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Tatiana [17]
1 year ago
9

roblem 10: In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following p

ressure 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.
Physics
1 answer:
miskamm [114]1 year ago
3 0

Answer:

W= -2.5 (p₁*0.0012) joules

Explanation:

Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then

In adiabatic compression, work done by mixture during compression is

W= \int\limits^f_i {p} \, dV  where f= final volume and i =initial volume, p=pressure

p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ

W= \int\limits^f_i {K/V^} \, dV

W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)

W=1/1-γ (p₁Vf-p₀Vi)

W= 1/1-1.40 (p₁*6/5 -p₀*0)  

W= -2.5 (p₁*6/5*0.001)   changing liters to m³

W= -2.5 (p₁*0.0012) joules

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Two very large, flat plates are parallel to each other. Plate A, located at y=1.0 cm, is along the xz-plane and carries a unifor
Dmitry [639]

Answer:

 E ≈ 1.70 10⁵ N/C

Explanation:

The electric field is a vector quantity, so we can calculate the field of each plate and then add them. To calculate the field of a plate let's use Gauss's law

       Φ = ∫ E. dA = q_{int} / ε₀

To apply this law we must create a Gaussian surface that takes advantage of the symmetry of the problem. The electric field lines on the surface are perpendicular, so the Gaussian surface that will be a cylinder with the base parallel to the plate.

On this surface the normal to the base (A) is parallel to the field lines whereby the scalar product is reduced to the ordinary product. The normal on the sides of the cylinder is perpendicular to the field, therefore, the product scale is zero.

        ∫I E dA = q_{int}  /ε₀

Let's look for the load under the cylinder, let's use the concept of load density

        σ =  q_{int} / A

         q_{int} = σ A

Let's write Gauss's law for this case

       E A =  q_{int} /ε₀  

       E A = σ A / ε₀

       E = σ / ε₀

As the field is emitted for each side of the plate the value to only one side is

      E = G / 2ε₀  

This expression is the same for each plate, now let's add the electric field at the requested point

     R = (0.50, 0.00, 0.00) cm

We see that this point is on the X axis, between the plates that are at the points y = -1.0 cm and y = 1.0 cm, as the plates are very large the test point is between them

The negative plate has an incoming field and the positive plate has an outgoing field, the test load is always positive. The field due to the negative plate goes to the left, the field through the positive plate goes to the left at this point whereby two are added

     E = E_ + E +

     E = σ1 / 2ε₀  + σ2 / 2ε₀  

     E = 1 / 2o (σ1 + σ2)

Let's calculate the value

     E = 1/2 8.85 10⁻¹² (1.00 10⁻⁶ + 2.00 10⁻⁶)

     E = 3 10⁻⁶ / 17.7 10⁻¹²

     E = 1,695 10⁵ N / C

     E ≈ 1.70 10⁵ N/C

6 0
2 years ago
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim
crimeas [40]

Answer:

Centripetal acceleration of the car is 17.4 m/s²

Explanation:

It is given that,

Radius of the track, r = 57 m

Speed of car, v = 31.5 m/s

We need to find the centripetal acceleration of the race car. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(31.5\ m/s)^2}{57\ m}

a=17.4\ m/s^2

So, the centripetal acceleration of the race car is 17.4 m/s². Hence, this is the required solution.

5 0
2 years ago
Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
lions [1.4K]

Answer:

a) Probability mass function of x

x P(X=x)

0 0.0602

1 0.0908

2 0.1700

3 0.2050

4 0.1800

5 0.1550

6 0.0843

7 0.0390

8 0.0147

b) Cumulative Distribution function of X

x F(x)

0 0.0602

1 0.1510

2 0.3210

3 0.5260

4 0.7060

5 0.8610

6 0.9453

7 0.9843

8 1.0000

The cumulative distribution function gives 1.0000 as it should.

Explanation:

Probability of arriving late = 0.43

Probability of coming late = 0.57

Let's start with the probability P(X=0) that exactly 0 people arrive late, the probability P(X=1) that exactly 1 person arrives late, the probability P(X=2) that exactly 2 people arrive late, and so on up to the probability P(X=8) that 8 people arrive late.

Interpretation(s) of P(X=0)

The two singles must arrive on time, and the three couples also must. It follows that P(X=0) = (0.57)⁵ = 0.0602

Interpretation(s) of P(X=1)

Exactly 1 person, a single, must arrive late, and all the rest must arrive on time. The late single can be chosen in 2 ways. The probabiliy that (s)he arrives late is 0.43.

The probability that the other single and the three couples arrive on time is (0.57)⁴

It follows that

P(X=1) = (2)(0.43)(0.57)⁴ = 0.0908

Interpretation(s) of P(X=2)

Two late can happen in two different ways. Either (i) the two singles are late, and the couples are on time or (ii) the singles are on time but one couple is late.

(i) The probability that the two singles are late, but the couples are not is (0.43)²(0.57)³

(ii) The probability that the two singles are on time is (0.57)²

Given that the singles are on time, the late couple can be chosen in 3 ways. The probability that it is late is 0.43 and the probability the other two couples are on time is (0.57)².

So the probability of (ii) is (0.57)²(3)(0.43)(0.57)² which looks better as (3)(0.43)(0.57)⁴ It follows that

P(X=2) = (0.43)²(0.57)³ + (3)(0.43)(0.57)⁴ = 0.0342 + 0.136 = 0.1700

Interpretations of P(X=3).

Here a single must arrive late, and also a couple. The late single can be chosen in 2 ways. The probability the person is late but the other single is not is (0.43)(0.57).

The late couple can be chosen in 3 ways. The probability one couple is late and the other two couples are not is (0.43)(0.57)². Putting things together, we find that

P(X=3) = (2)(3)(0.43)²(0.57)³ = 0.2050

Interpretation(s) P(X=4)

Since we either (i) have the two singles and one couple late, or (ii) two couples late. So the calculation will break up into two cases.

(i) Two singles and one couple late

Two singles' probability of being late = (0.43)² and One couple being late can be done in 3 ways, so its probability = 3(0.43)(0.57)²

(ii) Two couples late, one couple and two singles early

This can be done in only 3 ways, and its probability is 2(0.57)³(0.43)²

P(X=4) = (3)(0.43)³(0.57)² + (3)(0.57)³(0.43)² = 0.0775 + 0.103 = 0.1800

Interpretations of P(X=5)

For 5 people to be late, it has to be two couples and 1 single person.

For couples, The two late couples can be picked in 3 ways. Probability is 3(0.43)²(0.57)

The late single person can be picked in two ways too, 2(0.43)(0.57)

P(X=5) = 2(3)(0.43)³(0.57)² = 0.1550

Interpretations of P(X=6)

For 6 people to be late, we have either (i) the three couples are late or (ii) two couples and the two singles.

(i) Three couples late with two singles on time = (0.43)³(0.57)²

(ii) Two couples and two singles late

Two couples can be selected in 3 ways, so probability = 3(0.43)²(0.57)(0.43)²

P(X=6) = (0.43)³(0.57)² + 3(0.43)⁴(0.57) = 0.0258 + 0.0585 = 0.0843

Interpretation(s) of P(X=7)

For 7 people to be late, it has to be all three couples and only one single (which can be picked in two ways)

P(X=7) = 2(0.57)(0.43)⁴ = 0.0390

Interpretations of P(X=8)

Everybody had to be late

P(X=8) = (0.43)⁵ = 0.0147

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2 years ago
When a particle is a distance r from the origin, its potential energy function is given by the equation U(r)=kr, where k is a co
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Answer

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Step-by-step explanation:

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2 years ago
The brain receives messages in signals called nerve impulses. Which part of the ear first generates these nerve impulses?
liberstina [14]

That's the job of the tiny "hair cells", located in the <em>inner ear.</em>

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-- go into the big funnel of skin on the outside of the head, that thing we call the "ear"

-- go about an inch or two, down through a skinny dark tunnel inside the skull

-- at the end of the tunnel, hit a dead end, made of a wall of thin skin like a drum, called the "ear drum";  sound waves hit the ear drum and make it vibrate

-- on the other side of the ear drum, inside, is the chamber called the "middle ear".  In there are the three smallest bones in the body; the ear drum touches the first one and makes it vibrate; the first one touches the second one and makes it vibrate; the second one touches the third one and makes it vibrate;  then the third one touches another dead end made of thin skin.

-- the region on the other side of this wall of thin skin is the "inner ear";  it's a long skinny chamber, called the "cochlea",  wound up in a spiral and filled with liquid; the walls of the cochlea are lined with millions of tiny hairs, sticking out into the liquid; the vibrations make waves in the liquid, and the waves make the tiny hairs wave back and forth; each tiny hair is the end of a nerve that goes into the brain; when that hair wiggles, it sends a nerve "message" into the brain.  

-- there are two complete copies of this whole structure ... one on each side of your head.

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2 years ago
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