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hodyreva [135]
4 years ago
13

Your employer has the ability to protect you from cave-ins and other hazards by using adequately-designed protection systems in

excavations, but these are not required when an excavation is made entirely in stable rock or is less than how deep? A. 5FT B.10FT C.15FT
Engineering
1 answer:
Rus_ich [418]4 years ago
5 0

Answer:15

Explanation:

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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 liquid-phase reaction:
OLEGan [10]

Answer:

attached below

Explanation:

4 0
3 years ago
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
3 years ago
A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the
alekssr [168]

Answer:

answer

Explanation:

4 0
3 years ago
What is Back EMF? How does it limits the speed of a permanent magnet DC?
ss7ja [257]

Answer and Explanation:

The DC motor has coils inside it which produces magnetic field inside the coil and due to thus magnetic field an emf is induced ,this induced emf is known as back emf. The back emf always acts against the applied voltage. It is represented by E_b

The back emf of the DC motor is given by E_b=\frac{NP\Phi }{60A}

Here N is speed of the motor ,P signifies the number of  poles ,Z signifies the the total number of conductor  and A is number of parallel paths

As from the relation we can see that back emf and speed ar dependent on each other it means back emf limits the speed of DC motor

8 0
3 years ago
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