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2 years ago

Earth completes one full ____ on its axis every 24 hours

Engineering

2 answers:

mars1129 [50]2 years ago

7 0

Answer:

rotation

Explanation:

I just answered this!

gayaneshka [121]2 years ago

5 0

Rotation. i hope this helped:)

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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

Applications of fleming hand rule

Eva8 [605]

Answer:

Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.

Explanation:

Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.

Fleming hand rule is used in the case of electric motors and electric generators.

Fleming hand rule is used to determine the following:

1. Direction of torque

2. Angular velocity

3. Angular acceleration

4 0

3 years ago

How can we calculate the speed of the output gear in a simple gear train? Explain with the help of an example.

Snowcat [4.5K]

Answer:

$N_3=\dfrac{T_1}{T_3}N_1$

Explanation:

In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.

Lets take

$Speed\ of\ gear 1=N_1$

$Number\ of\ teeth\ of\ gear 1=T_1$

$Speed\ of\ gear 3=N_3$

$Number\ of\ teeth\ of\ gear 3=T_3$

All external matting gears will rotates in opposite direction with respect to each other.

So the speed of gear third can be given as follows

$\dfrac{T_1}{T_3}=\dfrac{N_3}{N_1}$

$N_3=\dfrac{T_1}{T_3}N_1$

3 0

3 years ago

(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th

arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$