Ask question

Ask question

All categories

3 years ago

13

Power plants release sulfur dioxide and nitrogen oxides into the atmosphere, where they combine with water vapor to form a. acid

precipitation b. ozone c. particulates d. carbon minoxide

2 answers:

Alla [95]3 years ago

6 0

Power plants release sulfur dioxide and nitrogen oxides into the atmosphere, where they combine with water vapor to form acid precipitation. The correct option among all the options that are given in the question is the first option or option "a". I hope the answer will help you and it is the one you were looking for.

adell [148]3 years ago

3 0

Answer:

Acid precipitation ( option a)

Explanation:

You might be interested in

A 62.2-kg person, running horizontally with a velocity of 3.80 m/s, jumps onto a 19.7-kg sled that is initially at rest. (a) Ign

jek_recluse [69]

Answer:

a) $v = 2.886\,\frac{m}{s}$ , b) $\mu_{k} = 0.014$

Explanation:

a) The final speed is determined by the Principle of Momentum Conservation:

$(62.2\,kg)\cdot (3.80\,\frac{m}{s} ) = (81.9\,kg)\cdot v$

$v = 2.886\,\frac{m}{s}$

b) The deceleration experimented by the system person-sled is:

$a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(2.886\,\frac{m}{s} \right)^{2}}{2\cdot (30\,m)}$

$a = -0.139\,\frac{m}{s^{2}}$

By using the Newton's Laws, the only force acting on the motion of the system is the friction between snow and sled. The kinetic coefficient of friction is:

$-\mu_{k}\cdot m\cdot g = m\cdot a$

$\mu_{k} = -\frac{a}{g}$

$\mu_{k} = -\frac{\left(-0.139\,\frac{m}{s^{2}} \right)}{9.807\,\frac{m}{s^{2}} }$

$\mu_{k} = 0.014$

3 0

4 years ago

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C.

salantis [7]

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_{1} = u_{g}$ = 2553.6 kJ/kg

$v_{1} = v_{g} = 0.4625 m^{3}/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_{2} = v_{g} = 0.4625 m^{3}/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_{2} = v_{g} = 0.4625 m^{3}/kg$ and temperature $T_{2} = 360^{o}C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})$

$u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$\frac{Q}{m} = \Delta u$

$\frac{Q}{m} = u_{2} - u_{1}$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

4 0

3 years ago

Air in human lungs has a temperature of 37.0°C and a saturation vapor density of 44.0 g/m³.

ioda

Answer:

(a). The maximum loss of water vapor by the person is $8.8\times10^{-2}\ g$

(b). The partial pressure of water vapor is $6.29\times10^{3}\ N/m^2$

Explanation:

Given that,

Temperature = 37.0°C

Volume of air = 2.00 L

Density of vapor = 44.0 g/m³

We need to calculate the maximum loss of water vapor by the person

Using formula of density

$m=\rho_{air}\times V_{air}$

Put the value into the formula

$m=44.0\times2.00\times10^{-3}$

$m=0.088\ g$

$m=8.8\times10^{-2}\ g$

(b). We need to calculate the partial pressure of water vapor

Using formula of pressure

$PV=nRT$

$P=\dfrac{nRT}{V}$

$P=\dfrac{\rho\times R\times T}{M}$

Put the value into the formula

$P=\dfrac{44\times8.314\times(37+273)}{18.01528}$

$P=6294.82\ N/m^2$

$P=6.29\times10^{3}\ N/m^2$

Hence, (a). The maximum loss of water vapor by the person is $8.8\times10^{-2}\ g$

(b). The partial pressure of water vapor is $6.29\times10^{3}\ N/m^2$

3 0

4 years ago

Does newton all three of his laws have to do with the egg drop

andrew11 [14]

All a of Newton's laws have to do with gravity, he just happens to use eggs to prove his point.

8 0

3 years ago

Suppose that you are standing on a train accelerating at 0.20g. What minimum coefficient of static friction must exist between y

Ilia_Sergeevich [38]

Acceleration = (0.2 x g) = 1.96m/sec^2.
<span>Accelerating force on 1kg. = (ma) = 1.96N. </span>
<span>1kg. has a weight (normal force) of 9.8N. </span>
<span>Coefficient µ = 1.96/9.8 = 0.2 minimum. </span>

<span>Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. </span>
<span>e.g. 75kg = (75 x g) = 735N. </span>
<span>Accelerating force = (735 x 0.2) = 147N</span>

5 0

4 years ago

Read 2 more answers