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2 years ago

Which two pieces of data indicate that Uranus resides in the outer region of the solar system

Physics

1 answer:

LuckyWell [14K]2 years ago

4 0

Answer:

Our solar system has total eight planets out of which four are inner planets and four are outer planets. The four outer planets are Jupiter, Saturn, Uranus and Neptune. The common characteristics of outer planets is that they are gaseous planets. They are larger on size than the inner rocky planets and are faraway from Sun. They have larger period of revolution around the Sun.

Uranus is a gaseous planet and lies far from Sun and hence has large period of revolution. It takes 84 Earth years to revolve around Sun. This data indicates that Uranus resides in the outer region of the Solar System.

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PLEASE WILL MARK BRAINLIEST

devlian [24]

HI!

Conventional vehicles use gasoline or diesel to power an internal combustion engine. Hybrids also use an internal combustion engine—and can be fueled like normal cars—but have an electric motor and battery, and can be partially or wholly powered by electricity. Also pollute less and save the drivers money.

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6 0

2 years ago

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A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a

AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

$q = -Kb\frac{dh}{dl}$

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25 m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

$V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day$

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

$= 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours$

4 0

2 years ago

Is 5g greater than 20.43 mg

worty [1.4K]

yes that's true 6g is larger

4 0

2 years ago

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How many volts would it take to push 1 amp through a resistance of 1 ohm?

ELEN [110]

V=I x R so V= 1 x 1 =1V

5 0

2 years ago

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Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f

mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , $h_{f} = 1344.8 kJ/kg$

Specific internal energy of saturated liquid at 300°C, $U_{f1} =$ 1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa, $U_{f2}$ = 340.49 kJ/kg

$(U_{fg} )_{2} = 2142.7 kJ/kg$

Let the dryness fraction at the second chamber = x

$U_{2} = U_{f2} + U_{fg2}$

$U_{2} = 340.49 + x2140.7$ Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, $u_{f2} = 0.00103 m^{3} /kg\\$

$u_{2} = u_{f2} + xu_{fg2}$

$u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\$

$u_{2} = \frac{v_{2} }{m_{2} }$

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C $S_{1} = S_{f} = 3.2548 kJ/kg-k\\$

$S_{2} = S_{f2} + xS_{fg2}$

From the steam table,

$S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k$

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

5 0

2 years ago

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