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3 years ago

8

An engine absorbs 2150 J as heat from a hot reservoir and gives off 750 J as heat to a cold reservoir during each cycle. What is

the efficiency of the engine?

1 answer:

Aleks [24]3 years ago

6 0

Input = 2,150 J; Output = 750 J
The efficiency of the engine = Output / Input = 750 J / 2,15 J = 0.3488
Answer: 0.3488 or 34.88 %

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The batteries in the figure above are connected in

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The answer is a, series circuit.

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Which element below has chemical properties similar to the chemical properties of fluorine.

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For number one the answer is Iodine because it is in the same group as fluorine. For number two the answer is Germanium for the same reason. For number three the answer is Aluminum for the same reason.

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4 years ago

What are the physical and chemical properties of sodium?

Hitman42 [59]

Answer:

Physical Properties of Sodium

Atomic number 11

Melting point 97.82°C (208.1°F)

Boiling point 881.4°C (1618°F)

Volume increase on melting 2.70%

Latent heat of fusion 27.0 cal/g

Lenntech Water treatment & purification

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Home Periodic table Elements Sodium

Sodium - Na

Chemical properties of sodium - Health effects of sodium - Environmental effects of sodium

Atomic number

11

Atomic mass

22.98977 g.mol -1

Electronegativity according to Pauling

0.9

Density

0.97 g.cm -3 at 20 °C

Melting point

97.5 °C

Boiling point

883 °C

Vanderwaals radius

0.196 nm

Ionic radius

0.095 (+1) nm

Isotopes

3

Electronic shell

[Ne] 3s1

Energy of first ionisation

495.7 kJ.mol -1

8 0

2 years ago

A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista

slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

$C' = \frac{1}{2}C$

The potential difference across the capacitor is given by

$V= \frac{Q}{C}$

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

$V' = 2 V$

Now we can analyze the electric field between the plates of the capacitor, which is given by

$E=\frac{V}{d}$

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

$E'=\frac{2V}{2d}=\frac{V}{d}=E$

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0

3 years ago

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

4 0

4 years ago