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3 years ago

In the equation for elastic potential energy below, what is represented by the symbol k? Ee = ½ × k × e²

Physics

1 answer:

yarga [219]3 years ago

5 0

Answer:

K is the spring/elastic constant measured in Newton per metre square (N/m²)

Explanation:

E = ½Ke²

In the equation above,

E is the energy in Joules (J)

K is the elastic/spring constant in Newton per metre square (N/m²)

e is the extension in metre (m)

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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

K donates, or transfers, one electron to bromine, which has 7 electrons. Both K and Br are now stable with 8 electrons. K become

Charra [1.4K]

K is cation by losing of electron whereas Br is anion due to accepting of electrons.

<h3 /><h3>Is charge appears when an atom lose or accept electron?</h3>

Yes, the positive ion appears on K and become cation whereas the negative ion bears on Br which make it anion because of losing and gaining of electron by these atoms. This transferring of electrons leads to formation of ionic bonds between them.

So we can conclude that K is cation by losing of electron whereas Br is anion due to accepting of electrons.

Learn more about ionic bond here: brainly.com/question/2687188

#SPJ1

4 0

1 year ago

What are the strength and direction of an electric field that will balance the weight of a 0.9 g plastic sphere that has been ch

Vladimir79 [104]

(a) The force exerted by the electric field on the plastic sphere is equal to
$F=qE$
where $q=-3.4 nC=-3.4 \cdot 10^{-9} C$ is the charge of the sphere and E is the strength of the electric field. This force should balance the weight of the sphere:
$F=mg =0.9 g$
where m is the mass of the sphere and g is the gravitational acceleration.

Since the two forces must be equal, we have:
$qE=mg$
and so we find the intensity of the electric field
$E= \frac{mg}{q}= \frac{0.9 \cdot 9.81 m/s^2}{3.4 \cdot 10^{-9} C} =2.6 \cdot 10^9 N/C$

(b) Now let's find the direction of the field. The electric force must balance the weight of the sphere, which is directed downward, so the electric force should be directed upward. Since the charge is negative, the force is opposite to the electric field direction, and so the direction of the electric field is downward.

4 0

3 years ago

Mercury is characterized by

cupoosta [38]

<span>It's close to the sun without much atmosphere, so it's characterized by </span><span>very extreme temperatures.

Happy studying ^_^</span>

8 0

3 years ago

If I turn on a light in a spaceship traveling 1C BACKWARDS, what happens to the photons? Speed -0-, or 1C in opposite direction?

Advocard [28]

If the spaceship's Physicist happens to be hanging out of one side
of the ship, and he measures the speed of the photons as they pass
him and leave the ship, he'll see them passing him at 'c' ... the speed
of light.

When those photons pass somebody who happens to be in their
path, and he decides to measure their speed, he'll see them move
past him at 'c' ... the speed of light.

It doesn't matter whether the observer who measures them is
moving, or at what speed.

And it doesn't matter what source the photons come from, or
whether the source is moving, or at what speed.

And it doesn't matter what the photons' wavelength/frequency is ...
anything from radio to gamma rays.

The photons pass everybody at 'c' ... the speed of light.

Yes, I hear you. That can't be true. It's crazy.
Maybe it's crazy, but it's true.

5 0

3 years ago