Aggregate blend composed of 65% coarse aggregate (SG 2.701), 35% fine aggregate (SG 2.625)
Compacted specimen weight in air = 1257.9 g, submerged weight = 740.0 g, SSD weight = 1258.7 g
Compacted specimen contains 5.0% asphalt by total weight of the mix with Gb = 1.030
Theoretical maximum specific gravity = 2.511
Bulk specific gravity of the aggregate __________
Bulk specific gravity of the compacted specimen__________
Percent stone __________
Effective specific gravity of the stone__________
Percent voids in total mix__________
Percent voids in mineral aggregate__________
Percent voids filled with asphalt__________

Answer:

When traveling at higher speeds (40 mph or faster), the most fuel-efficient way to keep the car cool is to follow these tips:

1. Use the vehicle's ventilation system: Instead of relying on air conditioning, use the car's ventilation system to circulate fresh air from outside. This helps to cool down the interior without putting extra load on the engine, thus saving fuel.

2. Close windows and sunroofs: To reduce wind resistance and drag, close all windows and sunroofs while driving at higher speeds. Open windows create drag, which can increase fuel consumption.

3. Park in the shade: Whenever possible, park your car in a shaded area to avoid excessive heating when it's not in use. This can help keep the car cooler and reduce the need for extra cooling when you start driving.

4. Use reflective sunshades or window tinting: Use reflective sunshades on your windshield and window tinting on side windows to reduce the amount of heat entering the car. This can help keep the interior cooler, reducing the need for excessive cooling while driving.

5. Maintain your vehicle: Regular maintenance, such as checking and replacing coolant, inspecting the radiator, and ensuring proper functioning of the engine cooling system, can help keep your car running efficiently and prevent overheating.

6. Plan your trips strategically: If possible, try to avoid driving during the hottest part of the day. By planning your trips to avoid peak temperatures, you can reduce the strain on your vehicle's cooling system and minimize the need for excessive cooling.

Remember that these tips are specifically focused on keeping the car cool while maintaining fuel efficiency at higher speeds. In certain circumstances, such as extremely hot weather, using the air conditioning sparingly may be necessary for passenger comfort, but it will increase fuel consumption.

Answer:

Emails and phone calls are both common forms of communication that are used in professional and personal settings. There are several similarities between email and phone calls:

1. Both are asynchronous forms of communication: Unlike instant messaging or face-to-face conversations, both emails and phone calls allow the sender or recipient to respond at their convenience. They don't require immediate attention or an instant response.

2. Both are written forms of communication: While phone calls rely on spoken words, emails are written. As a result, both can be used to convey detailed information and allow the sender to carefully consider their words before sending.

3. Both are forms of direct communication: Emails and phone calls both allow for direct communication between two parties. This can be beneficial for discussing sensitive information or resolving issues quickly.

4. Both can be used for formal and informal communication: Emails and phone calls can be used in both personal and professional contexts. They are both flexible forms of communication that can be adapted to fit different situations.

5. Both require attention to tone and etiquette: Just like with phone calls, emails require attention to tone and proper etiquette. Both forms of communication should be approached professionally and respectfully to ensure effective communication.

In conclusion, while there are differences between emails and phone calls, there are also similarities that make them useful communication tools. Both allow for direct, asynchronous communication and can be adapted to fit different situations.

Explanation:

The empty wheel tractor-scraper can maintain a speed of 15 mph on a 3% adverse grade as long as the rolling resistance is less than or equal to 1102 lb per ton.

Traction is basically the action of drawing or pulling something over a surface, especially a road or track.

From Figure 6.10, we can find the horsepower-to-weight ratio (HP/T) of the empty wheel tractor-scraper at a speed of 15 mph on a 3% adverse grade. For a 631G wheel tractor-scraper, the HP/T is approximately 0.045.

The rolling resistance (Rr) can be calculated using the equation:

Rr = (M x g x %grade) / V

Where M is the weight of the empty wheel tractor-scraper in tons, g is the acceleration due to gravity (32.2 ft/s²), %grade is the grade expressed as a decimal (3% = 0.03), and V is the speed in mph.

Assuming the weight of the empty wheel tractor-scraper is 46 tons (92,000 lb), we can calculate the maximum value of rolling resistance as follows:

Rr = (46 * 2000 * 32.2 * 0.03) / 15

Rr = 1102 lb

Thus, the answer is 1102 lb.

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Your question seems incomplete, the probable complete question is:

A wheel tractor-scraper is operating on a 3% adverse grade. Assume that no power derating is required for equipment condition, altitude, and temperature. Use equipment data from Figure 6.10. Disregarding traction limitations, what is the maximum value of rolling resistance (in lb per ton) over which the empty unit can maintain a speed of 15 mph?

On axonometric projection, all lines indicating height, width, and depth remain parallel.

In axonometric projection, which is a type of orthographic projection, the lines representing the three dimensions (height, width, and depth) of an object are drawn parallel to each other. This means that regardless of the distance or position of an object in relation to the viewer, the lines representing its dimensions do not converge or diverge.

Axonometric projection is commonly used in technical and architectural drawings to accurately represent three-dimensional objects on a two-dimensional surface. By keeping all lines parallel, it provides a consistent and proportional representation of the object's dimensions.

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Answer:

Option B

Explanation:

Atoms of ferromagnetic materials have north and south pole but these atoms are oriented in random directions due to which they do no exhibit magnetic properties until unless they are brought into influence of any external temporary or permanent magnetic field.

Under the influence of external magnetic force, the atoms of the ferromagnetic material get oriented in a particular direction.

Hence, option B is correct

Answer:

He probably tripped the wiring, when metal hits the electricity it creates a reaction that burns the wiring.

Explanation:

Answer:

yes they are naturally occurring substances.

Answer:

Explanation:where’s the question love?

Answer:they can cause detonation of the engine

Explanation:Answer and Explanation: High compression ratios are not used in spark-ignition engines as they can cause detonation of the engine. This detonation of an engine is commonly known as knocking. The engine knocking occurs especially when octane-rated fuel is utilized.

High compression ratios are not used in spark-ignition engines because they can cause irregular combustion, such as pre-ignition and knocking, which can reduce efficiency and damage the engine. Pre-ignition is when the fuel-air mixture ignites before the spark plug fires, and knocking is when the mixture detonates unevenly and creates pressure waves that can harm the engine components. These phenomena are more likely to occur when the compression ratio is high because the end-gas temperature and pressure are higher, which increases the reactivity of the mixture. To avoid pre-ignition and knocking, spark-ignition engines need to use high-octane fuels that have higher resistance to auto-ignition. However, high-octane fuels are more expensive and less available than low-octane fuels, which limits the use of high compression ratios in spark-ignition engines. Some possible ways to enable high compression ratios in spark-ignition engines are to use high-pressure gasoline direct injection, which can enhance the air-fuel mixing and flame speed, or to use fuels with high octane sensitivity, which can reduce the end-gas reactivity.

It is to be noted that all UHRS platforms are optimized for the popular kinds of internet browser applications.

The Universal Human Relevance System (UHRS) is a crowdsourcing platform that allows for data labeling for a variety of AI application situations.

Vendor partners link people referred to as "judges" to offer data labeling at scale for us. All UHRS judges are bound by an NDA, ensuring that data is kept protected.

A browser is a software tool that allows you to see and interact with all of the knowledgeon the World Wide Web. Web sites, movies, and photos are all examples of this.

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Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

\(z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}\)

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

\(50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

90.8777353 m = \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

ShopKey Pro presents repair information in a customized email template as well as auto repair software format or layout.

In regards to auto repair information tools, the solution that is known to be made available is the use of ShopKey® Pro with 1Search.

This is know  to be a kind of an exclusive auto repair software that helps to streamlines the search acts to bring closer a special  combination of OEM data as well as experience-based Real Fixes.

Therefore, based on the above, one can say that ShopKey Pro presents repair information in a customized email template as well as auto repair software format or layout.

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The correct answer is c. 24.Multiplexing enables a single T1 circuit to carry how many channels.

A T1 circuit has a total capacity of 1.544 Mbps, which is divided into 24 channels of 64 kbps each. This enables a single T1 circuit to carry up to 24 separate voice or data channels simultaneously through the use of multiplexing techniques.To elaborate further, T1 is a digital transmission technology used primarily in North America and Japan. It uses time-division multiplexing (TDM) to combine multiple voice or data signals into a single transmission path.In T1, each channel operates at a data rate of 64 kbps, which is sufficient for a single telephone conversation. The 24 channels are combined using TDM, where each channel is assigned a time slot in a repeating cycle. Each channel's data is sent in its assigned time slot, and the process repeats every 125 microseconds. This allows for the efficient use of the T1 circuit's total bandwidth.

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The concentration ratio for the PTC is approximately 1.48, and the length of the parabolic surface is approximately 5.2 meters.

To determine the concentration ratio and length of the parabolic surface for a Parabolic Trough Collector (PTC) with the given parameters, we can use the following formulas:

Concentration Ratio (CR) = Rim Angle / Aperture Angle

Length of Parabolic Surface (L) = Aperture^{2} / (16 * Focal Length)

First, let's calculate the concentration ratio:

Given:

Rim Angle (θ) = 80º

Aperture Angle (α) = 5.2 m

Concentration Ratio (CR) = 80º / 5.2 m

Converting the rim angle from degrees to radians:

θ_rad = 80º * (π / 180º)

CR = θ_rad / α

Next, let's calculate the length of the parabolic surface:

Given:

Aperture (A) = 5.2 m

Receiver Diameter (D) = 50 mm = 0.05 m

Focal Length (F) = A^{2} / (16 * D)

L = A^{2} / (16 * F)

Now we can substitute the given values into the formulas:

CR =\((80º * (π / 180º)) / 5.2 m\)

L = \((5.2 m)^2 / (16 * (5.2 m)^2 / (16 * 0.05 m))\)

Simplifying the equations:

CR ≈ 1.48

L ≈ 5.2 m

Therefore, the concentration ratio for the PTC is approximately 1.48, and the length of the parabolic surface is approximately 5.2 meters.

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It is correct to state that within the context of computer science, th e maximum number of outgoing flows a merge can have is Two (Option B)

Outgoing flows in computer science refer to the flow of data or control from a specific node in a system or process to other nodes or systems.

It refers to the movement of data or control signals out of a specific component or module and towards other components or modules within the same system or to other systems. Outgoing flows are a fundamental aspect of many computer systems and processes and are used to control and coordinate the actions of different components and systems.

They help to ensure that data and control signals are transmitted between different components of a system in an efficient and controlled manner.

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The area of the triangle when the base is 10 cm and the height is 7 cm is 35 cm.

To calculate the area of a triangle, the base, and height of the triangle are multiplied by 1/2.

Decide which side will serve as the triangle's base, then calculate the triangle's height from that base. After that, enter the height and base measurements you have into the formula.

A = kbh

36 = k (8)(9)

36 = k 72

36/72 = k 72/72

½ = k

A = (½)(10)(7)

A = (½)(70)

A = 35

Therefore, the area of the triangle is 35.

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Answer:

See explanation.

Explanation:

Since no figure was given I solved a problem that was similar to the one you described that I worked in my mechanics of materials class. The method should be very similar for your figure. See attached image for my work.

If it is subjected to the couple determine the maximum shear stress in regions AC and CB of the shaft. G st = 75 GPa. Than the answer will be 52Mpa.

We need to perform a two step process to obtain the maximum shear stress on the shaft. For the solid shaft,

P=2×pi×N×T/60 or T=60×p/2×pi×N

Where P=power transmitted by the shaft=50×10³W

N=rotation speed of the shaft in rpm=730rpm

Pi=3.142

T is the twisting moment

By substituting the values for pi, N and P, we get

T=654Nm or 654×10³Nmm

Also, T=pi×rho×d³/16 or rho=16×T/pi×d³

Where rho=maximum shear stress

T = twisting moment=654×10³Nmm

d= diameter of shaft= 40mm

By substituting T, pi and d

Rho=52Mpa

b. For a hollow shaft, the value for rho is unknown

T=pi×rho(do⁴-di⁴/do)/16

Rho=T×16×do/pi×(do⁴-di⁴)

Where

T= twisting moment=654×10³Nmm gotten above

do=outside shaft diawter=40mm

di= inside shaft diameter =30mm

Pi=3.142

Substituting values for pi, do, di and T.

Rho=76Mpa

Therefore, If it is subjected to the couple determine the maximum shear stress in regions AC and CB of the shaft. G st = 75 GPa. Than the answer will be 52Mpa.

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Answer:

Explanation:

Step one:

given data

V1=0.25m^3

T1=290k

P1=100kPa

V2=0.5m^2

T2=405k

P2=? final pressure

Step two:

The combined gas equation is given as

P1V1/T1=P2V2/T2

Substituting we have

(100*0.25)/290=P2*0.05/405

25/290=0.5P2/405

0.086=0.05P2/405

cross multiply

0.086*405=0.05P2

34.9=0.05P2

divide both sides by 0.05

P2=34.9/0.05

P2=698.3Kpa

Therefore the new pressure is 698.3Kpa when the gas is compressed

Using the de-goodman criteria and a design factor of 1.5, the shaft diameter will be 0.187 inch.

To find the shaft diameter, let's use the equation below:

`σa/σw = [1/(Ka)] [(E/2E')^(1/m)] `

We know that the load on the shaft is 50,000 lbf.

Using this, we can calculate the alternating stress on the shaft:

`σa = load / (π * (d^2)/4) `

where d is the shaft diameter, we need to find.

Alternating stress `σa = 50,000/(πd^2/4) = 64,041/d^2`

Factor `n = 1/1.5 = 0.67`

For the material of the shaft, Ka and m values can be obtained using Soderberg criterion:

`σw = (σm/Na)(σf/Nf) `where σm is the mean stress, Na and Nf are empirical factors, and σf is the fatigue strength at a specific number of cycles.

The ultimate strength of the shaft material is given as 140,000 psi.

The empirical factors Na and Nf can be calculated as:

`Na = (σm + σa)/σf and Nf = 2E6/10^6 = 2`

Substituting the values we get, `Na = (0 + 64041/d^2) / 60000 = 1.067/d^2 and Nf = 2`

We can now calculate σw using Soderberg criterion:

`σw = (σm/Na)(σf/Nf) `Mean stress σm can be taken as half of ultimate strength σut = 70000 psi.

Then, `σw = (70000/1.067d^2)(60000/2) = 17630/d^2`

Using De-Goodman criteria, substituting the values of σa/σw and factor m, we get:

`1.5 = [1/Ka] [(E/2E')^(1/m)] = [1/Ka] [(29*10^6)/(2 * 60,000)^(1/3)] `

We can simplify the above equation by substituting Ka and solving for the shaft diameter `d`. `d = 0.187 inch`

Hence, the shaft diameter is 0.187 inch.

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You need to configure a DNS A record for the intranet site's hostname to map it to the correct IP address.

What is DNS?
DNS, or Domain Name System, is an internet protocol used to translate domain names, such as www.example.com, into IP addresses, such as 192.168.2.1, that computers can understand and use to communicate with each other. DNS acts as a directory service, providing a link between domain names and their associated IP addresses, which are stored and managed in a distributed database system. DNS is essential for the functioning of the internet, allowing users to access websites and services without having to remember their numerical IP addresses. DNS works by querying a distributed network of DNS servers, which maintain a database of IP addresses and their corresponding domain names. When a DNS query is made, the DNS server looks up the requested IP address in its database and returns it to the user. DNS is an important part of the internet infrastructure and is used by many different types of services, such as email, web hosting, and streaming services.

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Answer:

Gasoline is injected directly into the cylinder.

Explanation:

In a direct injection system, the air and gasoline are not pre-mixed. Rather, air comes in using the intake manifold, while the gasoline is injected directly into the cylinder.

Answer:

a) n > 8600 bytes

b) 9500 bytes

c) The results calculated above a very sensitive to the ratio of the packet size to the packet header

d) the model is accurate because N > 9500 bytes

Explanation:

A) The filesize n B

no of bytes transferred = no of packets * packet size

                                      = ( n / 1000 ) * 1024

circuit switching

number of bytes transferred = file size + two packets

                                               = n + (2 * 1024)

comparing both equations

= n + (2048)  <  ( n/1000) * 1024

=  24 * n / 1000 > 2048

hence: n > 8600 bytes

B )  The filesize n B is the total latency incurred before the entire file arrives at the destination less for circuits than for packets

packet switching :

Latency = ( n/1000 ) * ( 1024 * 8 bits/4 Mbps ) + 3 * ( 1024 * 8 bits / 4 Mbps) s + (3 * 1 ms + 4 * 2 ms)

=  ( n * 2 * 1026 / 10^6 )ms + 6.144 + 11

Circuit switching

Latency = ( n * 8 bits/ 4 Mbps ) s + 4 * 2 ms + 2 * ( 4 * 1024 * 8/4 mbps + 3 * 1 ms + 4 * 2 ms )

= n * 2*10 n3 / 10^6 ms + 8 + 2 * ( 19.1920 )

= 2n / 10 n3 + 38.3840

therefore for n to meet the requirement

( n * 2 * 1024 / 10 n6 ) ms + 17.144 > 2n / 10 n3 + 38.384

hence N > 9500 BYTES

C ) The results calculated above a very sensitive to the ratio of the packet size to the packet header

D) The model is accurate because N > 9500 bytes

and It doesn't ignore any important considerations

19.97kbps is the maximum achievable data rate.

The maximum data rate in a noisy channel as = B*log_2^(1+S/N)

Therefore, we get S/N first by 20dB=10log_10^(S/N) ==> S/N=10^2=100

Substitude S/N  

we get the maximum bps as 3k*log2^(1+S/N)=

                                               3log2^101kbps = 19.97kbps.

The maximum achievable data rate is 19.97kbps.

What are Binary signals ?

        The binary signal has only two possible values, making it the simplest signal of any kind that may be used to carry messages. The binary numbers, or bits, 1 and 0, are used to represent these values. A simple method of generating a random binary signal is to take Gaussian white noise, filter the noise for the desired spectra and then convert the noise to a binary signal by taking the sign of the filtered signal.

        The data rate is the number of bits transmitted per second from one device to another or across a network. Typically, data rates are expressed in bits per second or bytes per second. Bit rate is defined as the transmission of a number of bits per second. Baud rate is defined as the number of signal units per second . Bit rate is also defined as per second travel number of bits. Baud rate is defined as the number of signal units per second. Baud rate is also defined as per second number of changes in signal. Baud rate can determine the amount of bandwidth necessary to send the signal.

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The maximum achievable data rate is 19.97kbps. The binary signal is the simplest signal that may be used to transmit messages because there are only two possible values for it.

The maximum data rate in a noisy channel as = B*log_2^(1+S/N)

where B is the bandwidth

S/N is the signal-to-noise ratio. Usually, S/N is given in "decibel", not just a ratio. decibel is calculated by dB=10log_10^(S/N)

Therefore, we get S/N first by 20dB=10log_10^(S/N) ==> S/N=10^2=100

Substitude S/N  

we get the maximum bps as 3k*log2^(1+S/N)=

                                              3log2^101kbps = 19.97kbps.

The maximum achievable data rate is 19.97kbps.

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The color code of a resistor (from band 1 through band 4) is brown-black-orange-gold. The approximate value of the resistor is 10 kohms (10,000 ohms). Therefore the correct option is option C.

The first band denotes the first digit of the resistance value, the second band is the second digit of the resistance value, the third band is the number of zeros in the resistance value, and the fourth band is the tolerance value.

The formula for calculating the resistance value of a resistor is given as, R = AB × 10^C± D Where, A = First digit of the resistance value B = Second digit of the resistance value C = Number of zeros in the resistance value D = Tolerance value

If we use the given color code of the resistor (brown-black-orange-gold), we get the following values: A = 1B = 0C = 1 (number of zeros = 3)D = ± 5%Now, applying the formula of resistance, we get,

R = AB × 10^C= 10 × 10^3= 10,000 ohms or 10 kohms (since 1 kohm = 1000 ohms) Therefore, the approximate value of the resistor is 10 kohms (10,000 ohms). Therefore the correct option is option C.

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When the process is in control but does not meet specification, it is referred to as a special cause error.

In statistical process control, a process is considered to be in control when it operates within the defined limits and shows only random variations. However, when a process is in control but does not meet the desired specifications, it indicates the presence of a special cause error.

Special cause errors are attributed to specific factors or events that cause the process to deviate from the expected outcome. These errors are typically unpredictable and require investigation and corrective action to bring the process back within the desired specifications.

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True. According to the International Building Code (IBC), a special inspection of formwork for concrete construction is required to be continuous. The IBC outlines guidelines and regulations to ensure the safety and stability of structures, including the inspection process for various construction elements.

For concrete construction, formwork plays a crucial role as it shapes and supports the concrete until it hardens.
The continuous special inspection involves monitoring the formwork installation, shoring, and bracing to ensure that it complies with the approved design and applicable codes. This inspection helps to identify any potential issues or deviations from the design, allowing for prompt corrective actions. The goal of continuous special inspection is to minimize the risk of structural failures and to maintain the integrity of the construction process.

In summary, the IBC requires continuous special inspection for formwork in concrete construction to ensure the safety, stability, and compliance of the structure with the approved design and relevant codes.

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Answer:

Static Model

Explanation:

In the radio communication system, multipath is the propagation phenomenon that causes diffusion or reflection in multiple different directions of a signal.

Multipath is a propagation mechanism that impacts the propagation of signals in radio communication. Multipath results in the transmission of data to the receiving antenna by two or more paths. Diffusion and reflection are the causes that create multiple paths for the signal to be delivered.

Diffraction occurs when a signal bends around sharp corners; while reflection occurs when a signal impinges on a smooth object. When a signal is received through more than one path because of the diffraction or reflection, it creates phase shifting and interference of the signal.

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