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3 years ago

What are difference between conic sectional and solids?

Engineering

1 answer:

Murljashka [212]3 years ago

3 0

Answer:

Conic Sections

a conic section is a curve which is obtained when a surface performs an intersection with a plane. The types of conic sections include hyperbola, parabola and ellipse. A circle can also be considered as a conic section.

Conic Solids on the other hand are the set of points on any segment between a region and a point which is not present in the plane of the base. They are solids with one base.

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The current through a 10-mH inductor is 10e−t∕2 A. Find the voltage and the power at t = 8 s.

NNADVOKAT [17]

Answer:

voltage = -0.01116V

power = -0.0249W

Explanation:

The voltage v(t) across an inductor is given by;

v(t) = L $\frac{di(t)}{dt}$ -----------(i)

Where;

L = inductance of the inductor

i(t) = current through the inductor at a given time

t = time for the flow of current

From the question:

i(t) = $10e^{-t/2}$ A

L = 10mH = 10 x 10⁻³H

Substitute these values into equation (i) as follows;

v(t) = $(10*10^{-3})\frac{d(10e^{-t/2})}{dt}$

Solve the differential

v(t) = $(10*10^{-3})\frac{-1*10}{2} (e^{-t/2})$

v(t) = -0.05 $e^{-t/2}$

At t = 8s

v(t) = v(8) = -0.05 $e^{-8/2}$

v(t) = v(8) = -0.05 $e^{-4}$

v(t) = -0.05 x 0.223

v(t) = -0.01116V

(b) To get the power, we use the following relation:

p(t) = i(t) x v(t)

Power at t = 8

p(8) = i(8) x v(8)

i(8) = i(t = 8) = $10e^{-8/2}$

i(8) = $10e^{-4}$

i(8) = 10 x 0.223

i(8) = 2.23

Therefore,

p(8) = 2.23 x -0.01116

p(8) = -0.0249W

7 0

3 years ago

Read 2 more answers

In addition to passing an ASE certification test, automotive technicians must have __________ year(s) of on the job training or

Lunna [17]

Two years or one year

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3 years ago

Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012

devlian [24]

The pressure drop of air in the bed is 14.5 kPa.

<u>Explanation:</u>

To calculate Re:

$R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}$

From the tables air property

$\mu_{394 k}=2.27 \times 10^{-5}$

Ideal gas law is used to calculate the density:

ρ = $\frac{2.2}{2.83 \times 10^{-3} \times 394.3}$

ρ = 1.97 Kg / $m^{3}$

ρ = $\frac{P}{RT}$

R = $\frac{R_{c} }{M}$ = 8.2 × $10^{-5}$ / 28.97× $10^{-3}$

R = 2.83 × $10^{-3}$ $m^{3}$ atm / K Kg

q is expressed in the unit m/s

$q=\frac{2.45}{1.97}$

q = 1.24 m/s

Re = $\frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}$

Re = 2278

The Ergun equation is used when Re > 10,

$\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}$

$\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24$ $+\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}$

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa

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3 years ago

The bulk density of a compacted soil specimen (Gs = 2.70) and its water content are 2060 kg/m^3 and 15.3%, respectively. If the

yaroslaw [1]

Answer:

the saturated density should be

Explanation:

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3 years ago

What is aviation? What is the difference between aviation and flight?

-Dominant- [34]

Aviation refers to flying using an aircraft, like an aeroplane. It also includes the activities and industries related to flight, such as air traffic control. The biggest of the many uses of aviation are in air travel and military aircraft . The main difference between aeronautics and aviation is their focus. Aeronautics involves the study of the science, design, and manufacture of flying vehicles while aviation involves flying or operating an aircraft and various activities surrounding mechanical flight and the aircraft industry

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