Ask question

Ask question

All categories

3 years ago

7

In an isochoric process, the pressure in a system changes from 10 kPa to 20 kPa at V= 5m3. Calculate the work done in this proce

ss according to the sign convention in the class.

1 answer:

Elodia [21]3 years ago

4 0

Answer:

work done in the process is 0 J.

Explanation:

Given :

Initial pressure of system , $p_i=10\ kPa.$

Final pressure of system , $p_f=20\ kPa.$

Volume of system , $V=5\ m^3.$

We know, In thermodynamics work done is defined as :

$W=P\Delta V$

Here , P is pressure and $\Delta V$ is change in volume.

Now, coming back to question , it state that the process is isochoric .

Therefore, change in volume , $\Delta V=0.$

Putting value of $\Delta V=0$ in above equation we get ,

W = 0 J.

Therefore , work done in this process is 0 J.

Hence, this is the required solution.

You might be interested in

Gravity on the surface of the moon is only 1/6 as strong as gravity on the Earth. What is the weight of a 19 kg object on the Ea

Dahasolnce [82]

The weight of anything in any place is

(mass of the thing) x (acceleration of gravity in that place).

-- On Earth, the acceleration of gravity is about 9.807 m/s²

Weight of 19 kg of mass is (19 kg) x (9.807 m/s²) = <em>186.3 newtons</em>

-- On the Moon, the acceleration of gravity is about 1.623 m/s²

Weight of the same 19 kg of mass is (19 kg) x (1.623 m/s²) = <em>30.8 newtons</em>

7 0

3 years ago

Ksivusya [100]

5.77 × $10^1^4$ Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

<h3>Given:</h3>

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

c = Speed of light = 3 × $10^8$ m/s

λ = Wavelength of light

∴ f = c / λ

f = $\frac{3*10^8}{520 * 10^-^9}$

= 5.77 × $10^1^4$ Hz

Therefore, 5.77 × $10^1^4$ Hz is the green photon's frequency .

Learn more about wavelength here:

brainly.com/question/10728818

#SPJ1

4 0

1 year ago

A 400-n block is dragged along a horizontal surface by an applied force as shown. the coefficient of kinetic friction is uk = 0.

gulaghasi [49]

The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.

We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force $F_f$ acting against the motion. Since their resultant must be zero, we have:
$F-F_f = 0$
The frictional force is
$F_f = \mu mg$
where
$\mu=0.4$ is the coefficient of kinetic friction
$mg=400 N$ is the weight of the block.

Substituting these values, we find the magnitude of the force F:
$F=F_f = \mu mg=(0.4 )(400 N)=160 N$

4 0

3 years ago

What factors cause a rock to become metamorphic

Lynna [10]

Pressure and heat. I hope this helps

3 0

2 years ago

Read 2 more answers

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c) </em> $\frac{4492.18}{v_{car} ^{2} }$

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = <em>6738.27 J</em>

<em></em>

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

3 0

3 years ago