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3 years ago

The United States, England, and Russia were members of the:

1 answer:

wel3 years ago

8 0

Allied powers
The United States, England, and Russia were members of the: <span> Axis Powers
Allied Powers
Triple Entente</span>

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Define an astronomical unit. Choose all that apply. a. 8.3 minutes Average distance from Earth to the Sun. b. Distance that ligh

Naddik [55]

Answer:

a. 8.3 minutes average distance from earth to the sun

d. 93 miles or 150 million km

Explanation:

The distance between the earth and the sun is defined as an astronomical unit (AU). It takes 8.3 minutes to go from earth to the sun at the speed of light. That distance has a length of 150 million Kilometers or 93 miles.

It is common to see in planet charts that distance to the sun are compared in astronomical units. In the case of Mars is 1.524 AU away from the sun.

6 0

3 years ago

A"car"initially"at"rest"experiences"a" constant"acceleration"along"a"horizontal" road."the"position"of"the"car"at"several" succe

marysya [2.9K]

In the process of peppering the question with those forty (40 !) un-necessary quotation marks, you neglected to actually show us the illustration. So we have no information to describe the adjacent positions, and we're not able to come up with any answer to the question.

7 0

3 years ago

What is the frequency of a microwave of wavelength 3cm?

navik [9.2K]

Frequency = (speed) / (wavelength)

Speed = 3 x 10⁸ m/s

Wavelength = 3 cm = 0.03 m

Frequency = (3 x 10⁸ m/s) / (0.03 m)

Frequency = (3 x 10⁸ / 0.03) (m / m-s)

Frequency = 1 x 10¹⁰ Hz (10 Gigahertz)

5 0

3 years ago

The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the

Airida [17]

Answer:

$\frac{I}{I_0}=113.68$

Explanation:

P = Acoustic power = 63 µW

r = Distance to the sound source = 210 m

Acoustic power

$P=IA\\\Rightarrow I=\frac{P}{A}\\\Rightarrow I=\frac{63\times 10^{-6}}{4\times \pi \times 210^2}$

Threshold intensity = $I_0=1\times 10^{-12}\ W/m^2$

Ratio

$\frac{I}{I_0}=\frac{\frac{63\times 10^{-6}}{4\times \pi \times 210^2}}{1\times 10^{-12}}\\\Rightarrow \frac{I}{I_0}=113.68$

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68

6 0

3 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

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5 0

3 years ago