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2 years ago

5

Please help been stuck on this for a couple minutes

1 answer:

AnnyKZ [126]2 years ago

3 0

Answer:

true

Explanation: Fire extinguishers do need water buckets on sand on the other hand sand is also true so that means the awnser is true!

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Differentiate between isohyetal method and arithmetical average method of rainfall

prohojiy [21]

Answer:

Explained below

Explanation:

The isohyetal method is one used in estimating Rainfall whereby the mean precipitation across an area is gotten by drawing lines that have equal precipitation. This is done by the use of topographic and other data to yield reliable estimates.

Whereas, the arithmetic method is used to calculate true precipitation by the way of getting the arithmetic mean of all the points or arial measurements that will be considered in the analysis.

7 0

3 years ago

What are the inputs and outputs of a sailboat?

bearhunter [10]

Answer:

sailing ships were the primary means of maritime trade and transportation; exploration across the seas and oceans was reliant on sail for anything other than the shortest distances. Naval power in this period used sail to varying degrees depending on the current technology, culminating in the gun-armed sailing warships of the Age of Sail.

7 0

1 year ago

15 POINTS! Help.

IRISSAK [1]

Answer: it would overload

Explanation:

4 0

2 years ago

Read 2 more answers

A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he

sergejj [24]

Answer:

T₂ =93.77 °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure ,P₂= 367+1 = 368 kPa

Lets take temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

${T_2}=T_1\times \dfrac{P_2}{P_1}$

Now by putting the values in the above equation we get

${T_2}=300\times \dfrac{368}{301}\ K$

The temperature in °C

T₂ = 366.77 - 273 °C

T₂ =93.77 °C

8 0

3 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago