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miv72 [106K]
3 years ago
5

What is the difference between POP3 and IMAP?

Engineering
1 answer:
stich3 [128]3 years ago
6 0
What is the difference between POP3 and IMAP?
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As you have learned, not all motor oils are the same. What are two things that make them different?.
saul85 [17]

Answer:

quality ingredients and exceeding industry standards

Explanation:

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2 years ago
Problema sobre programacion orientada a objetos!!
Alecsey [184]

Answer:If they are used to top-down programming or functional programming, which treats elements of code as precise mathematical functions, it takes .

Explanation:

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2 years ago
a vertical cylindrical container is being cooled in ambient air at 25 °C with no air circulation. if the initial temperature of
Sloan [31]

Answer:

the surface heat-transfer coefficient due to natural convection during the initial cooling period.  = 4.93 w/m²k

Explanation:

check attachement for answer explanation

7 0
3 years ago
Read 2 more answers
(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}]

\dot Q = 417.492\,\frac{BTU}{h}

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

Q = \dot Q \cdot \Delta t (2)

Where \Delta t is the heat transfer time, measured in hours.

If we know that \dot Q = 417.492\,\frac{BTU}{h} and \Delta t = \frac{1}{3}\,h, then the net energy transfer is:

Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

Q = 139.164\,BTU

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

7 0
2 years ago
A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
Kay [80]

Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³    = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

        2.33 × 10⁻³                  

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

5 0
3 years ago
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