What are the top 4 solar inventions, how they are used, and how they are better than the original way of powering them

If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respect

Natasha2012 [34]

Answer:

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance $x!$

= cos 75⁰ = $\frac{x^!}{10ft}$

$x! = 10 * cos 75^{o}$ = 2.59 ft

calculate distance x

= cos 75⁰ = $\frac{x}{30ft}$

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * $x!$ + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

= 4014.5 + 1950.35 + 2030.535

= 7995.385 ft. Ib ≈ 7.9954 kip.ft

6 0

4 years ago

Hi, everyone I'm a high school student in Texas. My engineering teacher is asking us to find an active engineer to complete a li

jenyasd209 [6]

Answer:ill look it up

Explanation:

6 0

3 years ago

ASAP correct answer plss When you are driving, if you see this traffic sign it means

vlabodo [156]

Answer:

C

Explanation:

5 0

3 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe

LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is $= 1.751 \times 10^{-5} \% per hr$

Explanation:

we know Arrhenius expression is given as

$\dot \epsilon =Ce^{\frac{-Q}{RT}$

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is $\dot \epsilon = 5.5\times 10^{-2}$ % per hr

At 800 degree C creep rate is $\dot \epsilon = 1$ % per hr

activation energy for creep is $\frac{\epsilon_{800}}{\epsilon_{700}}$ = $= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}$

$\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}$

$\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]$

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

$C = \epsilon e^{\frac{Q}{RT}}$

$=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr$

$\epsilon_{500} = C e^{\frac{Q}{RT}}$

$= 1.804 \times 10^{12} e{\frac{251758}{8.314(500+273}}$

$= 1.751 \times 10^{-5} \% per hr$

4 0

3 years ago

14. An engine is brought into the shop with a

Lostsunrise [7]

Answer:

B. To accurately measure spark advance, use a timing light that incorporates an

ignition advance meter. The spark advance cannot be determined by listening to the way the engine sounds.

8 0

2 years ago