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geniusboy [140]
4 years ago
6

"Hint: A backslash \ in a string acts as an escape character, such as with a newline \n. So, to print an actual backslash, escap

e that backslash by prepending another backslash. Ex: The following prints a single backslash: printf("\\");"
Engineering
1 answer:
Tcecarenko [31]4 years ago
7 0

Answer:

True

Explanation:

The \ is an escape character, is a way to indicate to the compiler that the following characters are special; if you want to print an actual \ you need to use two, one after the other one, in the printf function. But what exactly happens if you try to print just one \

#include <iostream>

using namespace std;

int main()

{

   printf ("\");

}

This will produce an error that prevents you from compiling:

main.cpp:15:13: warning: missing terminating " character

    printf ("\");

            ^

main.cpp:15:13: error: missing terminating " character

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This came off my car earlier today. What is it and what does it do?
patriot [66]

Answer:

get a new car or duck tape it back on or go get it fixed

Explanation:

that's just whut I wold do.

6 0
3 years ago
Read 2 more answers
find magnitude of the resultant force, if 30N,40N,50N and 60N forces are acting along the line joining the centr of a square to
viktelen [127]

Answer:

56 feet

Explanation:

c 20 n north

7 0
3 years ago
Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distanc
Andrei [34K]

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m

b) f = 1 GHz = 10⁹ Hz.

\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10}  e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10}  e=-1.5 \\\\z=\frac{-1.5}{log_{10}  e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm

4 0
3 years ago
The acceleration due to gravity at sea level is g=9.81 m/s^2. The radius of the earth is 6370 km. The universal gravitational co
solmaris [256]

Answer:

Mass of earth will be M=5.96\times 10^{24}kg

Explanation:

We have given acceleration due to gravity g=9.81m/sec^2

Radius of earth = 6370 km =6370\times 10^3m

Gravitational constant G=6.67\times 10^{-11}Nm^2/kg^2

We know that acceleration due to gravity is given by

g=\frac{GM}{R^2}, here G is gravitational constant, M is mass of earth and R is radius of earth

So 9.81=\frac{6.67\times 10^{-11}\times M}{(6370\times 10^3)^2}

M=5.96\times 10^{24}kg

So mass of earth will be M=5.96\times 10^{24}kg

3 0
3 years ago
Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a
Katyanochek1 [597]

Answer:

K_{I}=5.21 MPa\sqrt{m}

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}   ................................1

here Y is the geometrical function

so

Y = \frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}

Y = \frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}  

Y = \frac{0.8792}{0.89}  

Y = 0.9878

so put here value in equation 1

K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}    

K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319

K_{I} = 5210.45 × 10³  Pa\sqrt{m}  

K_{I} = 5.21 MPa \sqrt{m}

5 0
3 years ago
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