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3 years ago

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A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an

efficiency of 30%, and taken in 50 kJ from the hot steam per cycle. If a Carnot engine takes in the same amount of heat per cycle and operates at these temperatures, the work it can turn into is most likely to be:a) 15 kJ. b) 20 kJ. c) 10 kJ. d) 50 kJ.

1 answer:

Vanyuwa [196]3 years ago

6 0

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .

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Answer:

6.6 N

Explanation:

Let's take the direction of the force of 4.0 N as positive x-direction. This means that the force of 3.0 N is at 40 degrees above it. So the components of the two forces along the x- and y-directions are:

$F_{1x} = 4.0 N\\F_{1y} = 0$

$F_{2x} = 3.0 N cos 40^{\circ}=2.3 N\\F_{2y} = 3.0 N sin 40^{\circ} = 1.9 N$

So the resultant has components

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So the magnitude of the resultant is

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3 years ago

A helicopter is ascending vertically. a passenger accidentally drops her wallet out the sides of the helicopter when it is 160 m

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Answer:

(E)56.0 m/s

Explanation:

Height =h=-160 m

Because the wallet moving in downward direction

Time=t=7 s

Final speed of wallet=v=0

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$v^2-u^2=2gh$

Where $g=9.8 m/s^2$

Substitute the values

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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

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Answer:

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$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

<u>Now, gravitational force on particle 3 due to particle 1:</u>

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towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

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$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

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$F=18.58\times 10^{-11}\ N$

<em>For angle in counterclockwise direction from the +x-axis</em>

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

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