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2 years ago

What are the advantages of atmospheric pressure?

Physics

2 answers:

Harrizon [31]2 years ago

8 0

Answer:

ATMOSPHERIC PRESSURE HAS IMPORTANT EFFECTS ON WATER CHEMISTRY AND WEATHER CONDITION . IT AFFECTS THE AMOUNT OF GAS THAT CAN DISSOLVE IN WATER. LIKE OXYGEN CAN DISSOLVE IN HIGH PRESSURE MORE THAN LOWER AIR PRESSURE.

Andrew [12]2 years ago

4 0

Answer:

it effects the amount of gas that dissolve in water

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Match the type of boundary with it's characteristic

vodomira [7]

Answer:

Transform= not destroyed or created

Divergent= crust created

Convergent= crust destroyed

Explanation:

The plates move in the opposite or away from each other at a transforming plate boundary. The two platform borders are not produced or destroyed in this case. As both plates converge on each other and thus destroy the plates for converging plate boundaries. When the plate is divergent, both plates shift away from each other by opening up and solidification for a new crust.

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3 years ago

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How does the molecular structure of a water molecule affect its polarity?

MA_775_DIABLO [31]

Answer:

The polarity of water molecules means that molecules of water will stick to each other like when unlike charges attracts. This is called hydrogen bonding.

Polarity makes water a good solvent, gives it the ability to stick to itself (cohesion), stick to other substances (adhesion), and have surface tension (due to hydrogen bonding).

When the two hydrogen atoms bond with the oxygen, they attach to the top of the molecule. This molecular structure gives the water molecule polarity, or a lopsided electrical charge that attracts other atoms. The end of the molecule with the two hydrogen atoms is positively charged.

Explanation:

4 0

3 years ago

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A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched b

Firlakuza [10]

From the given information in the question, the correct option is Option 1: 14 cm.

A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = $\frac{1}{2}$ k $x^{2}$

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = $\frac{1}{2}$ k $x^{2}$

10 = $\frac{1}{2}$ k $10^{2}$

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = $\frac{1}{2}$ k $x^{2}$

20 = $\frac{1}{2}$ (0.2) $x^{2}$

40 = 0.2 $x^{2}$

$x^{2}$ = 200

x = $\sqrt{200}$

= 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

For more information, check at: brainly.com/question/1352053.

8 0

3 years ago

A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the

MissTica

Answer : The value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

$q=n\times c_v\times \Delta T$

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

$\Delta T$ = Change in temperature = 15.7 K

$c_v$ = heat capacity at constant volume of $N_2$ (diatomic molecule) = $\frac{5}{2}R$

R = gas constant = 8.314 J/mol.K

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

$q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K$

$q=286.2J$

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

As we know that, work done is zero at constant volume. So,

$\Delta U=q=286.2J$

Therefore, the value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

8 0

3 years ago

Read 2 more answers

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

3 years ago

What are the advantages of atmospheric pressure?​

What are the advantages of atmospheric pressure?