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RUDIKE [14]
3 years ago
7

I need help on what’s and input and output to make the flowchart

Engineering
1 answer:
e-lub [12.9K]3 years ago
3 0

Input: what is put in, taken in, or operated on by any process or system.

Output: the amount of something produced by a person, machine, or industry.

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D. projectile
Colt1911 [192]
Answer: parabola

Explanation:

•Parabolic Trajectory:

In conclusion, projectiles travel with a parabolic trajectory due to the fact that the downward force of gravity accelerates them downward from their otherwise straight-line, gravity-free trajectory.
5 0
3 years ago
On some late model cars, climate control systems wil create trouble codes for trouble shooting purposes.
Advocard [28]
I think it’s true because I mean that where you have air conditioner and stuff but as though it’s probably not codes but in a car there probably will be codes
8 0
3 years ago
A search will start from a visual lead<br> true<br> false
BlackZzzverrR [31]
True

An organized searching process will need to start from the visual lead area. Eye focus and eye movements from the path of travel in an organized pattern describes a visual search process.
4 0
2 years ago
Read 2 more answers
What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?
nasty-shy [4]

Answer:

0.14\ lb/in^{3}

Explanation:

Density is defined as mass ler unit volume, expressed as

\rho=\frac {m}{v}

Where m is mass, \rho is density and v is the volume. For a sphere, volume is given as

v=\frac {4\pi r^{3}}{3}

Replacing this into the formula of density then

\rho=\frac {m}{\frac {4\pi r^{3}}{3}}

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}

4 0
4 years ago
A condenser accepts steam from the turbine in problem 2 at a pressure of 2.34 kPa. Saturated water at the same pressure leaves t
vaieri [72.5K]

Answer:

The answer is "83.98, 1889.195, and 1889.195"

Explanation:

Given value:

\bold{P_{4}=2.34 \ kPa}

In point a:

The value of h_{f4}=83.915 \ \ \frac{Kj}{kg}\\

V_4=0.001002 \ \  \frac{Kj}{kg}\\\\U_4= 83.98 \ \ \frac{Kj}{kg}\\\\

In point b:

calculating heat leaves formula= h_3-h_{f4}

                                                      = 1973.11-83.915\\\\= 1889.195 \ \ \frac{KJ}{kg}

In point c:

calculating Heat transfer rate formula=m(h_3-h_4)

                                                              = 1(1889.195)\\\\                                     = 1889.19 \ \ kw.

7 0
3 years ago
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