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Lubov Fominskaja [6]

3 years ago

9

In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following pressure p,

in atm, as a function of volume V, in liters: p = p0 V-6/5. Here p0 is a constant of units atm⋅L6/5. show answer Incorrect Answer 50% Part (a) Write an expression for the work W done on the gas when the gas is compressed from a volume Vi to a volume Vf.

1 answer:

viktelen [127]3 years ago

7 0

Answer:

Explanation:

In case of gas , work done

W = ∫ p dV , p is pressure and dV is small change in volume

the limit of integration is from Vi to Vf .

= ∫ p dV

= ∫ p₀ $V^{-\frac{6}{5}$ dV

= p₀ $V^{-\frac{6}{5} +1}$ / ( $\frac{-6}{5} +1$ )

= - 5p₀ $V^{-\frac{1}{5}$

Taking limit from Vi to Vf

W = - 5 p₀ ( $V_f^\frac{-1}{5} - V_i^{\frac{-1}{5}$ ) ltr- atm.

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Could you please assist me, thank you kindly.

Bogdan [553]

In Newton's Third law of motion, the 'action' and 'reaction' forces act on different objects. That's why they don't cancel each other out and always result in zero force.

4 0

2 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

4 0

2 years ago

2. How can people know that they've learned?

photoshop1234 [79]

Answer: People know that they jap e learned by their mistakes or by someone ya being them something

4 0

3 years ago

horrorfan [7]

3) The work done is D. zero

4) The kinetic energy is B. 180 J

5) The potential energy is A. 120 J

6) The work done depends on B. position

7) The example of non-renewable energy is C. coal

8) The power expended is $3\cdot 10^4 W$

9) The efficiency is A. 100%

10) The velocity ratio is 5

Explanation:

3)

The work done by a force acting an object is given by:

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

When the force is applied perpendicular to the direction of motion,

$\theta=90^{\circ}$

Therefore, the work done is:

$W=Fd(cos 90^{\circ})=0$

4)

The kinetic energy of a body is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the girl in this problem, we have

m = 40 kg

v = 3 m/s

Therefore her kinetic energy is

$K=\frac{1}{2}(40)(3)^2=180 J$

5)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

$g=10 m/s^2$ is the acceleration of gravity

h is the heigth of the object relative to the ground

For the ball in this problem,

m = 0.4 kg

h = 30 m

So, the potential energy is

$PE=(0.4)(10)(30)=120 J$

6)

A conservative field is a field for which the work done by the field on an object does not depend on the path taken, but only on the initial and final position of the object.

Gravitational and electric fields are examples of conservative fields. In fact:

When an object is pulled down by gravity (free fall), the work done by the gravitational field only depends on the change in height $\Delta h$ between the two points, not on the path taken during the fall
When an electric charge is pushed by the electric field, the work done by the field depends only on the initial and final position of the charge in the field

For any conservative field, it is possible to define a "potential" function, which represents the energy per unit mass/charge, and depends only on the position of the object.

7.

Non-renewable energy sources are sources of energy whose rate of consumption is faster than the rate at which they are re-created. Examples of non-renewable sources are coal, oil, natural gas. These energy sources are consumed at a fast rate, while they take million of years to regenerate, so at the current rate they will eventually run out.
Renewable energy sources are sources of energy that replenish at faster rate than the rate at which it is consumed. Examples of renewable sources are solar energy, wind, hydroelectric power.

Therefore, the example of non-renewable energy in this case is

C. Coal

8.

For an object pushed by a force F and moving at a constant velocity v, the power expended is given by

$P=Fv$

where F is the force and v is the velocity.

for the rocket in this problem, we have:

F = 10 N is the force propelling the rocket

v = 3000 m/s is its velocity

Substituting into the equation, we find the power expended:

$P=(10)(3000)=30,000 W = 3\cdot 10^4 W$

9.

The efficiency of a machine is given by

$\eta = \frac{W_{out}}{W_{in}}$

where

$W_{in}$ is the energy in input to the machine

$W_{out}$ is the useful work in output from the machine

For a real machine, the useful work in output is always lower than the energy input, because part of the energy is "wasted" and converted into thermal energy due to the presence of internal frictions. However, for an ideal machine, all the input energy is converted into useful work, so

$W_{out}=W_{in}$

And therefore the efficiency is

$\eta=1$

which means 100%.

10.

The velocity ratio of a block and tackle system is the ratio between the distance moved by the effort and the distance moved by the load.

$VR=\frac{d_{eff}}{d_{load}}$

In a block and tackle system, the velocity ratio is also equal to the number of pulleys in the system.

For the system in the problem, there are 5 pulleys: therefore, this means that when the effort moves 5 metres, the load moves 1 metres, therefore the velocity ratio is

$VR=\frac{5}{1}=5$

Learn more about kinetic and potential energy:

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#LearnwithBrainly

5 0

3 years ago

3. A model rocket takes 0.05 seconds to speed up from rest to its maximum velocity of 80 m/s.

nikklg [1K]

Answer:

$1600 \frac{m}{s^2}$

Explanation:

Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

$a = \frac{change-in-velocity}{time} = \frac{Vf-Vi}{t}$

Where Vf is the final velocity of the object, (in our case 80 m/s)

Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)

and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.

Therefore we have:

$a = \frac{80 m/s - 0 m/s}{0.05 sec} = 1600 \frac{m}{s^2}$

Notice that the units of acceleration in the SI system are $\frac{m}{s^2}$ (meters divided square seconds)

7 0

2 years ago

Read 2 more answers