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1 year ago

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Cold air masses tend to originate from: Question 3 options: tropical areas an area where it's wintertime an area where it's summ

ertime polar areas

2 answers:

jek_recluse [69]1 year ago

7 0

Polar areas, Hope this helps

suter [353]1 year ago

6 0

Answer:

Polar areas

Explanation:

From what I know Arctic, Antarctic, and polar air masses are cold. Polar (cold), Arctic (very cold), Equatorial (warm and very moist), and Tropical (warm).

With the options you gave, you can immeditaly elimate 1. tropical areas and 2. an area where its summertime.

Your then left with two options 1. an area where its wintertime and 2. polar areas

I then concluded that cold air masses tend to originate from POLAR AREAS.

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In handball, who is the only player allowed in the goalie area?

lidiya [134]

Answer:

Only the goalie is allowed inside the goal crease. The only exception when another player is allowed in the goal area is when they take off from outside the goal area, and shoots or passes the ball before landing. To avoid interference with other players, the player must then exit the goal area as soon as possible.

Explanation:

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2 years ago

Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon

dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, $a=7.78\times 10^{11}\ m$

Mass of the sun, $M=1.99\times 10^{30}\ kg$

(a) Let T is the period of Jupiter's orbit. It is given by :

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$T^2=\dfrac{4\pi^2}{GM}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3$

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(b) We know that,

$1\ year=3.154\times 10^7\ s$

or

$1\ s=3.171\times 10^{-8}\ year$

$3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}$

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3 years ago

Read 2 more answers

How is most of the electricity we use at home generated?

Sveta_85 [38]

Nuclear power plants, wind farms, water farms, and geothermal heating

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3 years ago

Which shows the formula for converting from kelvins to degrees Celsius? °C = (9/5 × K) +32 °C = 5/9 × (K – 32) °C = K – 273 °C =

madam [21]

The freezing point of the water is 0 C , and it equals to 273 K

Then, To convert from Kelvins degrees to Celsius degrees we use the relation

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Also,

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3 years ago

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In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an

VashaNatasha [74]

Answer:

$V=3.475ft^3/s=3.48ft^3/s$

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

$P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})$

$P= 73.7*10^{3}ft.lbf/s$

we have other values such $h=400ft$ and $\gamma= 62.42lb/ft^3$ (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

$V=\frac{P}{\gamma h \eta_0}$

Where $\eta_0$ is the turbine efficiency, at which is,

$\eta_0 = \frac{P}{\gamma Vh}$

Replacing,

$V=\frac{73.7*10^{3}}{62.42*400*0.85}$

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With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

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converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

$m= 217lbm/s$

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2 years ago