Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount by which the star's thermal radiation increases the entropy of the entire universe each second. Assume that the star is a perfect blackbody, and that the average temperature of the rest of the universe is 2.73 K. Do not consider the thermal radiation absorbed by the star from the rest of the universe. J/K
Answers
Answer:
The value is [tex]\Delta s = 8.537 *10^{25 } \ J/K[/tex]
Explanation:
From the we are told that
The radius of the sphere is [tex]r = 6.32 *10^{8} \ m[/tex]
The temperature is [tex]T_x = 5350 \ K[/tex]
The average temperature of the rest of the universe is [tex]T_r = 2.73 \ K[/tex]
Generally the change in entropy of the entire universe per second is mathematically represented as
[tex]\Delta s = s_r - s_x[/tex]
Here [tex]s_r[/tex] is the entropy of the rest of the universe which is mathematically represented as
[tex]s_r = \frac{Q}{T_r}[/tex]
Here Q is the quantity of heat radiated by the star which is mathematically represented as
[tex]Q = 4 \pi * r^2 * \sigma * T^4_x[/tex]
Here [tex]\sigma[/tex] is the Stefan-Boltzmann constant with value
[tex]\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.[/tex]
=> [tex]Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4[/tex]
=> [tex]Q = 2.332 *10^{26} \ J[/tex]
So
[tex]s_r = \frac{2.332 *10^{26}}{2.73}[/tex]
=> [tex]s_r = 8.5415 *10^{25}\ J/K[/tex]
Here [tex]s_x[/tex] is the entropy of the rest of the universe which is mathematically represented as
[tex]s_x = \frac{Q}{T_x}[/tex]
=> [tex]s_x = \frac{2.332 *10^{26} }{5350}[/tex]
=> [tex]s_x = 4.359 *10^{22} \ J/K[/tex]
So
[tex]\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}[/tex]
=> [tex]\Delta s = 8.537 *10^{25 } \ J/K[/tex]
This question involves the concepts of entropy and the thermal radiation
The entropy of the entire universe is increased by "8.41 x 10²⁵ J/k
".
The increase in entropy is given as follows:
[tex]\Delta s = s-s_T[/tex]
where,
Δs = increase in entropy = ?
σ = Stefan-Boltzman's constant = 5.67 x 10⁻⁸ W/m².k⁴
A = surface area = 4πr² = 4π(6.32 x 10⁸ m)² = 5.01 x 10¹⁸ m²
Tr = Absolute temperature of the star = 5350 K
T = absolute temperature of the rest of the universe = 2.73 k
Q = thermal radiation energy
Q = [tex]\sigma A T_r^4=(5.67\ x\ ^{-8}\ W/m^2.k^4)(5.01\ x\ ^{18}\ m^2)(5350\ k)^4=2.3\ x\ 10^{26}\ J[/tex]
s = entropy of the universe = [tex]\frac{Q}{T}=\frac{2.3\ x\ 10^{26}\ J}{2.73 k}=8.42\ x\ 10^{25}\ J/k[/tex]
[tex]s_T[/tex] = entropy of the star = [tex]\frac{Q}{T_r}=\frac{2.3\ x\ 10^{26}\ J}{5350\ k}=4.3\ x\ 10^{22}\ J/k[/tex]
Therefore,
Δs = 8.42 x 10²⁵ J/k - 4.3 x 10²² J/k
Δs = 8.41 x 10²⁵ J/k
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Related Questions
12) Consider two identical bricks, each of dimensions 20.0 cm x 10.0 cm x 6.0 cm. One is stacked
on the other, and the combination is then placed so that they project out over the edge of a
table. What is the maximum distance that the end of the top brick can extend beyond the table
edge without toppling?
A) 7.5 cm
B) 10 cm
C) 12.5 cm
D) 15 cm
Answers
Answer:
7.5
Explanation:
An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?
Answers
Answer:
The fraction of the object that is below the surface of the water is ¹⁷/₂₀
Explanation:
Given;
specific gravity of the object, γ = 0.850
Specific gravity is given as;
[tex]specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3[/tex]
Fraction of the object's weight below the surface of water is calculated as;
[tex]= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}[/tex]
Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀
A ray moving in water at 55.5 deg
enters plastic, where it bends to
48.7 deg. What is the index of
refraction for the plastic?
( water n = 1.33, Air n = 1.00 )
Answers
Answer:
Refractive index of the plastic = 1.46
Explanation:
By Snell's law,
[tex]\frac{\text{sin}\theta _{2} }{\text{sin}\theta _{1}}=\frac{n_1}{n_2}[/tex]
Here, [tex]\theta _1[/tex] = Angle of incidence in medium 1 (Plastic)
[tex]\theta_2[/tex] = Angle of refraction in medium 2 (Water)
[tex]n_1[/tex] = Refractive index of medium 1 (Plastic)
[tex]n_2[/tex] = Refractive index of medium 2 (Water)
By substituting values in the formula,
[tex]\frac{\text{sin}(48.7)}{\text{sin}(55.5)}=\frac{1.33}{n_2}[/tex]
[tex]n_2=\frac{1.33\times \text{sin}(55.5)}{\text{sin}(48.7)}[/tex]
= 1.46
Therefore, refractive index of the plastic = 1.46
Describe the Rutherford model
Answers
Answer:
The Rutherford model was devised by the New Zealand-born physicist Ernest Rutherford to describe an atom. Rutherford directed the Geiger–Marsden experiment in 1909, which suggested, upon Rutherford's 1911 analysis, that J. J. Thomson's plum pudding model of the atom was incorrect.
Atomic theory year: 1911
Explanation:
Hope this helps, Merry Christmas, and have a good day
. A pendulum of length l = 9.8 m hangs in equilibrium and is then given velocity
v
=0.2
m/s at its lowest point. What is the amplitude of the subsequent oscillation?
Answers
Answer:
the amplitude of the sequence oscillation=
ASO = length × velocity
= 9.8 × 0.2 = 19.6
ASO = 19.6
A chimpanzee climbs a vine up and to the right for a total displacement of 50 m 50m50, start text, m, end text that makes a 60 ° 60°60, degree angle from the ground. What was the horizontal displacement of the chimpanzee in m mstart text, m, end text?
Answers
Answer:
25m
Explanation
Let the horizontal displacement formula be expressed as;
Dx = Dcos theta
theta is the angles subtended by the displacement
Given
D = 50m
theta = 60°
Required
horizontal displacement of the chimpanzee
Substitute into the formula
Recall that: Dx = Dcos theta
Dx = 50cos60°
Dx = 50(0.5)
Dx = 25m
Hence the horizontal displacement of the chimpanzee is 25m
Using the law of conservation of energy, what is the kinetic energy at e?
Answers
Answer:
Send the pic so I can see
A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleration α, the torque on the child is τ. The child moves to a position half way between the center and edge of the merry-go-round, and the angular acceleration increases to 2α. The torque on the child is now
Answers
Answer:
The torque on the child is now the same, τ.
Explanation:
It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:[tex]I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)[/tex]
So, τ = 3/2*m*r²*α (2)When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:[tex]I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)[/tex]
Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.
What is the lithosphere?
A. the outer layer of the Earth's crust
B. the inner core
C. the middle portion of the mantle
D. the outer core
Answers
Answer:
a. outer layer
Explanation:
lithosphere is right underneath the continental and ocean crust. it is approximately 100 km in deep and it is a brittle layer. It is broken into tectonic plates.
the inner core is located at the very center and its full of iron and nickel (so its not B)
on top of that is the outer core which is liquid (not D)
the middle portion of the mantle is the asthenosphere and mesosphere. they are right beneath the lithosphere. (not C)
so the best answer is A
How long( in hours, will it take for 500 000 C of charge to flow through a diode if it requires
0.05 Amp to operate it.
Answers
Answer:
277.78 hours
Explanation:
The formula for calculating the amount of charge is expressed as;
Q = It
I is the current
t is the time
Given
I =0.05A
Q = 50,000C
Required
Time t
Recall that: Q = It
t = Q/I
t = 50,000/0.05
t = 1,000,000secs
Convert to hours
1,000,000secs = 1,000,000/3600
1,000,000secs = 277.78 hours
Hence it will take 277.78 hours for the charge to flow through the diode
Which best describes the law of conservation of mass?
O The coefficients in front of the chemicals in the reactants should be based on the physical state of the products,
O Products in the form of gases are not considered a part of the total mass change from reactants to products
O When reactants contain both a solid and a liquid, the solid counts toward the overall mass and the liquid does not
O The mass of the reactants and products is equal and is not dependent on the physical state of the substances.
Answers
Explanation:
The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.
how many minutes in 1 hour
Answers
Answer:
60
Explanation:
Every 60 minutes is an hour
Answer:
60; minute in 1 HR is the answer
give 2 reasons why a person should move away from "isolating muscles"
Answers
Answer:
1) - They should move away because its cold
2) - You can get hypothermia......
Explanation:
A herdsman yelling out to a fellow herdsman heard his voice reflected by a cliff 4s later.What is
the velocity of sound in air if the cliff is 680m away
Answers
Answer:
v = 340 m/s
Explanation:
Given that,
A herdsman yelling out to a fellow herdsman heard his voice reflected by a cliff 4s later.
The cliff is 680 m away
We need to find the velocity of sound in air.
Velocity = distance/time
Distance = 2 × 680 = 1360 m
[tex]v=\dfrac{1360\ m}{4\ s}\\\\=340\ m/s[/tex]
So, the velocity of sound in air is 340 m/s.
52. Serves as an air passageway
a. Oropharynx
c. Nasopharynx
b. Laryngopharynx
d. Larynopharynx
Answers
Answer:
Nasopharynx
Explanation:
The nasopharynx is posterior to the nasal cavity and serves only as a passageway for air.
A race car starts from rest and accelerates down a track at a rate of 3.0 m/s2
How fast is the car moving after 10 seconds?
Answers
30 m/s
Explanation:We are given:
Initial Velocity (u) = 0 m/s
Acceleration of the Car (a) = 3 m/s²
Time Interval (t) = 10 seconds
Speed of the Car After 10 seconds:
From the First equation of motion:
v = u + at
replacing the given values
v = 0 + (3)(10)
v = 30 m/s
Hence, the car is moving at a velocity of 30 m/s after 10 seconds
Ishan is testing materials to see whether they will stick to a magnet. How can
he use his imagination in this experiment?
OA. To think of different materials to test
OB. To carefully record the results in a chart
O C. To do the test the same way each time
O D. To remember which materials are magnetic
Answers
Answer:
The Answer is A "To think of different materials to test"
Explanation:
APE X
A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction and round to 2 decimals)
Answers
Answer:
29.96m/s
Explanation:
Given parameters:
Initial speed = 25.5m/s
Acceleration = 1.94m/s²
Time = 2.3s
Unknown:
Final speed of the car = ?
Solution:
To solve this problem, we are going to apply the right motion equation:
v = u + at
v is the final speed
u is the initial speed
a is the acceleration
t is the time taken
Now insert the parameters and solve;
v = 25.5 + (1.94 x 2.3) = 29.96m/s
What's a Weber?
in electromagnetism
Answers
Answer:
In physics, the weber is the SI derived unit of magnetic flux. A flux density of one Wb/m2 (one weber per square metre) is one tesla.
Hope it helps !
Answer:
Weber unit of magnetic flux in the international system of units (SI), defined as the amount flux that, linking an electrical circuit of one turn (one loop of wire) , produces in it an electromotive force of one volt as the flux is reduced to zero as a uniform rate in one second .
it was named in honour of the 19th century German physicist Wilhelm Eduard Weber
The velocity of a wave is 500 m/s . The wavelength of the wave is 3.2 m . What is the frequency of the wave ?
Answers
Answer:
f=156.25Hz
Explanation:
v=fT
500=fx3.2
500=3.2f
divide both sides by 3.2
f=156.25Hz
A PERSON GETTING OUT OF MOVING BUS FALLS IN THE DIRECTION OF MOTION OF THE BUS. WHY?
Answers
Answer:
PLEASE MARK AS BRAINLIEST!!
Explanation:
A getting passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is because his feet come to rest on touching the ground and the remaining body continues to move due to inertia of motion.
Answer:
When the person steps on the ground, his feet do not move but his upper body moves in the direction of the bus due to inertia of its motion. Since his upper body moves in the forward direction and his lower body does not move, the person falls in the forward direction.
A small, 300 g cart is moving at 1.10 m/s on an air track when it collides with a larger, 4.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. What is the speed of the large cart after the collision (answer in m/s please)?
Answers
Answer:
0.0158m/s
Explanation:
Using the law of conservation of energy which states that the sum of momentum before collision is equal to the sum after collision. It is expressed mathematically as;
m1u1 + m2u2 = m1v1 + m2v1
m1 and m2 are the masses of the object
u1 and u2 are the initial velocities
v1 and v2 are the final velocities
Given
m1 = 300g = 0.3kg
u1 = 1.10m/s
m2 = 4.00kg
u2 = 0m/s (at rest)
v1 = 0.890
v2 = ?
Substitute the given values into the formula;
0.3(1.10) + 0 = 0.3(0.89) + 4v2
0.33 = 0.267 + 4v2
0.33-0.267 = 4v2
0.063 = 4v2
v2 = 0.063/4
v2 = 0.0158m/s
Hence the speed of the large cart after the collision is 0.0158m/s
According to the question:
Mass,
[tex]m_1 = 300 \ g = 0.3 \ kg[/tex][tex]m_2 = 4.00 \ kg[/tex]Final velocity,
[tex]u_1 = 1.10 \ m/s[/tex][tex]u_2 = 0 \ m/s[/tex]Initial velocity,
[tex]v_1 = 0.890 \ m/s[/tex][tex]v_2 = \ ?[/tex]By using the law of conservation, we get
→ [tex]m_1 u_1 +m_2 u_2 =m_1 v_1 +m_2 v_1[/tex]
By substituting the values, we get
→ [tex]0.3(1.10)+0 = 0.3(0.89)+4v_2[/tex]
→ [tex]0.33=0.267+4 v_2[/tex]
[tex]0.33-0.267 = 4 v_2[/tex]
[tex]0.063 =4v_2[/tex]
[tex]v_2 = \frac{0.063}{4}[/tex]
[tex]= 0.0158 \ m/s[/tex]
Thus the response above is appropriate.
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What is the centripetal force necessary to keep a 0.4 kg object rotating in a circular path
of radius 3m?
Answers
Answer:
ertdhrsg
Explanation:
gesrgsg
Two masses 1.2kg and 1.8kg are connected to the ends of a rod of length 2m. Find the moment of inertia about the axes, 1)going through the mid point of the rod. 2)going through the centre of mass of two masses.
Answers
Answers: 1) 3 kg m²
2) 2.88 kg m²
Explanation: Question 1
I = m(r)²+ M(r)²
I = 1.2 kg × (1 m )² +1.8 kg ×(1 m )²
∴ I = 3 kg m²
Question 2
ACCORDING TO THE DIAGRAM DRAWN FOR QUESTION 2
we have to decide where the center of gravity (G) lies and obviously it should lie somewhere near to the greater mass. (which is 1.8 kg). Since we don't know the distance from center of gravity(G) to the mass (1.8 kg) we'll take it as 'x' and solve!!
moments around 'G'
F₁ d ₁ = F₂ d ₂
12 (2-X) = 18 (X)
24 -12 X =18 X
∴ X = 0.8 m
∴ ( 2 - x ) = 1.2 m
∴ Moment of inertia (I) going through the center of mass of two masses,
⇒ I = m (r)² +M (r)²
⇒ I = 1.2 × (1.2)² + 1.8 × (0.8)²
⇒ I = 1.2 × 1.44 + 1.8 × 0.64
⇒ I = 1.728 + 1.152
⇒ ∴ I = 2.88 kg m²
∴ THE QUESTION IS SOLVED !!!
A 200 g air-track glider is attached to a spring. The glider is pushed 10.0 cm against the spring, then released. A student with a stopwatch finds that 10 oscillating take 12.0 s. What is the spring constant?
Answers
Answer:
5.5N/m
Explanation:
Calculation for What is the spring constant
First step is to calculate the time period
T = 12 second/10
T = 1.2 second
Now let calculate the spring constant using this formula
k=4π²m/T²
Where,
m=0.2kg
T=1.2second
k represent spring constant=?
Let plug in the formula
k=4π²×0.2kg/(1.2)²
k=39.48×0.2kg/1.44
k=7.90/1.44
k=5.48N/m
k=5.5N/m ( Approximately)
Therefore the spring constant will be 5.5N/m
The reason why you should always exercise your abdominal during an exercise session is :
(A) because you will be fully warmed up
(B) you can focus on this muscle better
(C) because is the abdominal muscle is needed in all exercises during a workout
(D) because most people do it that way
(E) it does not really matter if you exercise the abdominal first or last
Answers
A 250 N to the east and a 30 N force to the west act on an object. What is the net force on the object? (Remember ΣF also equals the sum of all forces)
Answers
Answer:
280N
Explanation:
As the statement put in parenthesis state-You need to add up all the forces acting on the object to find the net force, so over here-
250+30=280N
Give reason:
a) In 'coin on card' experiment a smooth card is used.
Answers
Answer:
Please mark as brainliest!!
Explanation:
In coin card experiment smooth card is used so that the card can slide easily from glass.
Answer:
In coin card experiment smooth card is used so that the card can slide easily from glass.
two asteroids crashed the crash caused both asteroids to change speeds scientist want to use the change
Answers
Answer:
The force each one experienced
Explanation:
Hope this helps :)
Find the total charge of a system consists 2X10^4 electrons.
Answers
ANSWER: The first thing to learn is how to convert numbers back and forth between scientific notation and ordinary decimal notation. The expression "10n", where n is a whole number, simply means "10 raised to the nth power," or in other words, a number gotten by using 10 as a factor n times:
105 = 10 x 10 x 10 x 10 x 10 = 100,000 (5 zeros)
108 = 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 100,000,000 (8 zeros)
Notice that the number of zeros in the ordinary decimal expression is exactly equal to the power to which 10 is raised.
If the number is expressed in words, first write it down as an ordinary decimal number and then convert. Thus, "ten million" becomes 10,000,000. There are seven zeros, so in powers of ten notation ten million is written 107.
A number which is some power of 1/10 can also be expressed easily in scientific notation. By definition,
1/10 = 10-1 ("ten to the minus one power")
More generally, the expression "10-n" (where n is a whole number) means ( 1/10 )n. Thus
10-3 = ( 1 / 10 )3 = 1 / ( 10 x 10 x 10) = 1/1000
10-8 = ( 1 / 10 )8 = 1/100,000,000
Scientific notation was invented to help scientists (and science students!)deal with very large and very small numbers, without getting lost in all the zeros. Now answer the following on a separate sheet of paper and check your answers by clicking on "Answers":
Explanation: