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MArishka [77]
2 years ago
6

Heyyy Powerpuff girls! what advantage does globe maps have over a flat map?

Chemistry
1 answer:
babunello [35]2 years ago
4 0

Answer:

The advantage of the globe is that it promotes visual accuracy. Students need to use a globe frequently if they are to form accurate mental maps. The advantage of the world map is that you can see the entire world at one time. The disadvantage is that world maps distort shape, size, distance, and direction.

Explanation:

miss girl- you could've just looked this up

(just make it look like you typed it)

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99 POINTS!!!!!! PLEASE HELP ASAP!!!!!
ohaa [14]
What you can do is organize them by color, what matter they are in room temperature, their molecular structure, or what kind of conductor in electricity and heat it is. I'm not sure what the format is supposed to look like but first just organize them all in different categories.
5 0
3 years ago
Read 2 more answers
What causes the jet stream to move farther south?
Anni [7]
<span>Jet streams are the major means of transport for weather systems.  A jet stream is an area of strong winds ranging from 120-250 mph that can be thousands of miles long, a couple of hundred miles across and a few miles deep.  Jet streams usually sit at the boundary between the troposphere and the stratosphere at a level called the tropopause.  This means most jet streams are about 6-9 miles off the ground. Figure A is a cross section of a jet stream.
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The dynamics of jet streams are actually quite complicated, so this is a very simplified version of what creates jets.  The basic idea that drives jet formation is this:  a strong horizontal temperature contrast, like the one between the North Pole and the equator, causes a dramatic increase in horizontal wind speed with height.  Therefore, a jet stream forms directly over the center of the strongest area of horizontal temperature difference, or the front.  As a general rule, a strong front has a jet stream directly above it that is parallel to it.  Figure B shows that jet streams are positioned just below the tropopause (the red lines) and above the fronts, in this case, the boundaries between two circulation cells carrying air of different temperatures.


7 0
3 years ago
The sodium-potassium exchange pump
trapecia [35]

Answer: The correct answer is option E

Explanation:

Sodium/potassium pump is a mechanism that involves the movement of sodium ions (Na+) out of a cell and potassium ions (K+) into a cell, thereby regulating concentration of ions on both sides of a typical cell membrane.

In this situation, the sodium-potassium pump is usually helps in the establishment of the resting potential. The potassium voltage channels normally closes before the membrane potential is brought to a resting level.

In summary, sodium/potassium pump helps to maintain a balance in the system.

6 0
3 years ago
Calculate the second volumes. <br> 7.03 Liters at 31 C and 111 Torr to STP
DIA [1.3K]

The second volume :    V₂= 0.922 L

<h3> Further explanation </h3><h3>Given </h3>

7.03 Liters at 31 C and 111 Torr

Required

The second volume

Solution

T₁ = 31 + 273 = 304 K

P₁ = 111 torr = 0,146 atm

V₁ = 7.03 L

At STP :  

P₂ = 1 atm

T₂ = 273 K

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

0.146 x 7.03 / 304 = 1 x V₂/273

V₂= 0.922 L

4 0
3 years ago
If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a
strojnjashka [21]

Answer : The final temperature of the mixture is, 22.14^oC

Explanation :

First we have to calculate the mass of ethanol and water.

\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g

and,

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ethanol = 2.42J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol = 35.5 g

m_2 = mass of water = 45.0 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 8.0^oC

T_2 = initial temperature of water = 28.6^oC

Now put all the given values in the above formula, we get:

35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.14^oC

Therefore, the final temperature of the mixture is, 22.14^oC

3 0
3 years ago
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