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lorasvet [3.4K]

3 years ago

15

A proposed ocean thermal-energy conversion (OTEC) system is a heat engine that would operate between warm water (25 degrees C) a

t the ocean's surface and cooler water (5 degrees C) 1,000 m below the surface. What is the maximum possible efficiency of the system?

1 answer:

Sphinxa [80]3 years ago

8 0

Answer:

$\eta=0.067$

Explanation:

Given that

Temperature of warm water ,T₁ = 25 °C

Temperature of cold water ,T₂ = 5 °C

The maximum possible efficiency of any heat is will be same as the efficiency of the Carnot heat engine.Therefore the maximum efficiency of OTEC system is given as follows

$\eta=1-\dfrac{T_2}{T_1}$

T should be Kelvin scale

Now by putting the values

$\eta =1-\dfrac{5+273}{25+273}$

$\eta=0.067$

Therefore the maximum efficiency will be 0.067 or 6.7%.

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What is the strength of an electric field that will balance the weight of a 3.3 g plastic sphere that has been charged to -7.6 n

sp2606 [1]

As the plastic sphere is charged, therefore it experience an electric force when placed in an electric fields and also experiences gravitational force acts downward so the electric force must act upward.

Let $F_{E}$ is electric force and $F_{G}$ is gravitational force.

If these forces are balanced, therefore $F_{E} = F_{G}$

or $\left | q \right |E=mg\\\\\ E=\frac{mg}{\left | q \right |}$

Given, $q=-7.6 nC=-7.6\times10^{-9} C$ and $m=3.3 g= 3.3\times10^{-3} kg$ .

Substituting these values in above equation we get,

$E=\frac{3.3\times10^{-3} kg\times9.8 m/s^{2} }{7.6\times10^{-9} C}\\\\E=4.2\times10^{6} N/C$

Thus, the magnitude of electric field is $4.2\times10^{6} N/C$ .

As the charge is negative, the electric field at the location of the plastic sphere must be pointing downward.

8 0

3 years ago

Now it's your turn

eduard

Answer:

The average force the golf club exerts on the ball is 600 N

Explanation:

Newton's second law of motion states that force, F, is directly proportional to the rate of change of momentum produced

F = m× (v₂ - v₁)/(Δt)

The given parameters of the motion of the ball are;

The mass of the ball, m = 45 g = 0.045 kg

The initial velocity of the ball, v₁ = 0 m/s

The speed with which the ball was hit by the golfer, v₂ = 40 m/s

The duration of contact between the golf club and the ball, Δt = 3 ms = 0.003 seconds (s)

By Newton's law of motion, the average force, 'F', which the golf club exerts on the ball is therefore, given as follows;

F = 0.045 kg × (40 m/s - 0 m/s)/(0.003 s) = 600 N

The average force the golf club exerts on the ball = F = 600 N.

5 0

3 years ago

Some people think that coloured filters add colour to white light. Why is this not correct

Elis [28]

Answer:

White is every colour.

Explanation:

The filters make whatever they are colour stronger.

4 0

3 years ago

Read 2 more answers

Convert 6.8 gallons into liters

poizon [28]

25.7408 liters. There are 3.78541 liters in one gallon

4 0

4 years ago

Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i

Kobotan [32]

The magnitude of <em>electrical</em> force on charge $q_{3}$ due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector <em>position</em> of each particle respect to origin are described below:

$\vec r_{1} = (-0.500, 0)\,[m]$

$\vec r_{2} = (+0.500, 0)\,[m]$

$\vec r_{3} = (0, +0.500)\,[m]$

Then, distances of the former two particles particles respect to the latter one are found now:

$\vec r_{13} = (+0.500, +0.500)\,[m]$

$r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{13} =\frac{\sqrt{2}}{2}\,m$

$\vec r_{23} = (-0.500, +0.500)\,[m]$

$r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{23} =\frac{\sqrt{2}}{2}\,m$

The resultant force is found by Coulomb's law and principle of superposition:

$\vec R = \vec F_{13}+\vec F_{23}$ (1)

Please notice that particles with charges of <em>same</em> sign attract each other and particles with charges of <em>opposite</em> sign repeal each other.

$\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13} +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23}$ (2)

Where:

$k$ - Electrostatic constant, in newton-square meters per square Coulomb.
$q_{1}$ , $q_{2}$ , $q_{3}$ - Electric charges, in Coulombs.
$r_{13}$ , $r_{23}$ - Distances between particles, in meters.
$\vec u_{13}$ , $\vec u_{23}$ - Unit vectors, no unit.

If we know that $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = 2\times 10^{-6}\,C$ , $q_{2} = 2\times 10^{-6}\,C$ , $q_{3} = 2\times 10^{-6}\,C$ , $r_{13} =\frac{\sqrt{2}}{2}\,m$ , $r_{23} =\frac{\sqrt{2}}{2}\,m$ , $\vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \right)$ and $\vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ , then the vector force on charge $q_{3}$ is:

$\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$

$\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\,[N]$

$\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]$

And the magnitude of the <em>electrical</em> force on charge $q_{3}$ ( $R$ ), in newtons, due to the others is found by Pythagorean theorem:

$R = 0.102\,N$

The magnitude of <em>electrical</em> force on charge $q_{3}$ due to the others is 0.102 newtons. $\blacksquare$

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

8 0

2 years ago