What is a good way to become a game developer without spending a lot of money?

What is computer with figure

GenaCL600 [577]

Answer:

A computer is an electronic device that accept raw data and instructions and process it to give meaningful results.

5 0

2 years ago

How many times do you need to click the format painter button to apply copy formats to multiple paragraphs one right after the o

IRISSAK [1]

U have to double-click for multiple paragraphs instead of single click

5 0

2 years ago

What are the hardware and software components of a computer

DaniilM [7]

Answer:

I. Hardware components of a computer includes monitor, speaker, central processing unit, motherboard, hard-drive, joystick, mouse, keyboard, etc.

II. Software components of a computer includes operating system, registry keys, antivirus, media player, word processor, etc.

Explanation:

The hardware component of a computer can be defined as the physical parts or peripherals that enables it to work properly. Some examples of hardware components are monitor, speaker, central processing unit, motherboard, hard-drive, joystick, mouse, keyboard, etc.

A software component of a computer comprises of software application or program that are used by the computer to manage or control software application, computer hardware and user processes. Some examples of software components are operating system, registry keys, antivirus, media player, word processor, etc.

In conclusion, the hardware components of a computer are the physical parts that can be seen and touched while the software components cannot be touched but rather are installed as a program.

8 0

2 years ago

What is the rate constant at 25.0 ∘c based on the data collected for trial b?

Alecsey [184]

<h2><u>Answer: </u></h2>

acetone + I2 + HCl ---> iodated acetone

Equation:

rate = k * [acetone]^x * [I2]^y * [HCl]^z

Once we know x, y, z, we can plug in any of the trials A->D and determine k NOTE: We can't use run E because temperature has an effect on rate. E was run at a different temperature.

The first thing to note is that do NOT have concentrations. We have volumes at a given molarity.

Table:

.001M I2.. ..050M HCl.. .1.0M acetone.. .water.. temp..time.. total vol

.. ....mL.. ... ... .. .mL.. .. ... .. .. mL.. .. .. .. ..mL.. ..°C.. .sec.. .. .. .L

A.. ...5.. .. .. .. .. ..10.. .. .. .. .. ..10.. .. .. .. ..25.. .. .25.. .130.. .. 0.05

B.. ..10.. .. .. ... .. 10.. .. .. ... .. .10.. .. .. ... ..20.. . .25.... 249.. ..0.05

C.. . 10.. .. .. .. .. .20.. .. .. .... .. 10.. .. ... ... .10... ..25.. ..128... .0.05

D.. . 10.. ... .. ... ..10.. .. ... .. ... 20.. .. .. .. .. 10.. .. 25.. ..131.. ..0.05

E.. ..10.. ... .. ... ..10.. ... ... ... ..10.. ... .. ... .20.. .. 42.8.. .38.. ..0.05

We can translate that into molarity in solution using this formula. (molarity pure ingredient * mL used / 1000 / total volume in liters)

.. .. ..I2.. .... HCl.. acetone.. temp.. ..rxn time

.. .. ..M.. .. ...M.. .... .M.. .... ..°C.. .. .. sec

A.. 0.0001.. 0.01..... 0.2.. .... .25... .... .130

B.. 0.0002.. 0.01.. .. 0.2.. .. .. 25.. .. .. .249

C.. 0.0002.. 0.02.. .. 0.2.. .. .. 25.. .. .. .128

D.. 0.0002.. 0.01.. .. 0.4.. .. . .25.. ... ...131

E.. 0.0002.. 0.01.. .. 0.2.. . ....42.8.. .. .. 38

From runs B and D, we can see that rate dropped by half .

As [I2] and [HCl] were held constant and [acetone] was doubled.

This means x=-1 in this equation

Rate = k * [acetone]⁻¹ * [I2]^y * [HCl]^z

Rate = k * [I2]^y * [HCl]^z

From runs A and B

[I2] doubles

[HCl] remains the same

[acetone] remains the same

rate doubles as [I2] doubles, rate doubles

y = 1 rate = k * [acetone]⁻¹ * [I2]¹ * [HCl]^z

And from runs B and C, we can see that , As [HC] doubles, (all else equal) the rate halves.

Z = -1

Rate = k * [I2] / ([acetone] * [HCl])

Rearranging

k = rate * [acetone] * [HCl] / [I2]

From any experimental run (A-D), we can calculate k.

using A to calc k... ..k = 2600 M²/sec

using B to calc k... . k = 2490 M²/sec

using C to calc k... . k = 2560 M²/sec

using D to calc k... . k = 2620 M²/sec

NOTE.. the problem statement said to use the data from run B to calc k.

Hence Final Answer:

k = 2490 M²/sec

4 0

3 years ago

Contrast the performance of the three techniques for allocating disk blocks (contiguous, linked, and indexed) for both sequentia

Pie

The allocation methods define how the files are stored in the disk blocks.
There are three main disk space or file allocation methods:
1.Contiguous Allocation-in this scheme,each file occupies a set of blocks on the disk. For example if a file requires x blocks and is given a block y as the starting location,then the blocks assigned to the file be :x,y+1,y+2,......y+x-1.
This means that given the starting block address and the length of the file(in terms of blocks required) we can determine the blocks occupied by the file.
Advantages
-both the sequential and direct accesses are supported
-this is extremely fast since the number of seeks are minimal because of contiguous allocation of file blocks.
2.linked allocation-in this scheme,each file linked list of disk blocks which need not be contiguous disk blocks can be scattered anywhere on the disk.
Advantages
it is very flexible in terms of file size.file size can be increased easily since the system does not have to look for a contiguous chunk
of memory.
this method does not suffer from external fragmentation and it makes it relatively better in terms of memory utilization.
3.Indexed Allocation-in this scheme,a special block known as the index block contains the pointers to all the blocks occupied by a file.Each file has its own index block.the entry in the index block contains the disk address of the block
Advantages
it supports direct access to the blocks occupied by the file and therefore provides fast access to the file blocks
it overcomes the problem of external fragmentation.

3 0

3 years ago

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