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3 years ago

In remote areas, your gps devices may lose reception. It’s a good idea to have a

2 answers:

4 0

Answer: MAP!!! it's always a good Idea to have a map nearby, cause you'll never know when you'll need it or like you said, lose reception

Serhud [2]3 years ago

3 0

Answer:

You will need a map, having a map is always very important when electronic gadgets don't work, but you can also take a compass with you.

Explanation:

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Hi im ***ar and im doing sculptural but what should it be about star wars or Marvel

Anarel [89]

Answer:

Cool I think u should do Marvel first

4 0

3 years ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

Which claim does president Kennedy make in speech university rice ?

mafiozo [28]

Answer: The United States must lead the space race to prevent future wars.

Explanation: Hope this helps

4 0

3 years ago

Read 2 more answers

A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the

alekssr [168]

Answer:

answer

Explanation:

4 0

3 years ago

Considering the analogy between electrical circuit and thermal circuit, show your approach to derive an expression for the therm

Simora [160]

Answer:

Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.

Explanation:

Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that $R_{th}$ represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:

Wall Thickness. More thickness, more thermal resistance.
Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
Cross-section Area. More cross-section area, less thermal resistance.

A expression to define the thermal resistance for the wall is as follows: $R_{th} =\frac{l}{Ak}$ , where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.

5 0

3 years ago