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4 years ago

15

A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of

heat output. Express your answer using two significant figures.

1 answer:

scoray [572]4 years ago

5 0

Answer:1200

Explanation:

Given data

Upper Temprature $\left ( T_H\right )=240^{\circ}\approx 513$

Lower Temprature $\left ( T_L\right )=20^{\circ}\approx 293$

Engine power ouput $\left ( W\right )=910 W$

Efficiency of carnot cycle is given by

$\eta =1-\frac{T_L}{T_H}$

$\eta =\frac{W_s}{Q_s}$

$1-\frac{293}{513}=\frac{910}{Q_s}$

$Q_s=2121.954 W$

$Q_r=1211.954 W$

rounding off to two significant figures

$Q_r=1200 W$

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A 0.366 kg metal cylinder is placed inside the top of a plastic tube, the lower end of which is sealed off by an adjustable plun

DochEvi [55]

Answer:

$a = 32.6 m/s^2$

Explanation:

As we know that pressure between the cylinder and plunger is increased by 1.59 times

So this will make a net force upwards on the cylinder which is given as

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now we will have

$\Delta P = P_2 - P_1$

Here initial pressure is given as

$P_1 = P_o + \frac{mg}{A}$

now new pressure is given as

$P_2 = 1.59 P_1$

so we have force on the cylinder given as

$F = P_2A - mg - P_oA$

$F = 1.59(P_0 + \frac{mg}{A})A - (mg + P_0A)$

$F = 0.59(1.01 \times 10^5 \times \pi(7.24 \times 10^{-3})^2 + 0.366(9.81))$

$F = 11.93 N$

now the acceleration is given as

$F = ma$

$11.93 = 0.366 a$

$a = 32.6 m/s^2$

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3 years ago

Why do ionic compounds make good conductors when dissolved in water?

sveticcg [70]

Answer: The correct answer is B- The ions are free to move in water and carry an electric charge.

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3 years ago

In an experiment, the 1 kg cart collides with a 3 kg cart but doesn’t stick to it. Instead, the 3 kg cart gets knock forward by

notka56 [123]

Explanation:

Given that,

Mass of the cart, $m_1=1\ kg$

Mass of the cart 2, $m_2=3\ kg$

Final speed of cart 2, $v_2=0.3\ m/s$

Final speed of cart 1 is 0 as it comes to rest.

Let us assume that the initial velocity of the cart 2 is 0. So using the conservation of linear momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1u_1+0=0+m_2v_2\\\\m_1u_1=m_2v_2\\\\u_1=\dfrac{m_2v_2}{m_1}\\\\u_1=\dfrac{3\times 0.3}{1}\\\\u_1=0.9\ m/s$

So, the initial velocity of the 1.0-kg cart is 0.9 m/s.

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3 years ago

Q.4. What is the kinetic energy of a 10 kg car that is moving 4 m/s?

kodGreya [7K]

An object of mass $10 \mathrm{~kg}$ is moving with a uniform velocity of $4 \mathrm{~ms}^{-1}$ . The kinetic energy possessed by the object is $80 \mathrm{~J}$ .

Given:

Mass of an object $=10 \mathrm{~kg}$

Velocity $=4 \mathrm{~ms}^{-1}$

Kinetic Energy $=1 / 2 \times$ Mass of Object $\times(\text { Velocity })^{2}$

$\Rightarrow$ Kinetic Energy $=1 / 2 \times 10 \times 4 \times 4$

$\Rightarrow$ Kinetic Energy $=\underline{80 \mathbf{J}}$

What is Kinetic Energy?

In physics, an object's kinetic energy is the energy it has as a result of its motion.
It is defined as the amount of work required to accelerate a body of a given mass from rest to a certain velocity.
The body retains its kinetic energy after gaining it during acceleration until its speed changes.
Kinetic energy is present in a speeding bullet, a walking human, and electromagnetic radiation such as light. The energy associated with the continual, random bouncing of atoms or molecules is another type of kinetic energy.

Learn more about kinetic energy brainly.com/question/12669551

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2 years ago

A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a p

Temka [501]

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m$

The image distance is 0.051 m

When u = 50 cm

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m$

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

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3 years ago