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3 years ago

Suppose you dissolve 12.8 g of one substance in 11 g of another. Is this reasonable? Explain

Physics

1 answer:

RUDIKE [14]3 years ago

5 0

Answer:

No, its not reasonable.

Explanation:

The substance that is to be dissolved is known as solute. The substance that is dissolving is known as solvent.

The amount of solvent in the mixture should be greater than that of solute.

Suppose we are taking a solvent in a beaker and we are continuously adding solute in it. Initially the solute dissolve quickly. At some point the solute stops dissolving in the solvent. This is known as saturation point of the solvent. After saturation point if solute is added further it does not dissolve in the solvent.

So, its not possible to dissolve 12.8 g of one substance in 11 g of another.

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An automobile starter motor has an equivalent resistance of 0.0510 Ω and is supplied by a 12.0 V battery with a 0.0090 Ω interna

Alisiya [41]

Answer:

a) 200A

b) 10.2V

c) 2.04kW

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

$I=\frac{V}{R}$

Where R is the equivalent resistance of the resistors in series

$R=0.0510+0.0090=0.0600[ohm]$

$I=\frac{12.0}{0.0600}=200A$

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

$V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V$

The power dissipated supplied to the motor is given by:

$P=I^2*R_m\\P=(200)^2*0.0510=2.04kW$

now solving adding a 0.0900 ohm resistor:

$I=\frac{12.0}{0.15}=80A$

$V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V$

$P=I^2*R_m\\P=(200)^2*0.0510=0.326kW$

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