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2 years ago

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3. The atmospheres of Mars and Venus are mostly carbon dioxide. Earth'satmosphere contains only a fraction of a percent of carbo

n dioxide. Whyis Earth's atmosphere so different from those of Mars and Venus?

1 answer:

maria [59]2 years ago

3 0

Answer:

Earth's atmosphere is a mixture of nitrogen (79%), oxygen (20%), and a small fraction of carbon dioxide, water vapours and other gases. However, the atmospheres on Venus and Mars mainly consist of carbon dioxide.

Explanation:

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Decibels I think that's the answer

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3 years ago

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2 years ago

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A real power supply can be modeled as an ideal EMF of 60 Volts in series with an internal resistance. The voltage across the ter

lara31 [8.8K]

Answer:

5 ohms

Explanation:

Given:

EMF of the ideal battery (E) = 60 V

Voltage across the terminals of the battery (V) = 40 V

Current across the terminals (I) = 4 A

Let the internal resistance be 'r'.

Now, we know that, the voltage drop in the battery is given as:

$V_d=Ir$

Therefore, the voltage across the terminals of the battery is given as:

$V= E-V_d\\\\V=E-Ir$

Now, rewriting in terms of 'r', we get:

$Ir=E-V\\\\r=\frac{E-V}{I}$

Plug in the given values and solve for 'r'. This gives,

$r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms$

Therefore, the internal resistance of the battery is 5 ohms.

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3 years ago

If the atoms of one object (initially neutral) have electrons rubbed off through friction with a second object, the first object

snow_lady [41]

Positive. The 1st object loses electrons and will thus have an imbalance of charge with loss of electrons.

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3 years ago

Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an

galben [10]

Answer:

$\triangle P=1.95*10^{-4}$

Explanation:

Mass $m=0.001$

Diameter $d=1.2m$

Length $l=10m$

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$Q=AV$

$V=\frac{Q}{\pi/4D^2}$

$V=\frac{0.001}{\pi/4(1.2)^2}$

$V=8.84*10^{-4}$

Generally the equation for Friction factor is mathematically given by

$F=\frac{64}{Re}$

Where Re

Re=Reynolds Number

$Re=\frac{pVD}{\mu}$

$Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}$

$Re=1040$

Therefore

$F=\frac{64}{Re}$

$F=\frac{64}{1040}$

$F=0.06$

Generally the equation for Friction factor is mathematically given by

$Head loss=\frac{fLv^2}{2dg}$

$H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}$

$H=19.9*10^{-9}$

Where

$H=\frac{\triangle P}{\rho g}$

$\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}$

$\triangle P=H*\rho g$

$\triangle P=1.95*10^{-4}$

6 0

2 years ago