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IrinaK [193]
2 years ago
12

The low-pressure area near Earth’s equator is filled by cool air moving in from ________. Btw this is science

Physics
1 answer:
bonufazy [111]2 years ago
5 0

Answer:

the north and south pole

Explanation:

this should be the correct answer

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What Is An Insulator Give An Example
kati45 [8]

plastics, Styrofoam hOPE THIS HELPS

8 0
3 years ago
Read 2 more answers
A type of Lead (Pb) ion has a + 4 oxidation number. Sulfur (S) has a — 2 oxidation number. What would be the chemical formula fo
serious [3.7K]

Answer: The chemical formula for the compound of these two elements is PbS_2

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here metal lead is having an oxidation state of +4 called as Pb^{4+} cation and sulphur non metal has oxidation state of -2 called as S^{2-}. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral PbS_2

The chemical formula for the compound of these two elements is PbS_2

4 0
3 years ago
.2, A car starting from rest has an acceleration of
riadik2000 [5.3K]

Answer: 2.5 m/s and 6.25 m

Explanation:

u = 0

a = 0.5 m/s²

t = 5 s

v = u + at

=  0 + 0.5 × 5

= <u>2.5 m/s</u>

s = ut + 1/2 at²

= 1/2 × 2.5 × 5

=<u> 6.25 m</u>

7 0
3 years ago
A beverage manufacturer wants to increase the solubility of carbon dioxide (CO2) in its carbonated drinks.
Brut [27]

D. Decreasing its temperature

Explanation:

Decreasing the temperature of the carbon dioxide gas to be dissolved in the carbonated drink will most likely increase the solubility of the gas in the drink.

Temperature has considerable effects on the solubility of gases in liquids.

  • Dissolution involves the surrounding of ions by water molecules, in this case, the carbon dioxide gas is to be surrounded by the liquid beverage medium.
  • Increasing pressure increases the rate at which gases are soluble. At high pressure, the gases are brought more in contact with the liquid medium.
  • Decreasing temperature aids gas solubility.
  • If the temperature of gases are increased,  they will not want to stay in solution as they gain a high amount of kinetic energy.
  • Therefore, it will increase their randomness and the urge to leave the solution.
  • Decrease in temperature and increase in pressure makes gas solubility to be fast.  

Learn more:

Rate of chemical reactions brainly.com/question/6281756

#learnwithBrainly

6 0
3 years ago
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
3 years ago
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