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3 years ago

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Which term describes the amount of heat needed to change 1 kg of a substance from solid to liquid at its melting point? A. Speci

fic heat of phase changes B. Specific heat capacity C. latent heat of fusion D. latent heat of vaporization

1 answer:

Lorico [155]3 years ago

6 0

It is called the specific latent heat of fusion. C.

This is the heat change required in changing 1kg of a solid to its liquid and it occurs at the melting point. It is not associated with temperature change.

If it is to convert 1kg of liquid to vapor at its boiling point it is called the latent heat of vaporization.<span />

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The first law of thermodynamics says that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another.

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3 years ago

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3 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

A tank of liquid (SG=0.80) that is 1 ft in diameter and 1.0 ft high is rigidly fixed to a rotating arm having a 2 ft radius. The

IgorLugansk [536]

Answer:

the pressure at B is 527psf

Explanation:

Angular velocity, ω = v / r

ω = 20 /1.5

= 13.333 rad/s

Flow equation from point A to B

$P_A+rz_A-\frac{1}{2} Pr_A^2w^2=P_B+rz_B-\frac{1}{2} pr^2_Bw^2\\\\P_B = P_A + r(z_A-z_B)+\frac{1}{2} pw^2[(r_B^2)-(r_A)^2]\\\\P_B = [25 +(0.8+62.4)(0-1)+\frac{1}{2}(0.8\times1.94)\times(13.333)^2[2.5^2-1.5^2] ]\\\\P_B = 25 - 49.92+551.79\\\\P_B = 526.87psf\\\approx527psf$

the pressure at B is 527psf

6 0

3 years ago