All categories

3 years ago

A cable in a motor hoist must lift a 700-lb engine. The steel cable is 0.375in. in diameter. What is the stress in the cable?

Engineering

1 answer:

UkoKoshka [18]3 years ago

7 0

Answer:43.70 MPa

Explanation:

Given

mass of engine $700 lb \approx 317.515 kg$

diameter of cable $0.375 in.\approx 9.525 mm$

$A=\frac{\pi d^2}{4}=71.26 mm^2$

we know stress( $\sigma$ ) $=\frac{load\ applied}{area\ of\ cross-section}$

$\sigma =\frac{317.515\times 9.81}{71.26\times 10^{-6}}=43.70 MPa$

You might be interested in

Make a proposal to add a small pizza shop to a historical part of town. How could it be designed to “fit” into the area?

spin [16.1K]

It could be designed because if we had a small little pizza shop it would save our problems of going through all the trouble just for pizza

7 0

3 years ago

Read 2 more answers

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

2 years ago

If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20M

Bess [88]

Answer:

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

C = W*log_2 ( 1 + P/N)

- Plug the values in:

C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

C = (20*10^6)*log_2 (25001)

C = (20*10^6)*14.6096

C = 292 Mbps

3 0

3 years ago

The host at the end of the video claims that ___________ is crucial to his success as a driver. A. Reaction time B. A safe space

Snezhnost [94]

Answer:

answer is C. his seat belt

5 0

2 years ago

Select the correct answer.

olya-2409 [2.1K]

Answer:

screw is the answer of the question

6 0

3 years ago

Read 2 more answers