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Ainat [17]
3 years ago
5

What is the formula that represents the simplest ratio of the atoms in the components

Physics
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

Empirical formula

<em>Hope</em><em> </em><em>it'll</em><em> </em><em>help</em><em>!</em>

<em>stay</em><em> </em><em>safe</em><em>:</em><em>)</em>

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Your partner has been working on your group's circuit and has left the work area. before you begin working on the circuit, what
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You should disconnect all wires from the circuit or make sure the switch is off or batteries are out
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when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why​
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A sealed beaker of what you are told is aqueous nickel sulfide was given to you by the local chemist and you put it on the windo
Masteriza [31]

Answer: D. It is a SUSPENSION

Explanation:

SUSPENSION

This is a combination of two or more single substances. The properties of the components involved are not however changed or lost as is the case with compounds.

For this reason this mixture can be separated due to sedimentation or filtering.

After a few days, this occurs in the aqueous nickel sulfide because the solid nickel sulfide is separating from the water.

7 0
3 years ago
heat engine operating between 100 °C and 700 °C has an efficiency equal to 40% of the maximum theoretical efficiency. How much e
ipn [44]

Answer:

Explanation:

Theoretical efficiency = T₁ - T₂ / T₁ where T₁ and T₂ is absolute temperature of hot and cold end of the heat engine.

= 600 / (273 + 700 )

= 600 / 973

= .6166

operating efficiency = 40% of .6166

= .4 x .6166

= .2466 = 24.66 %

efficiency = work output / heat input

= 5000 / heat input = .2466

heat input = 5000 / .2466

= 20275.75 J .

HEAT EXTRACED = 20275.75 J.

3 0
3 years ago
A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to
Nostrana [21]

Answer:

a) r=4.24cm d=1 cm

b) Q=5x10^{-10} C

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}

V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m

The distance must be the separation the r distance can be find also using

C=\frac{Q}{V_{ab}}

But now don't know the charge these plates can hold yet so

a).

d=0.01m

C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}

A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}

A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } }  \\r=42.55x^{-3}m

b).

C=\frac{Q}{V_{ab}}

Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C

5 0
3 years ago
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