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3 years ago

Yall pls help me out

Engineering

1 answer:

atroni [7]3 years ago

8 0

Answer:

M·L/T²

Explanation:

The given equation for T is presented as follows;

$T = \dfrac{M \cdot V^2 \cdot A}{L^3}$

Where;

m = Mass with dimension M

V = Velocity, having the dimensions L/T

A = Area, having the dimensions L²

l = Length, having the dimensions, L

Therefore;

$The \ dimensions \ of \ T = \dfrac{M \times (L/T)^2 \cdot L^2}{L^3} = \dfrac{M \times L^4/T^2 }{L^3} = M \cdot L/T^2$

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Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of

aleksandr82 [10.1K]

Answer:

Explanation:

Given

Temperature of solid $T=500\ K$

Einstein Temperature $T_E=300\ K$

Heat Capacity in the Einstein model is given by

$C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}$

$e^{\frac{3}{5}}=1.822$

Substitute the values

$C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})$

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2 years ago

Give two causes that can result in surface cracking on extruded products.

Andreas93 [3]

Answer:

1. High friction

2. High extrusion temperature

Explanation:

Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.

Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.

Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.

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2 years ago

Technician a says that if a tapered roller bearing is adjusted to loose

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The technician is true

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2 years ago

An energy system can be approximated to simply show the interactions with its environment including cold air in and warm air out

Elenna [48]

Answer: The energy system related to your question is missing attached below is the energy system.

answer:

a) Work done = Net heat transfer

Q1 - Q2 + Q + W = 0

b) rate of work input ( W ) = 6.88 kW

Explanation:

Assuming CPair = 1.005 KJ/Kg/K

Write the First law balance around the system and rate of work input to the system

First law balance ( thermodynamics ) :

Work done = Net heat transfer

Q1 - Q2 + Q + W = 0 ---- ( 1 )

rate of work input into the system

W = Q2 - Q1 - Q -------- ( 2 )

where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw

Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw

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W = 6.88 Kw

5 0

2 years ago

Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance h

eimsori [14]

Answer:

t = 6179.1 s = 102.9 min = 1.7 h

Explanation:

The energy provided by the resistance heater must be equal to the energy required to boil the water:

E = ΔQ

ηPt = mH

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η = efficiency = 84.5 % = 0.845

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t = time = ?

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H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

$t = \frac{1.362\ x\ 10^7\ J}{2205.45\ W}$

t = 6179.1 s = 102.9 min = 1.7 h

4 0

2 years ago