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Bezzdna [24]
3 years ago
9

Which type of mixture could this illustration represent?

Physics
2 answers:
Oksanka [162]3 years ago
4 0

Answer:

A homogeneous mixture

Explanation:

Helga [31]3 years ago
4 0

Answer: Option (B) is the correct answer.

Explanation:

A mixture in which solute particles are unevenly distributed in a solvent is known as a heterogeneous mixture.

For example, sand in water is a heterogeneous mixture.

Whereas a mixture in which solute particles are evenly or uniformly distributed in a solvent is known as a homogeneous mixture.

For example, salt in water is a homogeneous solution.

Thus, we can conclude that the given mixture shows that molecules are uniformly distributed in the solution.

Therefore, the given illustration represents homogeneous type of mixture.

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Describe two possibilities of what will happen to the universe in the future.
shutvik [7]
In our current view of the Universe there are 2 possible futures. One possiblity is that the Universe will come to an end in the opposite of a Big Bang called The Big Crunch. The other possibility is that we live in an Eternal Universe that will never come to end.

i hope this helps :)
4 0
3 years ago
Currents circulate in a piece of metal that is pulled through a magnetic field. What are these currents called?
MariettaO [177]
<h3><u>Answer;</u></h3>

Eddy currents

<h3><u>Explanation</u>;</h3>
  • <em><u>Eddy currents are currents which circulate in conductors like swirling eddies in a stream. These currents are induced by changing magnetic fields and flow in closed loops, perpendicular to the plane of the magnetic field. </u></em>
  • They are created as a result of change in the intensity or direction of a magnetic field on a conductor, this occurs either when a conductor is moving through a magnetic field, or when the magnetic field surrounding a stationary conductor is varying
4 0
3 years ago
Which is true of the greenhouse effect?
Aleksandr [31]

Answer:

c is correct option thanks to brainly

3 0
3 years ago
The current theory of the structure of the
Mariana [72]

Answers:

a) 2.82(10)^{21} kg

b) 1410 J

c) 36.62 m/s

Explanation:

<h3>a) Mass of the continent</h3>

Density \rho  is defined as a relation between mass m and volume V:

\rho=\frac{m}{V} (1)

Where:

\rho=2720 kg/m^{3} is the average density of the continent

m is the mass of the continent

V is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}

Finding the mass:

m=\rho V (2)

m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3}) (3)

m=2.82(10)^{21} kg (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy K is given by the following equation:

K=\frac{1}{2}mv^{2} (5)

Where:

m=2.82(10)^{21} kg is the mass of the continent

v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s is the velocity of the continent

K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2} (6)

K=1410 J (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass m=77 kg and the same kinetic energy as that of the continent 1413 J, we can find its velocity by isolating v from (5):

v=\sqrt{\frac{2 K}{m}} (6)

v=\sqrt{\frac{2 (1413 J)}{77 kg}}

Finally:

v=36.62 m/s This is the speed of the jogger

5 0
3 years ago
Two point charges lie on the x axis. A charge of + 2.30 pC is at the origin, and a charge of − 4.50 pC is at x=−11.0cm.
Radda [10]

r₁ = distance of point A from charge q₁ = 0.13 m

r₂ = distance of point A from charge q₂ = 0.24 m

r₃ = distance of point A from charge q₃ = 0.13 m

Electric field by charge q₁ at A is given as

E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C     towards right

Electric field by charge q₂ at A is given as

E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C    towards left

Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction

Electric field at A will be zero when

E₁ = E₂ + E₃

1.225 = 0.703 + E₃

E₃ = 0.522 N/C

Electric field by charge "q₃" is given as

E₃ = k q₃ /r₃²

0.522 = (9 x 10⁹) q₃/(0.13)²

q₃ = 0.980 x 10⁻¹² C = 0.980 pC

4 0
4 years ago
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