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AleksAgata [21]

2 years ago

5

Which type of wave interaction is shown by the water waves in the photo?

2 answers:

Neko [114]2 years ago

8 0

I think A but I’m not sure so

bija089 [108]2 years ago

5 0

Answer:

A. Reflection.

Explanation: You can see the reflection of the trees.

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

vladimir1956 [14]

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ $\frac{5.60}{2}gallons = 2.80 gallons$

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = $(M__1}) = Density*Volume$

$= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 11659.06grams$

On the other hand; the mass of water = $(M__2})= Density*Volume$

$= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 10577.95 grams$

Molarity = $\frac{massof solute*1000}{molarmassof solute*massofwater}$

= $\frac{11659.06*1000}{62.07*10577.95}$

= 17.757 m

≅ 17.76 m

∴ the boiling point of the solution is calculated using the boiling‑point elevation constant for water and the Molarity.

$\Delta T_{boiling} = k_{boiling}M$

where,

$k_{boiling}$ = 0.512 °C/m

$\Delta T_{boiling}$ = 100°C + 17.56 × 0.512

= 109.09 °C

6 0

3 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago

PLEASE HELP WITH TRUE OR FALSE QUESTIONS!!!!!

Elodia [21]

First one is true second one is False

3 0

3 years ago

If each of you pulls with equal force, the rope will not move (the two forces are equal in strength, but opposite in direction).

Vedmedyk [2.9K]

Answer:

i dont know

Explanation:

boooooooooooooooy

8 0

3 years ago

Please help ————//-//

grandymaker [24]

Probably b because soil can’t be the exact same and so that cancels out a and d

6 0

3 years ago

Read 2 more answers