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3 years ago

A solution is oversaturated with solute. Which could be done to decrease the oversaturation?

Physics

2 answers:

Gnoma [55]3 years ago

6 0

Dilute the solution, option a

salantis [7]3 years ago

3 0

Ans: Dilute the solution

Explanation:
To decrease the over-saturation, dilute the solution. Dilution is the process of decreasing the solute's concentration in the solution. It is usually done by mixing with more solvent. In other words, to dilute a solution means to add more solvent without the addition of more solute.

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Problem 24.3 The assembly is made from a steel hemisphere, rho st = 7. 80 Mg/m3 , and an aluminum cylinder, rho al = 2. 70 Mg/m3

AlladinOne [14]

Answer:

0.12 m

Explanation:

The mass of the steel hemisphere is:

$m_h=\rho_s\times \frac{2}{3}\pi r^2\\m_h = 7.8\times 10^3 \times \frac{2}{3}\pi (0.16)^2\\m_h=66.9 kg$

Mass of the aluminium cylinder is:

$m_c=\rho_a\times \pi r^2 h\\m_c=2.7\times 10^3 \times \pi (0.08)^2(0.18) \\m_c =9.7 kg$

mass center of steel hemisphere from the bottom:

$z_1=r-3r/8\\z_1=0.16-(3\times 0.16/8) =0.1 m$

mass center of aluminium cylinder from the bottom:

$z_2=r+h/2 \\z_2=0.16+0.18/2=0.25 m$

center of mass is

$Z=\frac{m_hz_1+z_2m_c}{m_s+m_c}\\Z=\frac{66.9\times 0.1+9.76\times 0.25}{66.9+9.76} = 0.12 m$

3 0

2 years ago

CAN SOMEONE HELP PLEASE

kolbaska11 [484]

Microplastics contain harmful chemicals that can accumulate in the food chain. The other chemicals released into the water when the plastic breaks down also has a negative impact on the sealife around. And with humans consuming contaminated fish and seafood, plastic pollution is certainly a human health issue too. microplastics were providing a medium to facilitate the transport of other toxic compounds such as heavy metals and organic pollutants.

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2 years ago

Why are collisions so important to physicists

Komok [63]

They study the causes

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2 years ago

A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl

Sergeu [11.5K]

Question: Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon $m_1$ and the mass of the basket $m_2$ , then according to newton's second law:

$(1). \:F = (m_1+m_2)a$ ,

where $a =0.265m/s^2$ is the upward acceleration, and $F = 688N$ is the net propelling force (counts the gravitational force).

Also, the tension $T$ in the rope is 79.8 N more than the basket's weight; therefore,

$(2). \:T = m_2g+79.8$

and this tension must equal

$T -m_2g =m_2a$

$(3). \:T = m_2g +m_2a$

Combining equations (2) and (3) we get:

$m_2a = 79.8$

since $a =0.265m/s^2$ , we have

$\boxed{m_2 = 301.13kg}$

Putting this into equation (1) and substituting the numerical values of $F$ and $a$ , we get:

$688N = (m_1+301.13kg)(0.265m/s^2)$

$\boxed{m_1 = 2295 kg}$

Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.

8 0

3 years ago

An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,

Sedbober [7]

Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

a. Internal energy and the relative specific volume at $s_1$ are determined from A-17: $u_1=214.07kJ/kg, \ \alpha_r_1=621.2$ .

The relative specific volume at $s_2$ is calculated from the compression ratio:

$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$