What should you consider when choosing the type of hearing protection you use?

Hi all, could you solve this please?<br> What is the value of the resistance R

Kazeer [188]

Answer:

try to 36v power and take 1a and intersect to 3

5 0

3 years ago

The post is made of Douglas fir and has a diameter of 100 mm. if it is subjected to the load of20 kN and the soil provides a fri

Solnce55 [7]

Answer:

Explanation:

Kindly go through the attachment for the step by step approach to the solution of this question.

5 0

3 years ago

andreev551 [17]

Question:

The question is not complete. The question to answer was not added. See below the possible question and the answer.

a. How many blocks are in the cache with this new arrangement?

b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.

C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?

Answer:

(a) Number of blocks = 512 blocks

(b) Tag is 18

(c) Total size of the cache = 8388608 bytes

Explanation:

a .

block size = 32 bytes

cache size = 16384 bytes

No.of blocks = 16384 / 32

No.pf blocks = 512 blocks

b.

Total address size = 32 bits

Address bits = Tag + Line index +block offset

Block Size = 32 bytes.

So block size = 25 bytes.

Hence Offset is 5

No . of Cache blocks = 512 blocks = 29 blocks

Hence line offset is 9

We know that Address bits = Tag + Line index +block offset

So , 32 =tag+9+5

tag = 32-(9+5)

So Tag is 18

c.

Data bits = 32 bits

Tag=18 bits

Valid bit is 1 bit

so Total cache size = 25+218+20

= 223

=8388608 bytes

7 0

4 years ago

Carbon dioxide (CO2) is used in industrial and research application, and is sometimes stored at very high pressure in rigid meta

12345 [234]

Answer:

a)m = 247.43 kg

b) m = 123.71 kg

Explanation:

a)

Given data:

volume =0.8 m^3

P = 18,000 kPa

T =35 degree C = 308 K

By ideal gas equation we have following relation

PV = mRT

where R is gas constant

$R = \frac{8.314}{44} = 0.18895 kJ/ kg K$

$m =\frac{PV}{RT}$

$m = \frac{18000\times 0.8}{0.18895 \times 308}$

m = 247.43 kg

b)

when pressure = 9000 kPa

from ideal gas equation

PV = mRT

where R is gas constant

$R = \frac{8.314}{44} = 0.18895 kJ/ kg K$

$m =\frac{PV}{RT}$

$m = \frac{9000\times 0.8}{0.18895 \times 308}$

m = 123.71 kg

3 0

3 years ago

You insert a dielectric into an air-filled capacitor. How does this affect the energy stored in the capacitor?.

umka2103 [35]

Answer:

Energy stored in the capacitor will increase first, and then it will decrease.

Explanation:

7 0

2 years ago