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Gnoma [55]
3 years ago
14

Which characteristic do arid and polar regions share?

Physics
2 answers:
Setler79 [48]3 years ago
8 0

few plants

Explanation:

Arid and polar regions of the world typically have few plant species. The plant species in such environment are called xerophytes. They are plants that can live in an area with little to no precipitation.

  • Arid regions of the world are about the most driest in the world with very high diurnal temperature range.
  • Polar regions are frigid and cold all year round.
  • Both arid and polar regions have little to no precipitation all year round.
  • They have poor soils to support plant growth.
  • This is why they both have few plants.

Learn more:

Temperate and tropical climate brainly.com/question/10856870

#learnwithBrainly

irinina [24]3 years ago
6 0

Answer:

B.

Explanation:

I just did the test.

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A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 31.0 nC , respecti
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Answer:

The charge on the third object is − 21.7nC

Explanation:

From Gauss's Law

Φ = Q/ε₀

where;

Φ is the total electric flux through the shell = − 533 N⋅m²/C

Q is the total charge Q in the shell = ?

ε₀ is the permittivity of free space = 8.85 x 10⁻¹²

From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

Q =  −4.7 X 10⁻⁹ C = -4.7nC

Q = q₁ + q₂ + q₃

− 4.7nC = − 14.0 nC + 31.0 nC + q₃

− 4.7nC − 17nC = q₃

− 21.7nC = q₃

Therefore, the charge on the third object is − 21.7nC

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PLS HELP!!!- The moon rotates around the Earth every 28 days. What is the frequency of the moon’s revolution?
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A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

4 0
3 years ago
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