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3 years ago

3) If an airplane is in flight cruising at constant velocity, what can be said about the net work

Physics

1 answer:

Crazy boy [7]3 years ago

6 0

Explanation:

<h3>In physics, constant velocity occurs when there is no net force acting on the object causing it to accelerate. In terms of airplane flight, the two main forces influencing its velocity forward are drag and thrust. At a constant altitude, when the force of thrust equals the opposing force of drag, then the airplane will experience uniform motion in one direction. This can be further explained by Newton’s First Law. </h3>

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If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pre

creativ13 [48]

Answer:

$-112.876J$

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

$16KNO_3(s) + 24C(s) + S_8(s) \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)$

Let us define $P - V$ work as;

$w_{pv} = - P_{external} \triangle Volume$

where $\triangle (Volume) = (V_{final} - V_{initial})$

External pressure is given as $1.00$ $atm$ , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence, $V_i = 0.$

To find the volume of the products, we need to first find the amount of moles of the product made from $2.40_gKNO_3,$ using the molar mass of $KNO_3$ which is 101.1032 g/mol

$2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3$

Now let us convert moles of $KNO_3$ into moles of $CO_2$ and $N_2$ using the stoichiometric ratios from our balanced equation of the reaction.

$0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2$

$0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2$

$K_2S$ is not factored into the volume calculation because it is a solid.

Now let us also convert the moles of $CO_2$ and $N_2$ into grams using their respective molar masses.

$0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2$

$0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2$

We will now proceed to convert grams into volume using the density values provided.

$1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2$

$0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2$

Summing up the two volumes, we get the final volume

$0.856L + 0.258L = 1.114L = V_f$

Plugging everything into the $w_{pv}$ equation, we get:

$w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm$

Finally, let us convert $L.atm$ into joules using the conversion rate of;

$1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J$

7 0

4 years ago

An astronaut is stationary in space. Which of these would have the greatest speed as observed by the astronaut?

yaroslaw [1]

Hello!

The answer that makes most sense to me is option A.

~ Hope I helped! ~

7 0

4 years ago

Read 2 more answers

Having _________is the most obvious difference between a eukaryotic and a prokaryotic cell, but there are other differences as w

8_murik_8 [283]

Answer:

1. d. a nucleus

2. b. Membrane-bound organelles

3. c. ten time smaller

4. b. animal cell

5. c. Bacteria

Explanation:

Hope this helps... :)

7 0

3 years ago

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

Kipish [7]

Answer:

The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

Explanation:

Given;

rock concert sound intensity level, β₁ = 120 dB

whisper sound intensity level, β₂ = 20 dB

The sound intensity level is given as;

$\beta = 10Log(\frac{I}{I_o} )\\\\$

where;

I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²

I is the sound intensity

Intensity of sound at rock concert ;

$120 = 10Log(\frac{I}{10^{-12}} )\\\\12 = Log(\frac{I}{10^{-12}} )\\\\10^{12} = \frac{I}{10^{-12}}\\\\I = 10^{12} * 10^{-12}\\\\I = 10^0\\\\I = 1 \ W/m^2$

The intensity of sound of a whisper;

$20 = 10Log(\frac{I}{10^{-12}} )\\\\2 = Log(\frac{I}{10^{-12}} )\\\\10^{2} = \frac{I}{10^{-12}}\\\\I = 10^{2} * 10^{-12}\\\\I = 10^{-10} \ W/m^2\\\\$

Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper

$\frac{I_{Concert}}{I_{whisper}} = \frac{1}{10^{-10}} \\\\\frac{I_{Concert}}{I_{whisper}} = 1 * 10^{10}\\\\I_{Concert} = 1 * 10^{10}*I_{whisper}$

Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

3 0

3 years ago

32. For each of the listed parts of a power plant, make a selection to indicate in what

k0ka [10]

Answer: a. boiler - I. part of a coal-fired power plant.

b. combustion chamber - I. part of a coal-fired power plant.

c. condenser - III. part of a coal-fired power plant and part of a nuclear power plant.

d. control rod - I. part of a coal-fired power plant.

e. generator- III. part of a coal-fired power plant and part of a nuclear power plant.

f. turbine - III. part of a coal-fired power plant and part of a nuclear power plant.

Explanation:

a. Boiler: It is used in the coal power plant to heat the water used for producing electricity. The boiler keeps the temperature of water or steam at a desired level.

b. Combustion chamber: It is used for combustion of coal and using thermal energy for electricity generation.

c. Condenser; It is used in both nuclear and coal power plants. It condense and receive from the turbine.

d. Control rod: It is used in the nuclear power plant for controlling the nuclear fission and fusion reactions of heavy metals.

e. Generator: It converts the thermal energy into electrical energy.

f. Turbine: The turbine blades move with the effect of steam or any form of energy and the energy so produce is used for generating electricity in generator.

7 0

3 years ago