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allochka39001 [22]
3 years ago
5

Smallest planet in our solar system

Physics
2 answers:
serg [7]3 years ago
7 0

Answer: pluto

Explanation:

pluto is the smallest

bulgar [2K]3 years ago
3 0
Mercury is the smallest planet in our solar system—only slightly larger than Earth's Moon.
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Why is genetic stability important?​
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A metal bar, pivoted at one end, oscillates freely in the absence of a magnetic field. But when it oscillates between the poles
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When the metal bar oscillates between the poles of a magnet, it experience a change in the magnetic flux ( no of magnetic field lines passing through the metal bar) as it enter or leaves the magnetic field of the poles. As we know that the change in magnetic field induces the electric current in the metal bar (conductor). By considering the lenz law which states that the direction of induced current in the conductor will be such that as to oppose the initial magnetic field that is producing it. This opposing force acting on the metal bar will damp the oscillations of the bar between the poles of a magnet.

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Imagine a particle that has three times the mass of the electron. All other quantities given above remain the same. What is the
melamori03 [73]

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It is only necessary to calculate the ratio Eh/Ee

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The kinetic energy of the heavy paricle is three times the kinetic energy of an electron

5 0
4 years ago
Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85% of the of the entering water is e
Natalka [10]

Answer:

a)

i) v'=\frac{17}{20}                                   ii) \frac{m_v}{m_f-m_v} =\frac{17}{83}

b) m_r=752963.55\ kg

Explanation:

Given:

fraction of water in wet sugar of m kg by mass, m'_w=\frac{1}{5} \times m

% of water evapourated from the total water after passing through the evapourator, m'_v=85\%

a)

amount of wet sugar fed to the evapourator, m_f=100\ kg

Now the mass of water present in the fed amount of sugar:

m_w=\frac{1}{5} \times m_f

m_w=\frac{100}{5}

m_w=20\ kg

Now the amount of water leaving from this total amount of water after passing through the evapourator:

m_v=m_v' \times m_w

m_v=\frac{85}{100} \times 20

m_v=17\ kg

i)

So, the fraction of of water leaving the evapourator:

v'=\frac{m_v}{m_w}

v'=\frac{17}{20}

ii)

Now the ratio of kg water vaporized/kg wet sugar leaving the evaporator.:

\frac{m_v}{m_f-m_v} =\frac{17}{100-17}

\frac{m_v}{m_f-m_v} =\frac{17}{83}

b)

amount of sugar fed per day, m_f=907185\ kg

<u>Now the mass of water in the given amount of sugar per day:</u>

m_w=\frac{m_f}{5}

m_w=\frac{907185}{5}

m_w=181437\ kg

Mass of water vapourized after passing through the evaporator:

m_v=\frac{85}{100}\times m_w

m_v=\frac{85}{100}\times 181437

m_v=154221.45\ kg

Now the mass of water still remaining in the sugar:

m_r=m_w-m_v

m_r=907185-154221.45

m_r=752963.55\ kg

is the mass of extra water to be evapourated.

8 0
3 years ago
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