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frosja888 [35]
3 years ago
8

Traffic at a roundabout moves

Engineering
1 answer:
sweet-ann [11.9K]3 years ago
3 0

Answer:

b-counter-clockwise

Explanation:

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iren [92.7K]

Answer:

7

Explanation:

A quotient is the answer to a division.For example,the quotient of 10 is 2 and 5 because 5÷10=2.

5 0
3 years ago
I need help!!!!!!!!!
katovenus [111]

Answer:

buy a new one

Explanation:

go on newegg and you will most likely find one there

8 0
3 years ago
Explain all the characteristics of computer<br>​
Sedaia [141]
They are as follows-

Speed
Accuracy
Storage
Versatility
Diligence
Automation
Reliability
Power of Remembering

Let us understand each characteristic in brief:-

Speed – Speed of a computer means the time it takes to complete any given task. The computer works at a very lightning speed. For example, a computer takes a second to calculate 3 million calculations rather than a human which takes years. Hence nowadays the speed of the computer is measured in terms of microseconds, Nanoseconds, and even in Pico seconds instead of seconds or milliseconds.


Accuracy – The computers are programmed and designed in such a manner that their results are almost 100% accurate. It performs all the tasks with the same accuracy. There are very few chances of uncertainty in results and even though an error occurs, it can happen either due to wrong input data or unreliable programs by a programmer. Such errors are referred to as Garbage In Garbage Out (GIGO) which means, if you provide wrong instructions to the computer then you get wrong results.


Versatility – Along with being accurate and diligent a computer is also a versatile device. It can perform several types of tasks at a time if they are reduced to a certain set of logical steps. A computer can be used for many purposes at different places such as booking air and rail tickets, weather forecasting, listening to music, and playing games. Also at a time, one can pay his utility bills and can make a monthly budget too. It can prepare documents, data sheets and also can make models of houses, dams, etc.
Diligence – The ability of a computer to perform tasks without getting tired is known as diligence. Computers are highly reliable, they do not get fed up, exhausted, or lack concentration. They can work for hours and hours unlike humans and can give accurate or error-free results. For example, humans will start feeling tired after 2 -3 hours and will lack concentration whereas a computer will keep on working until the results are achieved.
Automation – Automation means working automatically. A computer can work on its own without the intervention of the person using it. Programs can be made for the computer to perform the task and based on the sequence it executes the tasks and gives accurate results. In case an error occurs, it gives error messages too and registers the logs.
Reliability – High reliability of a computer depends on low failure rate and easy maintenance. The results given out by the computer are reliable as the accuracy is almost 100%.
Power of Remembering – The computer has the power of storing data or information for several years. It cannot lose the data on its own. A person can retrieve the data whenever required and the data remains the same after a number of years also. It gives the freedom to the user to decide on how much data to store and how much to remove.
5 0
3 years ago
A cylinder is to be cast out of aluminum. The diameter of the disk is 500 mm and its thickness is 20 mm. The mold constant 2.0 s
Nezavi [6.7K]

Answer:

a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min

b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Explanation:

Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²

first we find the volume and Area;

Volume V = πD²t / 4

Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³

Area A = 2πD²/ 4 + πDt

Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}

Area A = 392,699.08 + 31,415.93

Area A = 424,115 mm²

a)

Chvorinov’s rule

T(aluminium) = Cm (V/A) ²

T(aluminium) =  2.0 × (3,926,991 / 424,115) ²

T(aluminium) = 171.5 s = 2.86 min

∴ the minimum time (minutes) for the casting to solidify is 2.86 min

b)

For cast iron

Cm (mold constant = 1.488 sec/mm²)

Chvorinov’s rule

T(iron) = Cm (V/A) ²

T(iron) = 1.488 × (3,926,991 / 424,115) ²

T(iron) = 127.5720s = 2.13 min

Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

6 0
3 years ago
A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold
PIT_PIT [208]

Answer:

a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.

Explanation:

Maximum theoretical efficiency for a power cycle (\eta_{r}), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:

\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\% (1)

Where:

T_{C} - Temperature of the cold reservoir, in Kelvin.

T_{H} - Temperature of the hot reservoir, in Kelvin.

The maximum theoretical efficiency associated with this power cycle is: (T_{C} = 400\,K, T_{H} = 1200\,K)

\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%

\eta_{r} = 66.667\,\%

In exchange, real efficiency for a power cycle (\eta), no unit, is defined by this expression:

\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\% (2)

Where:

Q_{C} - Heat released to cold reservoir, in kilojoules.

Q_{H} - Heat gained from hot reservoir, in kilojoules.

W_{C} - Power generated within power cycle, in kilojoules.

A power cycle operates irreversibly for \eta < \eta_{r}, reversibily for \eta = \eta_{r} and it is impossible for \eta > \eta_{r}.

Now we proceed to solve for each case:

a) Q_{H} = 900\,kJ, W_{C} = 450\,kJ

\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 50\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

b) Q_{H} = 900\,kJ, Q_{C} = 300\,kJ

\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 66.667\,\%

Since \eta = \eta_{r}, the power cycle operates reversibly.

c) W_{C} = 600\,kJ, Q_{C} = 400\,kJ

\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%

\eta = 60\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

d) Since \eta >  \eta_{r}, the power cycle is impossible.

8 0
3 years ago
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