1. A farmer had 752 sheep and took one shot that got them all. How did he do it?

Engineering

1 answer:

Step2247 [10]3 years ago

3 0

Answer:

In number 752 subtract 5 and 2 from 7 i.e:7-5-2=0 it becomes zero.

You might be interested in

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

2 years ago

Brazing, Soldering and Adhesive Bonding are the types of • Liquid Solid System Welding Solid State Welding • Fusion Welding . No

olga_2 [115]

Answer: Solid state welding

Explanation: Solid state welding is the welding procedure which is based on the temperatures and pressure but without any liquid or vapor as aid for welding.This process is carried mainly cohesive forces and considering forces as less important. Brazing,soldering and adhesive is the process in which material are joint which the help of solid welding agent.Thus solid state welding is the correct option.

6 0

2 years ago

A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle

Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

$X_{s} =1.23(\frac{V_{line}^{2} }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms$