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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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telo118 [61]

Answer:This also means that Mercury's surface gravity is 3.7 m/s2, which is the equivalent of 38% of Earth's gravity (0.38 g). This means that if you weighed 100 kg (220 lbs) on Earth, you would weigh 38 kg (84 lbs) on Mercury.

Explanation:

5 0
2 years ago
If a cart of 2 kg mass has a force of 8 newtons exerted on it, what is its acceleration?
maria [59]
This is an example of the Newton`s Second Law:
F = m * a
a = F / m
F = 8 N, m = 2 kg.
a = 8 N : 2 kg
Answer:
a = 4 m/s²
3 0
3 years ago
the current in a hair dryer measures 24 amps. the resistance of the hair dryer is 11 ohms. what is the voltage?
Diano4ka-milaya [45]

ohm's law V=IR

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7 0
3 years ago
A horizontal 745 N merry-go-round of radius
Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

6 0
3 years ago
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igor_vitrenko [27]
C, at this moment it is directly 90 degrees to the right and is only travelling downwards
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