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Roman55 [17]
2 years ago
5

The distance from the dwarf planet Pluto to the sun is 4.43 × 109 km at perihelion and 7.37 × 109 km at aphelion. Find the eccen

tricity e of Pluto's orbit. (Round your answer to three decimal places.) e =
Physics
1 answer:
kaheart [24]2 years ago
3 0

Answer:

0.249

Explanation:

Perihelion = 4.43 x 10^9 km

Aphelion = 7.37 x 10^9 km

Let e be the eccentricity and a be the length of semi major axis.

The relation between the semi major axis, perihelion and aphelion s given by

Semi major axis = half of sum of perihelion and aphelion

a = \frac{4.43+7.37}{2}\times10^{9}

a = 5.9 x 10^9 km

The relation between the perihelion, semi major axis and the eccentricity is given by

Perihelion = a (1 - e)

4.43 x 10^9 = 5.9 x 10^9 (1 - e)

0.751 = 1 - e

e = 0.249

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A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou
attashe74 [19]

Answer:

Explanation:

Impulse = Force x time = change in momentum

F x t = m ( v - u )

In both cases, u are same and v=0

So change in momentum is same

hence , impulse is same.

F = Change in momentum / time

In case of air mattress , time increases

Hence average force decreases .

Option e is correct .

8 0
2 years ago
A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi
RSB [31]

Answer:

r = 4.44 m

Explanation:

 

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

         B = ρ g V

Now let's use Newton's equilibrium relationship

         B - W = 0

         B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

         σ = W / A

         W = σ A

The area of ​​a sphere is

           A = 4π r²

       W = W₁ + σ 4π r²

The volume of a sphere is

           V = 4/3 π r³

Let's replace

     ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

     P V = n RT

    P = ρ RT

    ρ = P / RT

 

    P / RT g 4/3 π r³ - σ 4 π r² = W₁

    r² 4π (P/3RT  r - σ) = W₁

Let's replace the values

     r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

     r² (11.81 r -0.060) = 13000 / 4pi

     r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

       r = 4.44 m

3 0
2 years ago
"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a
My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0
2 years ago
Steam at a pressure of 15 bar and a temperature of 320oC is contained in a large vessel. Connected to the vessel through a valve
Luda [366]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

7 0
2 years ago
Question
nataly862011 [7]

The density of a material is constant and it is given by the ratio of the mass to the volume of the material

The mass of the liquid and the full bottle ae

The mass of the liquid is <u>450 g</u>

The mass of the filled bottle is <u>530 grams</u>

<u></u>

The reason the above values are correct are as follows:

The given parameters are;

Volume of the bottle, V = Half litre

Mass of the bottle, m_b = 80 g

The volume of liquid in the bottle when filled, V = 500 cm³

The density of the olive oil with which the bottle is filled, ρ = 0.9 g/cm³

a. Required:

To calculate the mass of oil in the bottle

Solution:

The volume of oil in the bottle when the bottle is filled, V = 500 cm³

Density, \ \rho = \dfrac{Mass}{Volume}  = \dfrac{m}{V}

The mass of the liquid, m = ρ × V

∴ m = 0.9 g/cm³ × 500 cm³ = 450 g

The mass of the liquid, m = <u>450 g</u>

<u></u>

b. The mass of the oil in the bottle, m = grams

The mass of the full bottle, m_{filled} = m + m_b

∴ m_{filled} = 450 g + 80 g = 530 g

The mass of the full bottle, m_{filled} = <u>530 grams</u>

Learn more about density here:

brainly.com/question/18110802

4 0
2 years ago
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