All categories

3 years ago

Write a single statement that prints outsideTemperature with 2 digits in the fraction

Engineering

1 answer:

lord [1]3 years ago

3 0

Answer:

outsideTemperature=108.90

Explanation:

#include <iostream>

#include <ios>

#include <iomanip>

using namespace std;

int main()

{

double outsideTemperature = 108.90;

/* print outside temperature */

cout << setprecision(2) << fixed << outsideTemperature << endl;

return 0;

}

You might be interested in

You have just finished your OST takeoffs for a single-story home and found 175 LF of interior walls with 2x6 studs 14" OC. The h

zimovet [89]

Answer:

Total BF for the interior wall is 7.50BD

Explanation:

Given Data:

· Size of stud = 2” x 6”

· Height of Wall = 8 ft

· Top plates = 2

· Bottom Plate = 1

BF stands for board feet in lumber/wood terminology. It is the unit of volume.

1 BF (Board feet) = 1 ft x 1 ft x 1 inch

Since there are total three plates at top and bottom, we have to deduct their thickness from wall height to calculate height of stud.

Height of stud = 8’ – 3 x 2” = 7’6” = 7.5 ft

Board feet of one stud = 7.50 6/12 x 2 = 7.50 BD

Total BF for the interior wall is 7.50BD

7 0

3 years ago

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

$\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}$

$2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}$

$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$

b) Maximum shear stress in BC

$\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi$

Maximum shear stress in AB

$\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi$

8 0

4 years ago

If superheated water vapor at 30 MPa iscooled at constant pressure, it will eventually become saturated vapor, and with suffic

nirvana33 [79]

Answer:

False.

Explanation:

False. The pressure is above pressure at critical point (22.064 MPa.), the limit where pressure can prevent boiling.

3 0

3 years ago

An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir

Artyom0805 [142]

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

$\frac{A}{F}$ = $\frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}$

Formula Used:

$A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}$

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

$\omega = \frac{2 \pi\times N}{60}$

so,

$\omega = \frac{2 \pi\times 720}{60}$ = 75.39 rad/s

Using the given formula:

Damping is negligible, so, $\zeta = 0$

$\frac{A}{F}$ will give the tranfer function

Therefore,

$\frac{A}{F}$ = $\frac{1}{\sqrt{(4K - 120\ ^{2})}}$

$0.1\times 10^{-3}$ = $\frac{1}{\sqrt{(4K - 120\ ^{2})}}$

Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm

8 0

3 years ago

The best saw for cutting miter joints is the

ZanzabumX [31]

Answer:

The best saw for cutting miter joints is the backsaw.

Add-on:

i hope this helped at all.

6 0

3 years ago