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faust18 [17]
4 years ago
7

The army has cars and boats etc right

Engineering
1 answer:
pochemuha4 years ago
7 0

Answer:

Yes this is true we have anything except the flying fighter we call them in if we need them by support

Explanation:

The army is a good place to learn how to drive cool things how to drive a tank thats why the army owns all this army things helps us stay safe and protect the usa government

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Vital role of maritime english among seaferers
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Answer:

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8 0
3 years ago
On a piece of paper, sketch the x-y stress state and the properly oriented principal stress state. Use the resulting sketch to a
stealth61 [152]

Answer:

See explaination

Explanation:

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6 0
4 years ago
1. A team of students have designed a battery-powered cooler, which promises to keep beverages at a high-drinkability temperatur
Anit [1.1K]

Answer:

Minimum electrical power required = 3.784 Watts

Minimum battery size needed = 3.03 Amp-hr

Explanation:

Temperature of the beverages, T_L = 36^0 F = 275.372 K

Outside temperature, T_H = 100^0F = 310.928 K

rate of insulation, Q = 100 Btu/h

To get the minimum electrical power required, use the relation below:

\frac{T_L}{T_H - T_L} = \frac{Q}{W} \\W = \frac{Q(T_H - T_L)}{T_L}\\W = \frac{100(310.928 - 275.372)}{275.372}\\W = 12.91 Btu/h\\1 Btu/h = 0.293071 W\\W = 12.91 * 0.293071\\W_{min} = 3.784 Watt

V = 5 V

Power = IV

W_{min} = I_{min} V\\3.784 = 5I_{min}\\I_{min} = \frac{3.784}{5} \\I_{min} = 0.7568 A

If the cooler is supposed to work for 4 hours, t = 4 hours

I_{min} = 0.7568 * 4\\I_{min} = 3.03 Amp-hr

Minimum battery size needed = 3.03 Amp-hr

6 0
3 years ago
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
4 years ago
What is a Flame Front Generator?
Inessa [10]

Answer and Explanation:

Flame Front Generator: It is a ignition system which is very useful in flaring system .In this system the air and gases are mixed together and make a combustible air gas mixture. There is a flame front region where the combustion reaction takes place , it is the region where gases as like hydrogen and air mixed with each other and form combustible gases.

7 0
3 years ago
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