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3 years ago

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The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular clo

ud is in the form of water molecules, and the mass density of such a cloud is roughly 4.0×10−21 g/cm3 Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.

1 answer:

Rudik [331]3 years ago

5 0

To solve the problem we first need to define the nomenclature by which there is an X amount of water in our body. Let's call this a variable m, corresponding to the total mass in us and whose units are Kilograms.

$Amount_{water} = m (Kg)$

We understand that one millionth of this mass, a small fraction of it, is the equivalent of that of a Water Molecule.

Assuming this to have the same body of water in our body that amount should be approximately

$m_{water}=10^6$

Density of water molecule

$\rho = 10^{-18}kg/m^3$

$V = \frac{m*10^6}{10^{-18}}$

$V = m*10^{24}m^3$

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PLZ HELP I DONT GET IT

GaryK [48]

Answer:

0.1 L

Explanation:

From the question given above, we obtained the following data:

Initial volume (V₁) = 0.05 L

Initial Pressure (P₁) = 207 KPa

Final pressure (P₂) = 101 KPa

Final volume (V₂) =?

We can obtain the new volume (i.e the final volume) of the gas by using the Boyle's law equation as illustrated below:

P₁V₁ = P₂V₂

207 × 0.05 = 101 × V₂

10.35 = 101 × V₂

Divide both side by 101

V₂ = 10.35 / 101

V₂ = 0.1 L

Thus, the new volume of the gas is 0.1 L

6 0

3 years ago

*example included* Two uncharged spheres are separated by 3.50 m. If 1.30 ✕ 10¹² electrons are removed from one sphere and place

likoan [24]

Considering the Coulomb's Law, the magnitude of the Coulomb force is 3.1865 N.

<h3>Coulomb's Law</h3>

Charged bodies experience a force of attraction or repulsion on approach.

From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.

From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=k\frac{Qq}{d^{2} }$

where:

F is the electrical force of attraction or repulsion. It is measured in Newtons (N).
Q and q are the values of the two point charges. They are measured in Coulombs (C).
d is the value of the distance that separates them. It is measured in meters (m).
K is a constant of proportionality called the Coulomb's law constant. It depends on the medium in which the charges are located. Specifically for vacuum k is approximately 9×10⁹ $\frac{Nm^{2} }{C^{2} }$ .

The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.

<h3>This case</h3>

In this case, you know that:

The two uncharged sphere are separated by the distance of d= 3.50 m
The number of electrons are 1.30×10¹².
Electrons is elementary charge and charges on both the sphere is same. The value of electron is 1.602×10⁻¹⁹ C. This is, Q=q=1.30×10¹²×1.602×10⁻¹⁹ C= 2.0826×10⁻⁷ C

Replacing in Coulomb's Law:

$F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(2.0826x10^{-7} C)x(2.0826x10^{-7} C)}{(3.50 m)^{2} }$

Solving:

<u><em>F= 3.1865 N</em></u>

Finally, the magnitude of the Coulomb force is 3.1865 N.

Learn more about Coulomb's Law:

brainly.com/question/26892767

#SPJ1

7 0

1 year ago

In most circumstances, the normal force acting on an objectand

Lynna [10]

Answer:

Normal Force is usually perpendicular to the movement and static friction usually means that there is no movement.

Explanation:

The work donde by any force on an object is equal to the displacement of the object multiplied by the component of the force that is in the direction of the displacement.

Normal force is usually perpendicular to the movement, so there is no component in the direction of the displacement. This is why it is zero in most circumstances.

<em>Static</em> friction on the other hand, usually means that there is no movement at all (it's static). It means that there is no displacement between the object and ground (in most cases). If there is no displacement, there is no work.

4 0

3 years ago

The electric potential at the dot in the figure is 3160 V. What is charge q?

PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

$V = \frac{kQ}{r}$

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

$V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V$

Lower left charge's potential:

$V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V$

Add the two, and subtract from the total EP at the point:

$3160 + 1123.75 - 2247.5 = 2036.25$

The remaining charge must have a potential of 2036.25 V, so:

$2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}$

5 0

2 years ago

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

emmainna [20.7K]

Answer:

$5,2$

Explanation:

From the question we are told that:

Speed of light $C=3.0×10^8 m/s.$

Generally the equation for Average Speed is mathematically given by

$V_{avg}=\frac{d}{t}$

Where

d=Distance between the Earth and the sun

$d=1.5*10^11m$

Therefore

$t=\frac{d}{V_{avg}}$

$t=\frac{1.5*10^11m}{3.0×10^8 m/s.}$

$t=5*10^2s$

Since m and n is given in the form of

$m*10^n$

Therefore

$m=5 & n=2$

$5,2$

3 0

3 years ago