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Tanya [424]
3 years ago
11

Kirsin is learning about the outer planets what else would she know about a planet that is the second largest gas giant and has

the largest rings?

Physics
1 answer:
vodka [1.7K]3 years ago
7 0

Answer: <u>Saturn is the second largest gas giant and has the largest rings.</u>

Explanation:

  1. Saturn is the 2nd largest planet in the solar system after Jupiter.
  2. It is the 6th planet from the Sun and  located between Jupiter and Uranus.
  3. Saturn was named after "Saturnus", the roman god of agriculture.
  4. Saturn orbital period to move around sun is  29 and half earth years
  5. Saturn is called gas giant because it is not made of any solid material but light gases (hydrogen and helium).
  6. Saturn has largest rings. These are made up of pieces of ice, rocks and dust.
  7. The diagram is attached below:

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Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. The initial temperature of the gas is 28.0
mel-nik [20]

Answer:

63.5 °C

Explanation:

The expression for the calculation of work done is shown below as:

w=P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

Also,

Considering the ideal gas equation as:-

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 8.314 J/ K mol

So,

V=\frac{nRT}{P}

Also, for change in volume at constant pressure, the above equation can be written as;-

\Delta V=\frac{nR\times \Delta T}{P}

So, putting in the expression of the work done, we get that:-

w=P\times \frac{nR\times \Delta T}{P}=nR\times \Delta T

Given, initial temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.0 + 273.15) K = 301.15 K

W=1770 J

n = 6 moles

So,

1770\ J=6 moles\times 8.314\ J/ Kmol \times (T_2-301.15\ K)

Thus,

T_2=301.15\ K+\frac{1770}{6\times 8.314}\ K

T_2=336.63\ K

The temperature in Celsius = 336.63-273.15 °C = 63.5 °C

<u>The final temperature is:- 63.5 °C</u>

7 0
4 years ago
The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of
kolbaska11 [484]

Answer:

a) A = 4.0 m , b)   w = 3.0 rad / s , c)  f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

          x = A cos (wt + fi)

In the exercise we are told that the expression is

          x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

         A = 4.0 m

b) the frequency or angular velocity

         w = 3.0 rad / s

c) angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 3 / 2π

           f = 0.477 Hz

d) the period

frequency and period are related

           T = 1 / f

           T = 1 / 0.477

           T = 20.94 s

e) the phase constant

          Ф = 0.10 rad

f) velocity is defined by

          v = dx / dt

         

         v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

         v = A w

          v = 4 3

          v = 12 m / s

g) the angular velocity is

          w² = k / m

          k = m w²

          k = 1.2 3²

          k = 10.8 N / m

h) the total energy of the oscillator is

          Em = ½ k A²

           Em = ½ 10.8 4²

          Em = 43.2 J

i) the potential energy is

           Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

               x = 3.98 m

j) kinetic energy

           K = ½ m v²

for t = 00.1 ²

    v = A w sin 0.10

    v = 4 3 sin 0.10

    v = 1.98 m / s

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Answer:

I think the answer is 1 m per second.

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