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The energy provided each hour by heat to the turbine in an electric power plant is 9.0×10^12 J. If 5.4 × 10^12 J of energy is exhausted each hour from the engine as heat, what is the efficiency of this heat engine?

Answers

Answer 1

Answer:

Answer:

60 %

Explanation:

Efficiency is defined as the ratio of output power to the input power.

Input energy each hour = 9 x 10^12 J

Output energy each hour = 5.4 x 10^12 J

Efficiency = Output energy per hour / input energy per hour

Efficiency = (5.4 x 10^12) / ( 9 x 10^12) = 5.4 / 9 = 0.6

Efficiency in percentage form = 0.6 x 100 = 60 %

Answer 2

Answer:

Final answer:

The efficiency of a heat engine is calculated using the formula (Energy Input - Energy Output) / Energy Input. Given the figures, the efficiency of the engine is 40%, indicating that 40% of the input energy is converted into work.

Explanation:

The efficiency of a heat engine is determined by the ratio of work output to energy input. In the given scenario, the turbine in an electric power plant is supplied with 9.0 x 10^12 joules of energy, and 5.4 x 10^12 joules of energy is expelled as heat per hour. We can calculate the efficiency using the equation:

Efficiency = (Energy Input - Energy Output) / Energy Input

By substituting the given values, Efficiency = (9.0 x 10^12 J - 5.4 x 10^12 J) / 9.0 x 10^12 J = 0.4 or 40%

This means the heat engine of the power plant has a 40% efficiency, meaning 40% of the energy input is converted into work while 60% is discarded as waste heat.

Learn more about Thermal Efficiency here:

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Students run an experiment to determine the rotational inertia of a large spherically shaped object around its center. Through experimental data, the students determine that the mass of the object is distributed radially. They determine that the radius of the object as a function of its mass is given by the equation r = km², where k = 3. Which of the following is a correct expression for the rotational inertia of the object?

(A) m3
(B) 1.8 m3
(C) 3.6 m3
(D) 6 m3
(E) 9 m3

Answers

Answer:

Explanation:

$r=km^2\n$ = $3m^2$

Since the object is a solid sphere, the equation for rotational inertia is:

$I = (2)/(5)mr^2$

$I=(2)/(5)m(3m^2)^2=(2)/(5)*9m^5=3.6m^5$

Final answer:

The provided question seems to have a discrepancy as the calculated value of rotational inertia for a spherical object with a given mass-radius relationship is 4.5M³, which does not match any of the supplied answer choices.

Explanation:

The question is asking for the correct expression for the rotational inertia of a spherically shaped object with mass distribution given by the radius as a function of mass (r = km² where k = 3). The rotational inertia, or moment of inertia, for a solid sphere is given by the formula ⅒MR², where M is the mass of the sphere, and R is its radius. Considering that R is defined by r = km², we substitute R with km² in the formula:

I = ⅒M(km²)² = ⅒Mk²m⁴ = ⅒Mk²M²

Since k = 3, we further simplify the expression:

I = ⅒M(3M)² = ⅒(3²)M³ = ⅒ × 9M³ = 4.5M³

However, none of the options (A) to (E) match the value 4.5M³, which indicates there may be an error in the supplied options or an error within the initial assumptions or question parameters. It's important to recheck the given data and the calculation steps to ensure accuracy. If the question and the parameters are indeed accurate as stated, additional information or clarification would be necessary.

A 13.0-Ω resistor, 13.5-mH inductor, and 50.0-µF capacitor are connected in series to a 55.0-V (rms) source having variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.

Answers

Answer:

E = 0.13 J

Explanation:

At resonance condition we have

$\omega = \sqrt{(1)/(LC)}$

$\omega = \sqrt{(1)/((13.5 * 10^(-3)){50* 10^(-6))}}$

$\omega = 1217.2 rad/s$

now if the frequency is double that of resonance condition then we have

$x_L = 2\omega L$

$x_L = 2(1217.2)(13.5* 10^(-3))$

$x_L = 32.86 ohm$

now we have

$x_c = (1)/(2(1217)(50* 10^(-6)))$

$x_c = 8.22 ohm$

now average power is given as

$P = i_(rms)V_(rms)* (R)/(z)$

$P = (55)/(√((32.86 - 8.22)^2 + 13^2)))(55)* (13)/(√((32.86 - 8.22)^2 + 13^2))$

$P_(avg) = 50.67 Watt$

Now time period is given as

$T = (2\pi)/(\omega)$

so total energy consumed is given as

$E_(avg) = 50.67((2\pi)/(2(1217.2)))$

$E = 0.13 J$

Which two types of simple machines can be found in a bicycle?

Answers

Answer:

The correct answer is A) lever and wheel and axle

Explanation:

I took the quiz

hope this helps :)

Air contains 78.08% nitrogen, 20.095% oxygen, and 0.93% argon. calculate the partial pressure of oxygen if the total pressure of the air sample is 1.7 atm.a.

Answers

partial pressure in a mixture of two or more gases will be given by formula

$P_(partial)$ = mole fraction of gas * total pressure

now here mole fraction is same as percentage of gas in the mixture

Now mole fraction of oxygen is 0.20095 (20.095%)

now here pressure of oxygen in the mixture is given as

$P_(o_2) = 0.20095 * P_(total)$

$P_(o_2) = 0.20095 * 1.7$

$P_(o_2) = 0.342 atm$

so pressure due to oxygen in the mixture will be 0.342 atm

Answer:

20.095

Explanation:

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979.(a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° south of east. What was his total displacement?

(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air?

(c) What was his total displacement relative to the air mass?

Answers

Answer:

a) $s=35794.2\ m$

b) $v_w=3.53\ m.s^(-1)$

c) $s_w=56074.2\ m$

Explanation:

Given:

duration of flight, $t=169* 60=10140\ s$

velocity of flight, $v=3.53\ m.s^(-1)$

direction of flight, $45^(\circ)$ to the south of east

Now the total displacement:

$s=v.t$

$s=3.53* 10140$

$s=35794.2\ m$

Velocity of air, $v_a=2\ m.s^(-1)$

When the aircraft encounters a headwind in the opposite direction to the velocity of motion then the speed of the aircraft is lowered with respect to the ground.

But when the speed is observed with respect to the wind the reduced velocity of the aircraft is observed from an opposite moving wind having a magnitude equal to the difference in velocity of the aircraft. This results in no change in the apparent velocity of the aircraft.

Mathematically:

Velocity of the aircraft with respect to the ground:

$v_(g)=v-v_a$

$v_(g)=3.53-2$

$v_g=1.53\ m.s^(-1)$

Now the velocity of the aircraft with respect to the wind:

$v_w=v_g+v_a$

$v_w=1.53+2$

$v_w=3.53\ m.s^(-1)$

Now the total displacement with respect to the wind:

$s_w=v_w.t+v_a.t$

$s_w=3.53* 10140+2* 10140$

$s_w=56074.2\ m$

Water flows at speed of 4.4 m/s through ahorizontal pipe of diameter 3.3 cm . The gauge
pressure P1 of the water in the pipe is 2 atm .
A short segment of the pipe is constricted to
a smaller diameter of 2.4 cm
(IMAGE)
What is the gauge pressure of the water
flowing through the constricted segment? Atmospheric pressure is 1.013 × 10^5 Pa . The density of water is 1000 kg/m^3
. The viscosity
of water is negligible.
Answer in units of atm

Answers

Answer:

1.75 atm

Explanation:

Mass is conserved, so the mass flow before the constriction equals the mass flow after the constriction.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = h₂:

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Writing v₂ in terms of v₁:

P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ d₁²/d₂²)²

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (d₁/d₂)⁴

P₁ + ½ ρ v₁² (1 − (d₁/d₂)⁴) = P₂

Plugging in values:

P₂ = 2 atm + ½ (1000 kg/m³) (4.4 m/s)² (1 − (3.3 cm / 2.4 cm)⁴) (1 atm / 1.013×10⁵ Pa)

P₂ = 1.75 atm

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