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alexandr402 [8]
2 years ago
8

Gold and silicon are mutually insoluble in the solid state and form a eutectic system with a eutectic temperature of 636 k and a

eutectic composition of xsi=0.186. calculate the gibbs free energy of the eutectic melt relative to (a) unmixed liquid au and liquid si, and (b) unmixed solid au and solid si.
Physics
1 answer:
kupik [55]2 years ago
3 0
Yupp its c because my dad farted 
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A quantity y is to be determined from the equation y=(px)/q^2
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Answer:

heya answer option b

Explanation:

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4 0
1 year ago
A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are
Dahasolnce [82]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

4 0
2 years ago
A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri
MaRussiya [10]

Answer:

dz=19217687.07\ m

Explanation:

Given:

  • initial gauge pressure in the container, P_0=2.02\times 10^{5}\ Pa
  • atmospheric pressure at sea level, P_a=1.01\times 10^5\ Pa
  • initial volume, V_0=4.4\times 10^{-4}\ m^3
  • maximum pressure difference bearable by the container, dP_{max}=2.26\times 10^{5}\ Pa
  • density of the air, \rho_a=1.2\ kg.m^{-3}
  • density of sea water, \rho_s=1.2\ kg.m^{-3}

<u>The relation between the change in pressure with height is given as:</u>

\frac{dP_{max}}{dz} =\rho_a.g_n

where:

dz = height in the atmosphere

g_n= standard value of gravity

<em>Now putting the respective values:</em>

\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8

dz=19217.687\ km

dz=19217687.07\ m

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

7 0
1 year ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
1 year ago
2. Heavier football players tend to play on the front line. Why? <br> What law is it?
gogolik [260]

Answer: They are put in front for defense so so they can block the opponents from getting the ball

Explanation:

4 0
1 year ago
Read 2 more answers
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