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Lesechka [4]
3 years ago
14

Disconnecting means shall be capable of being locked in the open position. The provisions for locking do not have to reamin in p

lace. A. True B. False
Engineering
1 answer:
Lorico [155]3 years ago
7 0
The answer is a.true
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(d) Arches NP is known for its spectacular arches that develop in the jointed areas of the park. Placemark Problem 2d flies you
Whitepunk [10]

Answer:

☐ NE-SW

Explanation:

Based on the description, the rock direction is North East - South West (NE-SW). Rocks generally can expand or compress depending on the type and magnitude of stress applied on the rocks. However, if the applied stress is sufficiently high, cracks and fractures will be created on the rock and it can ultimately lead to the formation of particles.

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3 years ago
What tool is used to measure aggregates in laying tools<br>​
denpristay [2]

Answer:

Sieve Shaker

Explanation:

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2 years ago
Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x
sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

<u>To = 517.24° C</u>

5 0
3 years ago
Air enters a control volume operating at steady state at 1.2 bar, 300K, and leaves at 12 bar, 440K, witha volumetric flow rate o
topjm [15]

Answer:

Heat transfer = 2.617 Kw

Explanation:

Given:

T1 = 300 k

T2 = 440 k

h1 = 300.19 KJ/kg

h2 = 441.61 KJ/kg

Density = 1.225 kg/m²

Find:

Mass flow rate = 1.225 x [1.3/60]

Mass flow rate = 0.02654 kg/s

mh1 + mw = mh2 + Q

0.02654(300.19 + 240) = 0.02654(441.61) + Q

Q = 2.617 Kw

Heat transfer = 2.617 Kw

4 0
3 years ago
Identify the branch of study each student should pursue.
Paha777 [63]

Answer:

Explanation:

i took the test and screenshot it

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3 years ago
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