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ale4655 [162]
3 years ago
14

if you had 100 B size sheets and you cut them into A size sheets, how many sheets of A size paper would you have

Engineering
1 answer:
castortr0y [4]3 years ago
8 0

Answer:

200

Explanation:

A size sheets (also known as letter size) are 8.5 inches by 11 inches.

B size sheets (also known as ledger size) are 11 inches by 17 inches.

One B size sheet is twice as large as a A size sheet.  So if you have 100 B size sheets and cut each one in half, you'll get 200 A size sheets.

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Which branch of engineering studies the physical behavior of metallic elements?
olga55 [171]

Material engineering studies the physical behavior of metallic elements.

Answer: Option C

<u>Explanation: </u>

Material Engineering is the creation and learning about the materials at an atomic level. An engineers from this branch focus on material and model its characteristics using the computer.

Also, they combine the knowledge of solid-states, metallurgy, chemistry and ceramics to the application level. It also has a great role in building the future with the advancing study in nanotechnology, biotechnology, etc. Simply, these are meant to have vivid applications in future life.

6 0
3 years ago
Read 2 more answers
Define waves as it applies to electromagnetic fields
julsineya [31]

Waves in the electric and magnetic fields are known as electromagnetic waves. You must first understand what a field is, which is just a technique of giving each square inch of space a numerical value. You may see that as a temperature field, for instance, when you look at the weather predictions and they mention the temperature in several locations. Every location on Earth has a unique temperature that can be quantified. Everywhere on Earth has its own wind velocity, which is another form of field. This field differs somewhat from the temperature field in that the wind velocity has both a direction and a magnitude, whereas the temperature just has a magnitude (how hot it is). A vector is a quantity that has both magnitude and direction, hence a field that contains vectors at every location is referred to as a vector field. Vector fields include the magnetic and electric fields. We may examine what would happen if we placed a charged particle at any given position in space. If the charged particle were to accelerate, we would state that the electric field there is the direction in which the particle is moving. In general, positively charged particles will move in the electric field's direction, whereas negatively charged particles will move in the opposite way. Because it is a vector field, the magnetic field exhibits comparable behavior. We discovered in the 19th century that the same interaction, electromagnetism, really produces both electric and magnetic fields. Like an electromagnet, a changing electric field will produce a magnetic field, and a changing magnetic field will induce an electric field (like in a generator). If your system is configured properly, you may have an electric field that fluctuates, which in turn produces a magnetic field, which in turn induces another electric field, which in turn generates another magnetic field, and so on indefinitely. At the speed of light, this oscillation between a strong magnetic field and strong electric field spreads out indefinitely. In reality, light is an electromagnetic wave—an oscillation in the electromagnetic fields. An electric or magnetic field may exist without a medium since they exist in a vacuum, which implies that waves in these fields don't require a medium like sound to flow through.

5 0
2 years ago
Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
White raven [17]

Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0
4 years ago
Using the results of the Arrhenius analysis (Ea=93.1kJ/molEa=93.1kJ/mol and A=4.36×1011M⋅s−1A=4.36×1011M⋅s−1), predict the rate
uysha [10]

Answer:

k = 4.21 * 10⁻³(L/(mol.s))

Explanation:

We know that

k = Ae^{-E/RT} ------------------- euqation (1)

K= rate constant;

A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;

E = activation energy = 93.1kJ/mol;

R= ideal gas constant = 8.314 J/mol.K;

T= temperature = 332 K;

Put values in equation 1.

k = 4.36*10¹¹(M⁻¹s⁻¹)e^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}

k = 4.2154 * 10⁻³(M⁻¹s⁻¹)

here M =mol/L

k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)

 or

k = 4.21 * 10⁻³((L/mol)s⁻¹)

or

k = 4.21 * 10⁻³(L/(mol.s))

3 0
3 years ago
The "Big Dig" was the nickname of the civil engineering project that redesigned the highway Infrastructure for the city of
zheka24 [161]
Geotechnical since it’s geologicaly based
4 0
3 years ago
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