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Marysya12 [62]
3 years ago
5

A steady device in each cycle converts 155 J of work to thermal energy that is released into a reservoir at 340 K. Part A Calcul

ate the change in entropy of the environment that results from this transfer.
Physics
1 answer:
sukhopar [10]3 years ago
7 0

Answer:

change in entropy is 3.3034 × 10^{22}  

Explanation:

give data

thermal energy Q  = 155 J

temperature T = 340 K

to find out

change in entropy

solution

we know change in entropy formula that is

change in entropy = Q / ( K×T )   ..............1

here K is boltzmann constant that is 1.38 × 10^{-23} kg-m²/s²

put these value in equation 1 we get

change in entropy = Q / ( K × T )

change in entropy = 155 / (  1.38 × 10^{-23}  × 340 )

change in entropy = 3.3034 × 10^{22}  

so change in entropy is 3.3034 × 10^{22}  

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An atom undergoes nuclear decay, but its atomic number is not changed.
andreev551 [17]

Answer:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

  • Gamma rays have zero atomic and mass numbers.
  • When they cause decay, they cause no change to the mass and atomic numbers.
  • They simply produce gamma rays during such reactions and these rays are very energetic.
5 0
3 years ago
How do you tell a stars life expectancy based on an hr diagram?
e-lub [12.9K]
The X-axis of the H-R Diagram indicates the star's surface temperature in degrees Kelvin.  The Y-axis, on the other hand, indicates luminosity, or brightness.

Main sequence refers to a roughly diagonal, slightly S-curved line stretching between the upper-left and lower-right corners on which main sequence stars chart. They maintain a predictable relationship between luminosity and temperature: the brighter, the hotter. The upper-right quadrant of the H-R diagram is home to newly discovered red giants while the lower-left quadrant of the H-R Diagram belongs almost exclusively to white dwarfs.
8 0
3 years ago
Ball 1, with a mass of 100 g and traveling at 10 m/s, collides head-on with ball 2, which has a mass of 300 g and is initially a
qaws [65]
There is 4000 balls in the earth of the world why is that so hard for you
5 0
1 year ago
What distance is covered by an airplane traveling at a velocity of 660 miles per hour in 3.5 hours?
N76 [4]

As per the question, the velocity of the airplane [v] = 660 miles per hour.

The total time taken by airplane [t] = 3.5 hours.

We are asked to determine the total distance travelled by the airplane during that period.

The distance covered [ S] by a body is the product of velocity with the time.

Mathematically distance covered = velocity × total time

                                                      S = v × t

                                                        = 660 miles/hour ×3.5 hours

                                                        = 2310 miles.

Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.

4 0
2 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

5 0
2 years ago
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