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bulgar [2K]
2 years ago
6

Liquid X of volume 0.5m3 and density 900kgm-3 was mixed with liquid Y of volume 0.4m3 and density 800kgm-3. What was the density

of the mixture?​
Physics
1 answer:
masha68 [24]2 years ago
8 0

Answer:

Density of the mixture = 855.56kgm-3

Explanation:

Density = Mass / Volume

Volume of Liquid X = 0.5m³

Density of Liquid X = 900kgm-3

Mass of Liquid X = Density × Volume

= 900kgm-3 × 0.5m³ = 450kg

Volume of Liquid Y = 0.4m³

Density of Liquid Y = 800kgm-3

Mass of Liquid Y = Density × Volume

= 800kgm-3 × 0.4m³= 320kg

As X and Y are mixed, we add their masses and volumes together:

Mass = 770kg

Volume = 0.9m³

Now we can find the density of the mixture:

Density = 770kg / 0.9m³ = 855.56kgm-3 (rounded to the 2nd decimal)

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A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
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Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

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