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Ann [662]

Well basically in physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant, it is said to be conserved over time. This law means that energy can neither be created nor destroyed, it can only be transformed or transferred from one form to another.

Hope this helps (:

5 0

4 years ago

This equation is known as the ideal gas law, and it can be used to predict the behavior of many gases at relatively low pressure

Masja [62]

The correct answer is
<span>C) either the pressure of the gas, the volume of the gas, or both, will increase.

In fact, the ideal gas law can be written as
</span> $pV=nRT$
<span>where
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas

We can see that if the temperature T increases, then the term on the right in the equation increases, therefore the term on the left should increase as well. In order for this to be possible, at least one between p and V should increase, or also both of them. Therefore, the correct answer is C.</span>

6 0

3 years ago

Mars has a mass of about 6.24 Ã 1023 kg, and its moon phobos has a mass of about 9.2 Ã 1015 kg. if the magnitude of the gravitat

Effectus [21]

Given: Mass of Mars Mm = 6.24 x 10²³ Kg;

Mass of Phobos Mp = 9.2 x 10¹⁵ Kg

Gravitational Force F = 4.47 X 10¹⁵ N

Gravitational constant G = 6.67 X 10⁻¹¹ N m²/Kg²

Radius r = ?

Formula: F = GMmMp/r² derive r

r = √GMmMp/F

r = √(6.67 x 10⁻¹¹ N m²/Kg²)(6.24 x 10²³ Kg)(9.2 x 10¹⁵ Kg)/4.47 x 10¹⁵ N

r = √(6.67 x 10⁻¹¹ N m²/Kg²)5.74 x 10³⁹ Kg²/4.47 x 10¹⁵ N

r = √8.57 x 10¹³ m

r = 9,257,429 m

or r = 9.26 x 10⁶ m

8 0

4 years ago

An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the

Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

.

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

$(n1)*(sin(theta1))=(n2)*(sin(theta2))$

Then we will find the refraction angle relative to the vertical this way:

$(n1/n2)*(sin(theta1))=sin(theta2)$

$(1/1.33)*(sin(45))=sin(theta2)$

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

$tan(32.12)=(X/30cm)$

$X=(tan(32.12))*(30cm)$

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

$tan(32.12)=(24cm/Y)$