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elena-s [515]
3 years ago
14

A plane, opaque, surface M has the following properties: gray, diffuse, absorptivity = 0.7, surface area = 0.5 m2 , temperature

maintained at 500°C. Surface M experiences incoming radiant energy at 10,000 W/m2 . Find the energy absorbed per unit time. Parts d – f will also refer to surface M.
Engineering
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

The rate of energy absorbed per unit time is 3500W.

Explanation:

From the question, we were given the following parameters;

Plane, opaque, gray, diffuse surface

â = 0.7

Surface area, A = 0.5m²

Incoming radiant energy, G = 10000w/m²

T = 500°C

Rate of energy absorbed is âAG;

âAG = 0.7 × 0.5 × 10000

âAG = 3500W.

The energy absorbed is measured in watts and denoted by the symbol W.

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A driver counts 21 other vehicles using 3 EB lanes on one section of I-80 between her rented car and an overpass ahead. It turne
SVEN [57.7K]

Answer:

vehicle density =  28.205 veh/mile

flow rate =  0.909 veh/hr

Explanation:

given data

count n = 21

distance = 0.78 miles

speed = 52 mph

solution

we get here vehicle density that is express as

vehicle density = n ÷ distance    ...............1

vehicle density = ( 21 + 1 )  ÷ 0.78

vehicle density  k =  28.205 veh/mile

and

now we get here flow rate that is express as

flow rate = k × vs    .................2

flow rate = 28.205 × ( 52 × 0.00062 ÷ 1m )

flow rate =  0.909 veh/hr

6 0
2 years ago
A thin flat plate of surface area, 500 cm², is located between two fixed plates such that the gap below the top plate is 20 mm a
kodGreya [7K]

Answer:0.3166N

Explanation:

Given data

Area \left ( A\right )=500 cm^2

Gap below top plate\left ( y_1\right )=20 mm

Gap above bottom plate\left ( y_2\right )=30 mm

SAE 30 oil viscosity =0.38 N-s/m^2

Velocity of middle plate\left ( v\right )=1 m/s

There will viscous force on middle plate i.e. at above surface and below surface

Viscous force\left ( F\right )=\mu \frac{Av}{y}

Net Force on plate F =\mu Av\left [\frac{1}{y_1} +\frac{1}{y_2}\right ]

F=0.38\times 500\times 10^{-4}\left [\frac{1}{20\times 10^{-3}} +\frac{1}{30\times 10^{-3}}\right ]

F=31.66\times 10^{-2}=0.3166 N

this is force by oil on plate thus we need to apply atleast 0.3166N to move plate

8 0
3 years ago
P=3000W<br> f1=0.9<br> f2=0.5<br> A=15mm2<br> travel speed=?
iris [78.8K]

The Travel Speed at which the welding operation can be accomplished is given as 7.147mm/s. See the computation below.

<h3>What is travel speed?</h3>

The travel speed of the welding flame or gun across the workpiece is simply measured in millimeters per minute.

Travel speed, along with voltage and amperage, is one of three factors in arc welding that affect the amount of heat input.

<h3>What is the calculation that justifies the above answer is?</h3>

We are given the following:


Melting temperature = 1650 K

k = (3.33 * 10⁻⁶) * (1,650)²

k = 9.066 J/mm^3

The formula for velocity here is given as:

V = f1 * f2 * Rh / (k*A)

V = 0.9 * 0.6 * 3000 / (9.066*25)

V = 7.147 mm/s

Hence, the Travel Speed at which the welding operation can be accomplished is given as 7.147mm/s.

Learn more about welding operation at;
brainly.com/question/22494632
#SPJ1

7 0
2 years ago
The radiation meter is showing radiation 2x as much as background. Is this a hot zone? If so why or why not?
kakasveta [241]

Answer:

This is not a clear indication of the hot zone as the information of the radioactivity of the background is not provided clearly.

Explanation:

According to IAEA as well as NRCP, the hot area is defined on the basis of the radioactivity reading it shows instead of contrast or comparative reading from the background. The value of radiation activity which will be required to declare an area as hot zone is if it is greater than 0.1 mSv/h or 1.5091\times 10^{29} kg^{-1} s^{-1}.

5 0
2 years ago
for an electromotive force to be induced across a vertical loop from the field of an infinite length line of fixed current in th
Sever21 [200]

Answer:

The correct answer to the following question will be "a_{x} or a_{y}".

Explanation:

  • Since along that same z-axis none electromagnetic field would be triggered as being in the same orientation loop movement of them across different line portions would allow some caused emf/voltage to be canceled. And the only logical choice seems to be either x or y-axes.
  • The magnetic field of fluctuation should indeed be changed and changed across both X as well as Y directions.

So that the above is the appropriate choice.

3 0
2 years ago
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