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N76 [4]
3 years ago
12

You come across an open container which is filled with two liquids. Since the two liquids have different density there is a dist

inct separation between them. Water fills the lower portion of the container to a depth of 0.209 m which has a density of 1.00 × 103 kg/m3. The fluid which is floating on top of the water is 0.300 m deep. If the absolute pressure on the bottom of the container is 1.049 × 105 Pa, what is the density of the unknown fluid? The acceleration due to gravity is g = 9.81 m/s2 and atmospheric pressure is P0 = 1.013 × 105 Pa.
Physics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

526.57 Pa

Explanation:

P ( pressure at the bottom of the container) = 1.049 × 10^5 pa

Using the formula of pressure in an open liquid

Pw ( pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²

Pw = 1000 × 9.81 ×0.209 = 2050.29 Pa

P( atmospheric pressure) = 1.013 × 10^5 Pa

Pl ( pressure due to the liquid) = ρ(density of the liquid) × h (depth of the liquid) × g

Subtract each of the pressure from the absolute pressure at the bottom

P(bottom) - atmospheric pressure

(1.049 × 10^5) - (1.013 × 10^5) = 0.036 × 10^5 = 3600 Pa

subtract pressure due to water from the remainder

3600 - 2050.29 = 1549.71 Pa

1549.71 =  ρ(density of the liquid) × h (depth of the liquid) × g

ρ (density of the liquid) = 1549.71 / (h × g) = 1549.71 / (0.3 × 9.81) =526.57 Pa

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(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of
NISA [10]

Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

F = \dfrac{kq_1q_2}{r^2}

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}

Force due to Q₃ :

F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}

8.3. The net force on the particle at Q₂ is the vector

\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}

Its magnitude is

\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}

where we subtract 180° because \vec F terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.

8 0
2 years ago
In comparing molar specific heat for gases under constant pressure CP and constant volume CV, we conclude that (more than one co
Sauron [17]

Answer:

b. Specific heat increases as the number of atoms per molecule increases.

c. Specific heat at constant pressure is higher than at constant volume.

d. Monatomic gases behave like ideal gases.

Explanation:

Specific heat of the gas at constant pressure is usually higher than that of the volume.

i.e.

Cp - Cv = R

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However, monoatomic gases are gases that exhibit the behavior of ideal gases. This is due to the attribute of the intermolecular forces which plays a negligible role. Nonetheless, the case is not always true for all temperatures and pressure.

Similarly, the increase in the number of atoms per molecule usually brings about an increase in specific heat. This effect is true as a result of an increase in the total number associated with the degree of freedom from which energy can be separated.

Thus, from above explanation:

Option b,c,d are correct while option (a) is incorrect.

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