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Alona [7]
3 years ago
8

If you know that the universe is estimated to be 14 billion years old . what is the age of the universe in seconds ???? (knowing

that : the solar year = 365.25days )
Physics
2 answers:
frosja888 [35]3 years ago
7 0

Answer:

dont read this

Explanation:

because

Svetlanka [38]3 years ago
6 0

Answer:

<em>The estimated age of the universe is</em> 4.42*10^{17}\ seconds

Explanation:

<u>Time Units Conversion</u>

Sometimes we need to express magnitudes in different units, depending on the specific situation handled in the question or problem at hand. Magnitudes like length, time, mass, temperature, pressure, etc., can be expressed in several different units.

The most common units for time are years, months, days, hours, minutes, and seconds. There are 3,600 seconds in one hour and 24 hours in one day.

We are given the estimated age of the universe as 14 billion solar years. If one solar year is 365.25 days long, then the age, expressed in days is

14,000,000,000 * 365.25 days

Some quantities need to be expressed in scientific notation. This is the case since the numbers are too high

14,000,000,000 * 365.25 = 5.1135*10^{12}\ days

There are 24*3600=86400 seconds in one day, so

5.1135*10^{12}\ days=5.1135*10^{12} * 86400 =4.42*10^{17}\ seconds

The estimated age of the universe is 4.42*10^{17}\ seconds

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Read 2 more answers
Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

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