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RSB [31]
3 years ago
13

The volumes of a sample of lead and a sample of platinum are the same: 2 cm3. The density of lead is 11.3 g/cm3 and the density

of platinum is 21.5 g/cm3.
Which is an accurate description of their relative masses?
A. The masses of the lead and platinum samples are the same.
B. The mass of the lead sample is greater than the mass of the platinum sample.
C. The mass of the platinum sample is greater than the mass of the lead sample.
Physics
2 answers:
Vlad1618 [11]3 years ago
4 0

The answer is C. The mass of the platinum sample is greater than the mass of the lead sample. As I explained in a previous answer, if they are the same volume, but one is heavier, then it must be more dense. In this particular example, the platinum is more dense than the lead, and therefore has more mass.

JulsSmile [24]3 years ago
4 0

Answer:

The mass of the platinum sample is greater than the mass of the lead sample (option C.)

Explanation:

Density is a magnitude that allows you to measure the amount of mass in a given volume of a substance. Then, the expression for the calculation of density is the ratio between the mass of a body and the volume it occupies:

d=\frac{m}{V}

From this expression it can be deduced that the density is inversely proportional to the volume: the smaller the volume occupied by a given mass, the greater the density.

The density of lead is 11.3 \frac{g}{cm^{3} } . Then it is possible to apply a rule of three as follows: if in 1 cm³ of lead there is 11.3 g of substance, in 2 cm³ how much mass is there?

mass=\frac{2cm^{3}*11.3 g }{1cm^{3} }

mass=22.6 g

The density of platinum is 21.5 \frac{g}{cm^{3} }. Then it is possible to apply a rule of three as follows: if in 1 cm³ of platinum there is 21.5 g of substance, in 2 cm³ how much mass is there?

mass=\frac{2cm^{3}*21.5 g }{1cm^{3} }

mass=43 g

Then, you can see that <u><em>the mass of the platinum sample is greater than the mass of the lead sample (option C.)</em></u>

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For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

Z_{R} = R

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Z_{C} = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = Z_{R} + Z_{L} + Z_{C}

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

v_{C} = I×Z_{C}

Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

Z_{C} = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

v_{R} = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

v_{L} = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

v_{C} = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

v_{R} = (12.35∠1.147+220t)V

v_{L} = (10.95∠2.718+220t)V

v_{C} = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

v_{R} = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

v_{L} = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

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3 years ago
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