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Oksanka [162]
3 years ago
10

Which of the following is an idea that robert hutchings goddard introduced in his “further developments” to his research "a meth

od of reaching extreme altitudes"?
a. lunar energy
b. nuclear energy
c. solar energy
d. wave energy
Physics
1 answer:
asambeis [7]3 years ago
6 0
<span>Wave energy is an idea that Robert Hutchings Goddard introduced in his “Further Developments” to his research "A Method of Reaching Extreme Altitudes"    </span>
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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

\texttt{ }

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

\large {\boxed {PV = nRT} }

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

P_1V_1 = P_2V_2

5.00 \times 2.00 = 3.00 \times V_2

V_2 = 10 \div 3

V_2 = 3\frac{1}{3} \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_A = -P_2(V_2 - V_1)

W_A = -3.00(3\frac{1}{3} - 2.00)

W_A = \boxed{-4 \texttt{ L.atm}}

\texttt{ }

<h3>Step B:</h3>

<em>Using the same method as above:</em>

P_2V_2 = P_3V_3

3.00 \times 3\frac{1}{3} = 2.00 \times V_3

V_3 = 10 \div 2

V_3 = 5 \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_B = -P_3(V_3 - V_2)

W_B = -2.00(5 - 3\frac{1}{3})

W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}

\texttt{ }

<em>Finally we could calculate the total work done and heat added as follows:</em>

W = W_A + W_B

W = -4 + (-3\frac{1}{3})

W = -7\frac{1}{3} \texttt{ L.atm}

W = -7\frac{1}{3} \times 101.33 \texttt{ J}

\boxed{W \approx -743 \textt{ J}}

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

\boxed{Q = 743 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Minimum Coefficient of Static Friction : brainly.com/question/5884009
  • The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0
4 years ago
As the distance between two charged objects increases, the strength of the electrical force between the objects
GuDViN [60]

Answer:

I believe the answer is It increases

4 0
3 years ago
How many significant figures does the following number have? <br><br> 9876
Agata [3.3K]

Answer:

Four

Explanation:

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2 years ago
Define fundamental unit ? write its two example. ​
stealth61 [152]

the International System of Units, the fundamental units are: The meter (symbol: m), used to measure length. The kilogram (symbol: kg), used to measure mass. The second (symbol: s), used to measure time. The ampere (symbol: A), used to measure electric current.

8 0
3 years ago
James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m
Over [174]

Answer:

4.1\cdot 10^8 N

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

p=p_0 + \rho g h

where

p_0 =1.01\cdot 10^5 Pa is the atmospheric pressure

\rho = 1000 kg/m^3 is the water density

g=9.8 m/s^2 is the gravitational acceleration

h=11,000 m is the depth

Substituting,

p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2

And so, the inward force exerted on it is

F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N

8 0
3 years ago
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