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3 years ago

Mining runoff is _____ water that comes from the mining of rock.

Physics

2 answers:

Anvisha [2.4K]3 years ago

5 0

B is the answer mining runoff is purified water

PolarNik [594]3 years ago

3 0

A CONTAMINATED WATER

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How much total heat transfer is necessary to lower the temperature of 0.175 kg of steam from 125.5 °C to −19.5 °C, including the

GaryK [48]

Answer:

Explanation:

The heat required to change the temperature of steam from 125.5 °C to 100 °C is:

$Q_1 = ms_{steam} (125.5^0C - 100^0C) \\ \\ Q_1 = 0.175 \ kg ( 1520 \ J/kg.K ) (25.5^0 \ C) \\ \\ Q_1 = 6783 \ J$

The heat required to change the steam at 100°C to water at 100°C is;

$Q_2 = mL_v \\ \\ Q_2 = (0.175 \ kg) (2.25*10^6 \ J/kg ) \\ \\ Q_2 = 393750 \ J$

The heat required to change the temperature from 100°C to 0°C is

$Q_3 = ms_{water} (100^) \ C) \\ \\ Q_3 = (0.175\ kg)(4186 \ J/kg.K) (100 ^0c ) \\ \\ Q_3 = 73255 \ J$

The heat required to change the water at 0°C to ice at 0°C is:

$Q_4 = mL_f \\ \\ Q_4 = (0.175 \ kg)(3.34*10^5 \ J/kg) \\\\ Q_4 = 58450 \ J$

The heat required to change the temperature of ice from 0°Cto -19.5°C is:

$Q_5 = ms _{ice} (100^0 C) \\ \\ Q_5 = (0.175 \ kg)(2090 \ J/kg.K)(19.5^0C) \\ \\ Q_5 = 7132.125 \ J$

The total heat required to change the steam into ice is:

$Q = Q_1 + Q_2 + Q_3 + Q_4 +Q_5 \\ \\Q = (6788+393750+73255+58450+7132.125)J \\ \\ Q = 539325.125 \ J \\ \\ Q = 5.39*10^5 \ J$

The time taken to convert steam from 125 °C to 100°C is:

$t_1 = \frac{Q_1}{P} = \frac{6738 \ J}{835 \ W} = 8.12 \ s$

The time taken to convert steam at 100°C to water at 100°C is:

$t_2 = \frac{Q_2}{P} =\frac{393750}{834} =471.56 \ s$

The time taken to convert water to 100° C to 0° C is:

$t_3 = \frac{Q_3}{P} =\frac{73255}{834} = 87.73 \ s$

The time taken to convert water at 0° to ice at 0° C is :

$t_4 = \frac{Q_4}{P} =\frac{58450}{834} = 70.08 \ s$

The time taken to convert ice from 0° C to -19.5° C is:

$t_5 = \frac{Q_5}{P} =\frac{7132.125}{834} = 8.55 \ s$

5 0

3 years ago

When finding net force, why must you know the directions of the forces acting on an object?

Mariulka [41]

<h3>One reason we should know what the directions of the forces acting on an object is so we can know if we have to add or subtract. same = add together, opposite = subtract from each other. Also if we don't pay attention to the direction the Net force will be wont be accurate. There will be factors that will upset the calculation.So we must know the direction of the two forces because we have to know if we are adding or subtracting and if the answer is accurate. </h3>

<em>I hope this helps!.</em>

8 0

3 years ago

Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi

Nina [5.8K]

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

Hence, the time constance of (b) is 0.592 s

8 0

2 years ago

An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelengt

nadya68 [22]

Answer:

I = 4.189 mA V = 0.338 V

Explanation:

In order to do this, we need to apply the following expression:

I = Is[exp^(qV/kT) - 1] (1)

However, as the junction of the diode is illuminated, the above expression changes to:

I = Iopt + Is[exp^(qV/kT) - 1] (2)

Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

I ≅ Iopt

Therefore:

I ≅ I₀Aλq / hc (3)

Where:

I₀A = Area of diode (radiation)

λ: wavelength

q: electron charge (1.6x10⁻¹⁹ C)

h: Planck constant (6.62x10⁻³⁴ m² kg/s)

c: speed of light (3x10⁸ m/s)

Replacing all these values, we can get the current:

I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)

I = 4.189x10⁻³ A or 4.189 mA

Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

I = Is[exp^(qV/kT) - 1]

k: boltzman constant (1.38x10⁻²³ J/K)

4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]

4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]

465,444.44 + 1 = exp(38.65V)

ln(465,445.44) = 38.65V

13.0508 = 38.65V

V = 0.338 V

6 0

3 years ago

You treat 9.540 g of the mixture with the acid and isolate 9.355 g of nacl. what is the weight percent of each substance in the

Mila [183]

The weight percentage of sodium carbonate in the mixture is = 67.71%

The weight percentage of sodium bicarbonate in the mixture is = 32.57%

<h3>What does "molar mass" ?</h3>

The total mass throughout grams of all the atoms needed to form a molecule per mole is what makes up the molar mass, also known as the molecular weight. Grams per mole is the unit measuring molar mass.

<h3>According to the given information:</h3>

The interaction between sodium carbonate and hydrochloric acid has the following equation:

Na₂CO₃ + HCl --> NaCl + CO₂ + H₂O

The reaction's balanced equation is,

Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O -------(2).

When sodium hydrogen carbonate as well as hydrochloric acid react, the following equation results.

NaCO₃ + HCl --> NaCl + CO₂ + H₂O ----------------(3)

The Molar mass :

Na₂CO₃ = 106g/mol

NaCO₃ = 84g/mol

NaCl = 58.5g/mol

The mass of the mixture is Na₂CO₃/NaCO₃ = 9.540g

The molar masses of the constituent elements that make up the compounds are added to determine the molar weight of the substances. In both chemical equations (2) and (3), the unitary technique is employed to calculate the mass of the mixture based on the number of moles (3).

The mass of the sodium carbonate and sodium bicarbonate is calculated with the help of the mass of NaCl formed by the mixture of Na₂CO₃/NaCO₃ with HCl.

The mixture has the following percentage of sodium carbonate:

% of Na₂CO₃ = $\frac{x \times 106}{9.540} \times 100\\$

$=\frac{0.061 \times 106}{9.540} \times 100$

= (6.46/9.540)*100

= 67.71%

The weight percentage of sodium carbonate in the mixture is = 67.71%

The mixture has the following percentage of NaHCO3:

% of NaHCO₃ = $\frac{y \times 84}{9.540} \times 100$

= $=\frac{0.037 \times 84}{9.540} \times 100$

= (3.108/9.540)*100

= 32.57%

The weight percentage of sodium bicarbonate in the mixture is = 32.57%

To know more about molar mass visit:

brainly.com/question/12127540

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I understand that the question you are looking for is:

A mixture of sodium carbonate and sodium hydrogen carbonate is treated with aqueous hydrochloric acid. the unbalanced equations for the resulting's reactions are:

Na2CO3 (s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)

NaHCO3(s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)

You treat 9.540g of Na2CO3/NaHCO3 mixture with an excess of aqueous HCl and isolate 9.355g of NaCl. What is the weight percent of each substance in the mixture?

5 0

1 year ago