What was the earliest self-powered road vehicle?

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, $\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )$

Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\$

$0.01 = C \exp(\frac{-Q}{1073R} )$ .........................(1)

Creep rate at 700⁰C

$\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}$

$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

Take the natural log of both sides

$ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol$

3 0

3 years ago

The level of water in a dam is 6 m. The rectangular gate ABC is pinned at point B so it can rotate freely about this point. When

olchik [2.2K]

Answer:

The reaction at support B

Rb= 235440N

The reaction at support C

RC= 29430N

Explanation : check attachment

6 0

3 years ago

Create a series of eight successive displacements that would program a robot to move in an octagonal path that is as close as yo

Komok [63]

Answer:

bts biot bts biot jungkukkk

jungkukkkbiot

Explanation:

bts biot bts biot jungkukkk

jungkukkkbiot

5 0

3 years ago

The interior wall of a building is made from 2×4 wood studs, plastered on one side. If the wall is 13 ft high, determine the loa

Elanso [62]

Answer:

load = 156 lb/ft

Explanation:

given data

interior wall of a building = 2×4 wood studs

plastered = 1 side

wall height = 13 ft

solution

we get here load so first we get wood stud load and that is

we know here from ASCE-7 norm

dead load of 2 x 4 wood studs with 1 side plaster = 12 psf

and we have given height 13 ft

so load will be = 12 psf × 13 ft

load = 156 lb/ft

7 0

3 years ago

2)Prueba de presión Cuando a una persona se le somete a una prueba de presión, por lo general se le indica que, al llegar el rit

Olin [163]

Usando la ecuación de regresión que modela la frecuencia cardíaca máxima, podemos obtener el valor predicho de los problemas dados así:

La ecuación lineal que modela la frecuencia cardíaca máxima permitida en función de la edad del paciente está relacionada con la fórmula:

m = - 0,875x + 190

x = edad del paciente; m = máx. frecuencia cardíaca permitida

1.) Frecuencia cardíaca máxima permitida para una persona de 50 años:

Sustituye x = 50 en la ecuación:

m = -0,875 (50) + 190

m = 146,25

Por lo tanto, la frecuencia cardíaca máxima permitida es de aproximadamente 146 latidos / min.

2.) Edad para una persona con frecuencia cardíaca máxima de 160 latidos / min:

Sustituye m = 160 en la ecuación:

160 = -0,875x + 190

0,875x = 190 - 160

0,875x = 30

x = 30 / 0,875

x = 34,28

Por tanto, la edad de la persona sería de unos 34 años.

Más información: brainly.com/question/25395533

3 0

3 years ago