The 1400-kg mass of a car includes four tires, each of mass (including wheels) 34 kg and diameter 0.80 m. Assume each tire and wheel combination acts as a solid cylinder.
A. Determine the total kinetic energy of the car when traveling 92 km/h .
B. Determine the fraction of the kinetic energy in the tires and wheels.

Explanation:

Choose a small section of the book to measure. This section should be near the spine of the book, where the pages are tightly bound together.

Hold the ruler perpendicular to the book's spine, so that it is parallel to the pages you want to measure.Carefully close the book so that the pages you want to measure are sandwiched between the ruler and the spine.

Read the measurement on the ruler where the top and bottom of the pages meet. This measurement represents the thickness of the entire stack of pages.

Estimate the number of pages in the stack by dividing the height of the stack by the measurement you just took. For example, if the stack is 2 centimeters high and the measurement you took is 0.1 centimeters, then you can estimate that the stack contains approximately 20 pages.

Finally, divide the thickness of the entire stack by the number of pages in the stack to estimate the thickness of one paper. For example, if the stack is 0.1 centimeters thick and contains 20 pages, then the estimated thickness of one paper is 0.005 centimeters or 0.05 millimeters.

Answer:

40.6789km

243.098km

Explanation:

Answer:5.45X10^3m

Explanation:So use the formula,v= fλ

3X10^8=5.5X10^4λ what Im saying is divide both and u should get 5454.54m but do sig figs to get answer

Answer:

5.45X10^3m

Explanation:

Answer: They have different rigidities.

Explanation:

The statement "It is always necessary to use more precise instruments when redoing an experiment to get better results" is false. While using more precise instruments can often lead to improved accuracy and reliability in scientific experiments, it is not always necessary or the sole factor that determines better results. Several other factors contribute to the quality of experimental outcomes, and the use of more precise instruments is just one aspect.

Precision in instruments refers to the level of detail and accuracy with which measurements can be made. It allows for smaller increments of measurement and reduces the potential for errors. When redoing an experiment, using more precise instruments can help reduce measurement errors and increase the level of detail captured. This can be particularly important in experiments that require precise measurements, such as those involving small quantities or sensitive reactions.

However, there are situations where the precision of instruments may not be the primary concern or where it may not significantly impact the results. For example, in experiments where the variables being measured have a large magnitude or natural variability, the use of extremely precise instruments may not yield significantly different results compared to instruments with slightly lower precision. In such cases, other factors like experimental design, sample size, controls, and methodology may have a more significant impact on the quality of results.

Furthermore, there may be instances where the cost, availability, or practicality of using more precise instruments outweighs the potential benefits. Precise instruments are often more expensive, require specialized training for operation, and may have limited availability in certain settings. In such cases, researchers may need to make trade-offs between precision and other factors like cost, feasibility, or time constraints.

In conclusion, while using more precise instruments can generally improve the quality of experimental results, it is not always necessary or the sole determinant of better outcomes. Factors such as experimental design, sample size, controls, methodology, and the nature of variables being measured also play significant roles in obtaining accurate and reliable results. Scientists need to consider the specific requirements and constraints of their experiments and make informed decisions about instrument precision based on a comprehensive evaluation of these factors.

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The cart A experienced the greatest impulse.

The given parameters;

Impulse is defined as the product of the applied force  and time of action of the applied force.

J = Ft

where;

J is the impulse

F is the applied force

t is the time of action

The impulse experienced by cart A;

J = 2 x 2 = 4 Ns

The impulse experienced by cart B;

J = 1 x 3 = 3 Ns

Thus, the cart A experienced the greatest impulse.

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Adjustment of the lens by the ciliary body is accommodation.

The correct option is D.

The lens is stretched and its refractive strength is changed by the zonule, a collective term for the suspensory ligaments that the ciliary bodies anchor. Relaxation of the zonule allows for accommodation for nearby items.

A section of the central portion of the eye's wall. The ring-shaped muscle that modifies the curvature of the lens whenever the eye focuses is part of the ciliary body, which is located behind the iris. Additionally, it creates the transparent liquid that fills the area between the iris and the cornea.

There are muscles, blood vessels, and epithelium in the ciliary body.

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Answer:

When you take the compass away from the bar magnet, it again points north. So, we can conclude that the north end of a compass is attracted to the south end of a magnet. ... In Experiment 2, when you move the north pole of a magnet toward the south pole of the other magnet, the two magnets attract.

Answer:

Gravity. An object is moving across a surface, but it does not gain or lose speed.

Explanation:

The basic idea. Physicists see gravity as one of the four fundamental forces that govern the universe, alongside electromagnetism and the strong and weak nuclear forces.

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The three different types of meteorites originate from various locations on their parent bodies. Because only a small portion of a parent body contains both stone and iron, stony-iron meteorites are uncommon.

Pallasites were most likely formed in a relatively small area within these differentiated asteroids, which may account for their scarcity. Only about 45 known pallsites exist among the many thousands of identified meteorites.

More than 95% of meteorites that are seen to crash to Earth are made of stone. The two types of them are chondrites and achondrites. The majority of both types also contain metallic iron in small, dispersed grains, but silicate minerals make up the majority of both.

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According to Lenz's law, the induced current in a circuit always flows in a direction that opposes the original change in the external magnetic flux.  Option (a) oppose is the correct answer.

When there is a change in the magnetic field passing through a closed circuit, it induces an electromotive force (EMF) that causes an electric current to flow. Lenz's law states that the direction of this induced current is such that it creates a magnetic field that opposes the change in the magnetic field that caused it. This principle of opposing the change is a manifestation of the conservation of energy. The induced current acts to counteract the original change in the magnetic flux, thereby maintaining the stability of the system.

Option (a) oppose is the correct answer.

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Explanation:

a) Fixed points are the temperatures at which a thermometer is calibrated. They can refer either to the actual temperatures used for calibration, or the thermometer readings at those temperatures.

__

b) Fundamental interval is the difference between the fixed points. As with fixed points, it can refer either to the difference in actual temperature, or the difference in the corresponding thermometer readings.

Answer:The height that you can jump has an inverse relationship to gravitational field strength. So if the field strength doubles, the height

Explanation:

Answer:

Work is said to be done when there is change in the position of the object due to the force applied on it.

Explanation:

Hope this helped! Nya~ :3

Therefore, to focus on an object 0.540 m in front of the lens, the lens-to-film distance should be 37.6 mm. This means that we need to move the lens closer to the film by 2.4 mm (40.0 mm - 37.6 mm) in order to bring the image of the closer object into sharp focus.

When a camera lens is focused on a distant object, the distance between the lens and the film (or digital sensor) is equal to the focal length of the lens. This is because the lens is designed to bring parallel rays of light to a focus at a specific distance from the lens, which is called the focal length.

In this case, we are given that the lens-to-film distance for the distant object is 40.0 mm. This means that the focal length of the lens is also 40.0 mm, assuming that the lens is a thin lens with negligible thickness.

To focus on an object 0.540 m in front of the lens, we need to adjust the lens-to-film distance to bring the image of the object into sharp focus on the film. The formula that relates the lens-to-film distance, the object distance, and the focal length of the lens is:

1/f = 1/d_o + 1/d_i

where f is the focal length, d_o is the object distance, and d_i is the image distance (which is equal to the lens-to-film distance for a thin lens).

We can rearrange this equation to solve for d_i:

1/d_i = 1/f - 1/d_o

d_i = 1 / (1/f - 1/d_o)

Plugging in the values we know, we get:

d_i = 1 / (1/40.0 mm - 1/0.540 m)

d_i = 37.6 mm

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Answer:

Density= 2.78 g/cm³

Relative density=2.8

Explanation:

To calculate the density of the cube we have to use the formula ρ=mass/volume

ρ stands for density.

So now we don't have the volume of the cube and to find the volume of the cube we have to use the formula a³

3³= 9 cm³

Now plug in the values. ρ= 25 g/9 cm³

ρ= 2.78 g/cm³

To find the relative density, we have to use the formula ρsample/ρH20

The sample means the density of the substance earlier. We do not know the density of water but it is constant at 997 kg/m³.

Now we have to make the units same so you change the unit of the density of cube to kg/m³

So, 25/1000= 0.025 kg

9/100×100×100 (because cm³ which means that there should be 3 meters to change the unit and to conver cm to meter we need to divide by 100 so 9cm/100, 9cm²/100×100, 9cm³/100×100×100)

=0.000009 m³

The new density= 0.025 kg/ 0.000009 m³

= 2777.78 kg/m³

Now plug the values into the formula:

relative density= 2.777.78 kg/m³ / 997 kg/m³

=2.8

There is no unit since kg/m³ and kg/m³ cancels

The ball will fly 0.201 m above its original position when released.

To answer this question, we need to use the conservation of energy principle. When the spring is compressed by 0.150 m, it has potential energy stored in it. When the spring is released, this potential energy is converted to kinetic energy, which is then transferred to the ball.

a) To find the upward speed of the ball, we need to use the formula for potential energy stored in a spring: PE = (1/2)kx^2

where PE is the potential energy, k is the spring constant, and x is the compression of the spring. Substituting the given values, we get:
PE = (1/2)(895 N/m)(0.150 m)^2 = 16.02 J

This potential energy is converted to kinetic energy as the spring is released. The formula for kinetic energy is: KE = (1/2)mv^2

where KE is the kinetic energy, m is the mass of the ball, and v is its upward speed. Substituting the given values and equating the two energies, we get:
(1/2)mv^2 = 16.02 J
v^2 = (2 x 16.02 J) / 0.30 kg
v = 6.23 m/s

Therefore, the upward speed that the spring can give to the ball when released is 6.23 m/s.

b) To find the height above the original position that the ball will reach, we can use the formula for gravitational potential energy: PE = mgh

where h is the height above the original position. At the maximum height, the ball will have zero kinetic energy and all of its potential energy will be converted to gravitational potential energy. Equating the two energies, we get:
(1/2)mv^2 = mgh
h = (v^2 / 2g)

Substituting the given values, we get:
h = (6.23 m/s)^2 / (2 x 9.81 m/s^2) = 0.201 m

Therefore, the ball will fly 0.201 m above its original position when released.

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If a sleeve bearing fails in a small, fractional horsepower motor, B. the bearings must be replaced.

Sleeve bearings, also known as plain bearings, are commonly used in small fractional horsepower motors to support the motor shaft and reduce friction. Over time, due to wear and tear or insufficient lubrication, sleeve bearings can fail, resulting in increased friction, heat, and potentially motor malfunctions. When a sleeve bearing fails in a motor, simply replacing the bearings is the appropriate course of action.

By replacing the worn-out or damaged sleeve bearings, the motor can regain proper support and smooth rotation, reducing friction and preventing further damage to the motor components. It is essential to choose high-quality bearings that are compatible with the motor specifications to ensure optimal performance and longevity.

It is important to note that other options listed in the question, such as replacing the entire motor (option A), using a larger capacitor (option C), or reversing the motor rotation (option D), are not necessary or effective solutions specifically for addressing sleeve bearing failures. These options may be applicable in different scenarios but are not directly related to resolving bearing issues.

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The work done by the spring on the block is -1 Joule.

In the given question, we are given that a block of mass 2 kg is attached to a spring whose spring constant is k = 8 N/m. The block slides on an incline with θ = 37º and uk = 0.45. Also, the block starts at rest with the spring in relaxed position.

We have to calculate the work done by the spring on the block if the block has slid a distance of 0.5 m down the incline.

Using energy conservation, we know that the potential energy of the spring gets converted into kinetic and potential energies of the block in the presence of friction. Therefore, work done by the spring = - change in potential energy of the spring

The work done by the spring on the block can be calculated by finding the change in the potential energy of the spring. Using the formula for the potential energy of the spring, we get:

Potential energy of the spring = \(1/2 k x²\)

where k is the spring constant and x is the displacement of the spring from its relaxed position. Here, the spring is stretched by a distance x = 0.5 m when the block slides down the incline. Therefore, the potential energy of the spring at this position is:

Potential energy of the spring = 1/2 k x²

= 1/2 × 8 × (0.5)²= 1 Joule (approx)

Now, using the work-energy principle, we know that the work done by all the forces acting on the block is equal to the change in its kinetic energy. Therefore, the work done by the spring on the block can be calculated as follows:

Total work done on the block = Change in kinetic energy of the block + Work done by friction

where the change in kinetic energy of the block is given by:

Change in kinetic energy of the block = 1/2 m v²

where m is the mass of the block and v is its final velocity. Here, the block starts from rest, so its initial velocity is zero. Also, we know that the block slides a distance of 0.5 m down the incline. Therefore, the height of the incline is given by:\(h = x sin θ= 0.5 sin 37º= 0.3 m\)

Now, using the conservation of energy, we can write:

Potential energy at the top of the incline = Kinetic energy at the bottom of the incline + Potential energy lost due to friction + Work done by the spring

\(mgh = 1/2 m v² + μk N s\)

- 1 + 1= 1/2 m v² + μk

m g h - 1 + 1

where m is the mass of the block, g is the acceleration due to gravity, μk is the coefficient of kinetic friction, N is the normal force acting on the block, s is the distance moved by the block, and h is the height of the incline. Here, the normal force acting on the block is given by:

N = m g cos θ= 2 × 9.8 × cos 37º= 15.53 N (approx)

Therefore, substituting the given values, we get:

\(2 × 9.8 × 0.3 = 1/2 × 2 × v² + 0.45 × 15.53 × 5 - 1 + 1\)

= 1 × v² + 6.9875 v - 6.37375

where we have taken the positive direction to be downwards.

Solving this equation for v, we get:

v = -1.965 m/s (approx)

Therefore, the change in kinetic energy of the block is given by:

Change in kinetic energy of the block = 1/2 m v²= 1/2 × 2 × (-1.965)²= 3.8385 J (approx)

Finally, substituting the given values, we get:

Work done by the spring = - Change in potential energy of the spring= - 1 J (approx)

Therefore, the work done by the spring on the block is -1 J (approx).

The work done by the spring on the block is -1 Joule.

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Given :

An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west.

It starts at the 14-inch mark, crawls to the 20-inch mark, then moves to the 16-inch mark.

To Find :

The total distance the ant traveled.

Solution :

Total distance travelled by ant = (distance between 14 and 20 inch mark) +

                                                      (distance between 20 and 16 inch mark)

Total distance = (20-14 ) + ( 20-16) = 6 + 4 = 10 inch.

Therefore, total distance the ant traveled is 10 inch.

Hence, this is the required solution.

The fraction of light from the Sun intercepted and reflected by the Earth is approximately 4.26 x 10⁻⁵.

To calculate the fraction of light from the Sun intercepted and reflected by the Earth, we need to compare the cross-section of the Earth to the area of a sphere centered on the Sun with a radius equal to the radius of Earth's orbit.

The cross-section of the Earth can be calculated as the area of a circle with radius REarth, which is option (c) R Earth.

The area of a sphere centered on the Sun with a radius r is given by 4πr², where r is the radius of the Earth's orbit. Therefore, the area of the sphere centered on the Sun with a radius equal to the radius of Earth's orbit is 4π(149.6 x 10⁶ km)²= 2.83 x 10²³ m².

The ratio of the cross-section of the Earth to the area of the sphere is A1/A2 = πREarth² / 4πr² = (REarth/r)². Using the radius of Earth's orbit in meters, r = 149.6 x 10⁹ m, and the radius of Earth, REarth = 6,371 km = 6.371 x 10⁶ m, we get A1/A2 = (6.371 x 10⁶ m / 149.6 x 10⁹ m)² = 4.26 x 10⁻⁵.

Therefore, by calculating we can say that the fraction of light is approximately 4.26 x 10⁻⁵.

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Answer:

They prolly dead lol.

Explanation:

XD THEY DEAD!!

The force pushing down on the table is 147,000 pounds.

     
Explanation:

   
To calculate the force pushing down on the table, we need to determine the area of the table in square inches, and then multiply that by the pressure exerted by the atmosphere.

   
The area of the table is 50 inches x 200 inches = 10,000 square inches.

     
The pressure exerted by the atmosphere is 14.7 pounds per square inch.

   
So the force pushing down on the table is:

10,000 square inches x 14.7 pounds per square inch = 147,000 pounds.

If normal atmospheric pressure is 14.7 pounds/sq in at the surface of the earth. The force pushing down on a table measuring 50 inches wide by 200 inches long is 147,000 pounds.

This is because the pressure is defined as force per unit area, and the area of the table is 50 inches x 200 inches = 10,000 square inches. So, if the normal atmospheric pressure at the surface of the earth is 14.7 pounds/square inch, then the force pushing down on the table is simply pressure x area = 14.7 pounds/square inch x 10,000 square inches = 147,000 pounds.

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The eccentricity of the object's orbit can be determined by using the ratio of its maximum angular speed to its minimum angular speed.

Let's denote the maximum angular speed as ω_max and the minimum angular speed as ω_min. We are given that the ratio of these two speeds is n:

n = ω_max / ω_min

The angular speed (ω) is related to the angular momentum (L) and the moment of inertia (I) of the object by the equation:

L = Iω

Since the object moves in an inverse square centripetal force field, the angular momentum (L) is conserved. Therefore, we can write:

L_max = L_min

Iω_max = Iω_min

The moment of inertia (I) can be expressed as the product of the mass (m) and the square of the distance (r) from the object to the axis of rotation:

I = mr^2

Substituting this into the equation above, we get:

m(r^2)ω_max = m(r^2)ω_min

Canceling out the mass (m) and the square of the distance (r^2), we obtain:

ω_max = ω_min

This implies that the maximum and minimum angular speeds are equal, contradicting the given ratio n = ω_max / ω_min. Therefore, there must be an error in the question or the provided information.

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Answer:

In a second class lever, the load is located between the effort and the fulcrum. If the load is closer to the fulcrum than the effort, then less effort will be required to move the load. If the load is closer to the effort than the fulcrum, then more effort will be required to move the load.

Explanation:

from what i learned, if its farther away from the load, its easier to lift, like a wheel barrel

Answer:The balloon's negative charges are attracted to the positive charges in the can, and so the can rolls toward the balloon.

Explanation:

Answer:

mass

Explanation:

Let's calculate the values based on the given information:

The area of the wooden block can be calculated by multiplying the length and width of one of its faces:

Area = Length * Width

Area = 40 cm * 20 cm

Area = 800 cm²

To convert to square inches, we can use the conversion factor 1 inch = 2.54 cm:

Area in square inches = Area in square centimeters / (2.54 cm/inch)²

Area in square inches = 800 cm² / (2.54 cm/inch)²

Area in square inches ≈ 124.03 in²

The volume of the wooden block can be calculated by multiplying its length, width, and height:

Volume = Length * Width * Height

Volume = 40 cm * 20 cm * 10 cm

Volume = 8000 cm³

To convert to cubic inches, we can use the conversion factor 1 inch = 2.54 cm:

Volume in cubic inches = Volume in cubic centimeters / (2.54 cm/inch)³

Volume in cubic inches = 8000 cm³ / (2.54 cm/inch)³

Volume in cubic inches ≈ 488.19 in³

The density of the wooden block can be calculated by dividing its mass by its volume:

Density = Mass / Volume

Density = 5 kg / 8000 cm³

To convert to grams per cubic centimeter (g/cm³), we can use the conversion factor 1 kg = 1000 g:

Density in g/cm³ = Density in kg/cm³ * 1000 g/kg

Density in g/cm³ = (5 kg / 8000 cm³) * 1000 g/kg

Density in g/cm³ ≈ 0.625 g/cm³

To convert to pounds per cubic inch (lb/in³), we can use the conversion factor 1 kg = 2.20462 lb and 1 inch = 2.54 cm:

Density in lb/in³ = Density in kg/cm³ * (2.20462 lb/kg) / (2.54 cm/inch)³

Density in lb/in³ = (5 kg / 8000 cm³) * (2.20462 lb/kg) / (2.54 cm/inch)³

Density in lb/in³ ≈ 0.036 lb/in³

Pressure is defined as force divided by area. In this case, we need more information to calculate the pressure. If the block is subjected to a specific force, we can divide that force by the appropriate surface area to find the pressure.

The pressure on the top surface of the wooden block depends on the force applied to it. Without information about the applied force, we cannot calculate the pressure.

Similarly, the pressure on the bottom surface of the wooden block depends on the force applied to it. Without information about the applied force, we cannot calculate the pressure.

The force on the top surface of the wooden block depends on the pressure applied and the surface area. Without information about the pressure or force applied, we cannot calculate the force.

The force on the bottom surface of the wooden block depends on the pressure applied and the surface area. Without information about the pressure or force applied, we cannot calculate the force.

Without the values for forces on the top and bottom surfaces, we cannot determine the difference between them.

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1. The area of the wooden block can be calculated using the formula for the surface area of a rectangular prism: SA = 2(lw + lh + wh), where l, w, and h are the length, width, and height of the block, respectively. Using the given dimensions, we can find the surface area in cm²:

SA = 2(40 × 20 + 40 × 10 + 20 × 10)

SA = 2(800 + 400 + 200)

SA = 2(1400)

SA = 2800 cm²

To convert cm² to in², we can use the conversion factor 1 in² = 6.45 cm². So, the area in in² is:

2800 ÷ 6.45 = 434.96 in² (rounded to two decimal places)

2. The volume of the wooden block can be calculated using the formula for the volume of a rectangular prism: V = lwh. Using the given dimensions, we can find the volume in cm³:

V = 40 × 20 × 10

V = 8000 cm³

To convert cm³ to in³, we can use the conversion factor 1 in³ = 16.39 cm³. So, the volume in in³ is:

8000 ÷ 16.39 = 487.61 in³ (rounded to two decimal places)

3. The density of the wooden block can be calculated using the formula: density = mass/volume. The mass of the block is given as 5 kg. To convert this to grams, we can use the conversion factor 1 kg = 1000 g. So, the mass in grams is:

5 kg × 1000 g/kg = 5000 g

Using the volume calculated in part 2, we can find the density in g/cm³:

density = 5000 g/8000 cm³

density = 0.625 g/cm³

To convert g/cm³ to lb/in³, we can use the conversion factor 1 g/cm³ = 0.0361 lb/in³. So, the density in lb/in³ is:

0.625 g/cm³ × 0.0361 lb/in³/g/cm³ = 0.0226 lb/in³

4. The pressure on the wooden block is given by the formula: pressure = force/area. To find the pressure, we need to know the force acting on the block. Since the block is simply resting on the tabletop, the force acting on it is due to its weight. Using the formula for weight: w = mg, where w is weight, m is mass, and g is the acceleration due to gravity (9.8 m/s²).

To find the weight in newtons (N), we can use the conversion factor 1 kg = 9.8 N. So, the weight of the block is:

5 kg × 9.8 N/kg = 49 N

Using the area of the block's base (40 cm × 20 cm = 800 cm²), we can find the pressure in N/cm²:

pressure = 49 N/800 cm²

pressure = 0.06125 N/cm²

To convert N/cm² to psi, we can use the conversion factor 1 psi = 6894.76 N/m². So, the pressure in psi is:

0.06125 N/cm² × (1 m²/10,000 cm²) × (1 psi/6894.76 N/m²) = 0.0089 psi (rounded to four decimal places)

5. The pressure on the top surface of the wooden block is the same as the pressure calculated in part 4: 0.06125 N/cm² or 0.0089 psi.

6. To find the pressure on the bottom surface of the block, we can use the formula: pressure = force/area. Since the bottom surface has the same area as the top surface, the pressure will also be the same: 0.06125 N/cm² or 0.0089 psi.

7. The force acting on the top surface of the wooden block is simply its weight, which we calculated to be 49 N in part 4.

8. The force acting on the bottom surface of the wooden block is also its weight, which we calculated to be 49 N in part 4.

9. The force on the bottom surface is equal in magnitude to the force on the top surface.

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