79 * e² / K.E

To find the closest distance of approach d that an alpha particle will come to a gold nucleus (Z=79), we can use the concepts of energy conservation and the Coulomb force.

1. An alpha particle (with two protons and two neutrons) is given an initial nonrelativistic velocity at a far distance from a gold nucleus (Z=79). At this far distance, we can assume that the potential energy due to the Coulomb force is negligible.

2. As the alpha particle approaches the gold nucleus, it experiences an electrostatic repulsive force due to the Coulomb interaction between the alpha particle's charge (2e) and the gold nucleus's charge (79e), where e is the elementary charge.

3. The total mechanical energy of the system is conserved. The initial kinetic energy (KE) of the alpha particle is converted to potential energy (PE) due to the Coulomb force as the particle approaches the gold nucleus.

4. At the closest distance d, the alpha particle's KE is minimum (zero), and its PE is maximum. The initial KE is equal to the maximum PE.

5. The Coulomb potential energy formula is:

PE = k * q₁ * q₂ / r,

where k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between them.

Here, q₁ = 2e, q₂ = 79e, and r = d.

6. Using the conservation of energy:

Initial KE = Maximum PE, .
[tex]\frac{k * q₁ * q₂}{r}[/tex] = K.E.

Rearranging for d:

d = k * q₁ * q₂ / K.E. = k * 79 * e² / K.E.

By following these steps and solving for d, the closest distance the alpha particle will come to the gold nucleus can be found.

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The center of gravity of the additional mass should be located at the weighted average of the individual centers of mass. This location ensures that the object or system remains in balance and maintains its stability.

1. Identify the object: First, identify the object or system for which you need to find the center of gravity.

2. Locate the individual masses: Locate the individual masses within the object or system, including the additional mass.

3. Calculate the individual centers of mass: Determine the individual centers of mass for each component of the object, including the additional mass, based on their geometric shapes and density distributions.

4. Calculate the total mass: Add up the masses of all the components, including the additional mass, to obtain the total mass of the object or system.

5. Determine the center of gravity: Calculate the weighted average of the individual centers of mass by multiplying each center of mass by its corresponding mass and then dividing the sum by the total mass.

The center of gravity of the additional mass should be located at the weighted average position calculated in Step 5.

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The phase angle at 93% power factor is 23.98 degrees, and the capacitive vars required to rectify the power factor to 93% is 58,712.7 VAR (capacitive).

To begin, we need to calculate the current (I) of the motor using the formula:
I = (horsepower x 746) / (sqrt(3) x voltage)
I = (100 x 746) / (sqrt(3) x 2400)
I = 144.84 amps
Next, we need to calculate the apparent power (S) of the motor using the formula:
S = sqrt(3) x voltage x I
S = sqrt(3) x 2400 x 144.84
S = 397,327.7 volt-amperes (VA)
Now, we can calculate the real power (P) of the motor using the formula:
P = S x power factor
P = 397,327.7 x 0.75
P = 297,995.3 watts
At 75% power factor, the phase angle (θ) is:
θ = arccos(power factor)
θ = arccos(0.75)
θ = 41.41 degrees
To calculate the capacitive vars (cvars) needed to correct the power factor to 93%, we can use the formula:
cvars = S x (tan(arccos(desired power factor)) - tan(arccos(actual power factor))))
cvars = 397,327.7 x (tan(arccos(0.93)) - tan(arccos(0.75)))
cvars = 58,712.7 VAR (capacitive)
At 93% power factor, the phase angle (θ) is:
θ = arccos(power factor)
θ = arccos(0.93)
θ = 23.98 degrees
Therefore, the phase angle at 93% power factor is 23.98 degrees and the capacitive vars needed to correct the power factor to 93% is 58,712.7 VAR (capacitive).

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The person would weigh 73.35 N on the moon.

The weight is equal to mass times the acceleration due to gravity.

On Earth, the acceleration due to gravity is approximately 9.81 m/s2, but on the moon, it is only 1.62 m/s2.

So, to calculate the weight on the moon, we multiply the mass (45 kg) by the acceleration due to gravity on the moon (1.62 m/s2).
Therefore, 45 kg x 1.62 m/s2 = 73.35 N.
The weight is equal to mass times the acceleration due to gravity.

Hence,  A person with a mass of 45 kg weighs 72.9 N on the moon.

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The views given in figure 1, it appears that view (c) shows the correct orientation of the needle for the noted current direction.

The correct orientation of the needle for the noted current direction in figure 1, we need to use the right-hand rule. If we point our right thumb in the direction of the current flow (from positive to negative), the direction in which our fingers curl represents the direction of the magnetic field around the wire.

Looking at the views in figure 1, we can see that the current flows from left to right. Therefore, the correct orientation of the needle for this current direction would be perpendicular to the wire, pointing either up or down depending on the direction of the magnetic field.

The views given in figure 1, it appears that view (c) shows the correct orientation of the needle for the noted current direction.

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The concentration of the solution as mass percentage, % (m/m), is an option (b) 14.6 %.

To express the concentration of a solution as % (m/m), we need to know the mass of the solute and the mass of the solution. In this case, we have a solution that contains 35 g of KBr dissolved in 205 g of water.

To calculate the concentration of the solution as the mass percentage, we need to divide the mass of KBr by the total mass of the solution and then multiply by 100:

% (m/m) = (mass of KBr / mass of solution) x 100

The mass of the solution is the sum of the mass of KBr and the mass of water:

mass of solution = mass of KBr + mass of water
mass of solution = 35 g + 205 g
mass of solution = 240 g

Now we can calculate the % (m/m) concentration of the solution:

% (m/m) = (35 g / 240 g) x 100
% (m/m) = 0.146 x 100
% (m/m) = 14.6 %

Therefore, the concentration of the solution that contains 35 g of KBr dissolved in 205 g of water as % (m/m) is 14.6 %.

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The magnetic field at point P is zero.

When two wires are placed parallel to each other and carry currents in opposite directions, they generate a magnetic field. The direction of the magnetic field at point P depends on the direction of the currents in the wires. According to the right-hand rule, the magnetic field produced by a current-carrying wire points in the direction perpendicular to both the direction of the current and the direction from the wire to the point in question.

, the currents in the wires have the same magnitude but opposite directions. Therefore, the magnetic fields produced by each wire cancel each other out along the horizontal axis, but add up along the vertical axis. The resulting magnetic field points upward, perpendicular to the plane of the wires and in the direction of direction 1.


When two wires have currents with the same magnitude but opposite directions, their magnetic fields are also opposite in direction. At point P, which is equidistant from both wires, the magnetic fields produced by each wire cancel each other out due to their opposite directions. This results in a net magnetic field of zero at point P.

The magnetic field at point P is zero because the magnetic fields produced by the wires with currents in opposite directions cancel each other out.

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The work done on the block is 2,000 J.

The energy converted into thermal energy is 1,000 J.

The work done on the block is calculated by applying the following formula.

W = F x d

where;

W = 200 N x 10 m

W = 2,000 J

The energy converted into thermal energy is equal o work done by friction force.

W = 100 N x 10 m

W = 1,000 J

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In the situation described, where a 90-kg man and a 60-kg boy are pushing on each other with one hand extended out in front and neither is moving, the magnitudes of the two forces exerted by their hands are equal.

The situation is based on Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction.

So, the force exerted by the man's hand on the boy's hand is equal in magnitude but opposite in direction to the force exerted by the boy's hand on the man's hand. Although their masses are different, the forces they exert on each other must balance to keep them from moving.

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To determine the direction of the electric field at point E, we need to understand the basic concept of an electric field.

An electric field is a vector quantity that shows the direction and magnitude of the force experienced by a positive test charge placed in the field.

The electric field lines originate from positive charges and terminate at negative charges. In this case, we don't have information about the charges or their locations.

However, if you provide more context or a diagram related to the points mentioned (G, B, H, C, F), I will be able to give you a more accurate and helpful answer.

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The magnitude of the magnetic field is 0.228 T when a straight wire of length 1.25 m carrying a 75 A current at a 30-degree angle with a uniform magnetic field experiences a force of 5.0 N.

When a current-carrying wire is placed perpendicular to a uniform magnetic field, a force is exerted on the wire. In this problem, the wire is at an angle of 30 degrees to the magnetic field, which means that the force on the wire will be less than if the wire were perpendicular to the field.

Using the formula for the force on a current-carrying wire in a magnetic field, F = BILsin(theta), where B is the magnetic field strength, I is the current, L is the length of the wire, and theta is the angle between the wire and the magnetic field, we can solve for B.

Rearranging the equation to solve for B, we get B = F / (ILsin(θ)). Plugging in the given values, we get B = 5.0 N / (75 A x 1.25 m x sin(30 degrees)) = 0.228 T (teslas).

Therefore, the magnitude of the magnetic field is 0.228 tesla.

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To compare the kinetic energy of the two objects, we can use the formula KE = 1/2mv^2, where KE is kinetic energy, m is mass, and v is velocity.

For the truck, its mass is 22,000 kg and velocity is 130 km/h. We need to convert velocity to meters per second, which is 36.11 m/s. Plugging in the values, we get KE = 1/2 * 22,000 kg * (36.11 m/s)^2 = 14,930,557 J.

For the astronaut, its mass is 81.5 kg and velocity is 28,000 km/h. We need to convert velocity to meters per second, which is 7,777.78 m/s. Plugging in the values, we get KE = 1/2 * 81.5 kg * (7,777.78 m/s)^2 = 22,414,774,038 J.

As we can see, the kinetic energy of the astronaut in orbit is significantly greater than that of the truck moving at high speed. This is because kinetic energy is directly proportional to both mass and velocity, and the astronaut has a much higher velocity despite having a smaller mass.

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Car 2 has twice the kinetic energy.

Mass of the car 1, m₁ = m

Mass of the car 2, m₂ = 2m

v₁ = v₂ = v

Kinetic energy,

KE= 1/2 mv²

KE ∝ m

So,

KE₂/KE₁ = m₂/m₁

KE₂/KE₁ = 2m/m = 2

Therefore,

Kinetic energy of car 1 is twice that of the car 2.

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