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3 years ago

I need unseen passage

Engineering

2 answers:

monitta3 years ago

8 0

Answer:

this unseen passage__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Semmy [17]3 years ago

6 0

Answer:

An unseen passage is invisible on ground, but might be underground.

Explanation:

I'm sorry but if your question was clear, I would be able to answer it.

Hope this helps!

You might be interested in

In 1828, the diameter of the U.S. dime was changed to approximately 18 mm. What is this diameter when expressed in micrometers

neonofarm [45]

The diameter of an 18 mm dine when expressed in micrometers is; 18000 micrometers

<h3>Conversion of Units</h3>

We are given the diameter of dime at 18 mm.

Now, from conversion of units between mm(millimeter) and micrometer, we know that;

1 millimeter = 1000 micrometer.

Thus, by proportion, 18 millimeter will be;

18 millimeters × 1000 micrometers/1 millimeter

>> 18000 micrometers

In conclusion, the diameter of the 18mm fine in micrometers is; 18000 micrometers

Read more about conversion of units at; brainly.com/question/8421470

7 0

2 years ago

suppose we number the bytes in a w-bit word from 0 (less significant) to w/8-1 (most significant). write code for the followign

sammy [17]

Solution:

typedef unsigned char *byte_pointer;

static int int_of_bit (byte_pointer x, int loc)

{

return(x[loc] << loc*8);

}

static int replace_byte(unsigned int a, int loc, unsigned int b)

unsigned int a_loc = int_of_bit((byte_pointer) &a , loc);

unsigned int b_loc = (b << loc*8);

a -= a_loc;

a += b_loc;

return a;

}

Explanation:

This takes two ints in hex format, one with 8 bits (0x00000000) and one with 2 bits (0x00) then places the 2 bit hex into the 8 bit at a given location.

EX: replace_byte(0x00000000, 1, 0xFF) return 0x0000FF00

First thing first, it looks like you have some mixup:

" one with 8 bits (0x00000000) and one with 2 bits (0x00)*- what you mean is nibble not bits. Each hex character (0-9, A-F) represent 4 bits (16 possible combination), and is called a "nibble".

0x0000000 is 4 bytes. 0x00 is 2 bytes.

Next, you say "takes two ints in hex format" - your function takes two ints in any format. You're just choosing tp express them in hexidecimal. C++ doesn't care if you specify numbers in hex, decimal, octal, binary, etc.

8 0

3 years ago

During the reaction, 3.50 μmol of HCl are produced. Calculate the final pH of the reaction solution. Assume that the HCl is comp

lyudmila [28]

Answer:

The pH of the solution will be equal to 5.46

Explanation:

The dissociation reaction of HCl is equal to:

HCl → H+ + Cl-

To solve the exercise we must first convert the µmoles to moles using the following conversion factor:

3.5µmoles x $\frac{1 mol}{1x10^{6} umol}$ = 3.5x $10^{-6}$ moles

Assuming a liter of solution, we can calculate the molar concentration by:

M = $\frac{Number of moles}{Liter of solution}$

Replacing:

M = $\frac{3.5x10^{-6}moles }{1 L}$ = 3.5x $10^{-6}$ moles/L

As this acid dissociates completely, the concentration of protons and chloride will be equal to 3.5x $10^{-6}$ moles/L

The pH will be equal to:

pH = -log[H+]

Replacing:

pH = -log[3.5x $10^{-6}$ ] = 5.46

8 0

4 years ago

52 points+Brainliest for correct answer

pentagon [3]

Divide by 6 and divide the caw of squaw squaw

4 0

2 years ago

Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=50

inn [45]

Answer:

hello your question is incomplete attached below is the complete question

answer :

Slopes : B = 180 mm , C = 373 mm

Deflection: B = 0.0514 rad , C = 0.077 rad

Explanation:

Given data :

I = 500(10^6) mm^4

E = 70 GPa

The M / EI diagram is attached below

Deflection angle at B

∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI

= 1800 / ( 500 * 70 ) = 0.0514 rad

slope at B

ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI

= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm

Deflection angle at C

∅C = ∅CA = [ 1800 + 300*3 ] / EI

= 2700 / ( 500 * 70 )

= 2700 / 35000 = 0.077 rad

Slope at C

ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]

= 13050 / 35000 = 373 mm

3 0

3 years ago