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prohojiy [21]
3 years ago
5

Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro

py of the hot object
A.Will always have a greater magnitude than the change in entropy of the cold object

B.Could be either positive or negative

C.Will always have a smaller magnitude than the change in entropy of the cold object

D.None of the other answers is correct
Physics
1 answer:
aniked [119]3 years ago
5 0

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

S = \frac{Q}{T}

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy \Delta S_{hot} is smaller than \Delta S_{cold}

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

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posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g
            y- 0 = 10.0²/2 9.8
            y - 0 = 5.10 m
            
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
             y₂ = 5.1 + 44
             y₂ = 49.1 m
Let's use the other equation to find the time
              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
What force must be used to do 224 Joules of work on an object over a distance of 32 meters?
cestrela7 [59]
7.625 Newtons

work = force× distance
Newtons is an accepted value for force

so take the total 224 joules and decide by distance 32 meters to find force in Newtons
4 0
3 years ago
What is cocanve mirror?​
pochemuha

Answer:

A mirror that has a reflecting surface that is recessed inward is called concave mirror

8 0
3 years ago
Read 2 more answers
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the mete
disa [49]

Answer:

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.

(a) Determine the position of wire 3.

b) Determine the magnitude and direction of current in wire 3

Explanation:

a) F_{net} \text {on wire }3=0

\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm

position of wire = 50 - 1.2

= 48.8cm

b)  F_{net} \text {on wire }1=0

\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A

Direction ⇒ downward

5 0
3 years ago
Gravitational Force: Two small objects, with masses m and M, are originally a distance r apart, and the magnitude of the gravita
Alona [7]

Law of universal gravitation:

F = GMm/r²

F = gravitational force, G = gravitational constant, M & m = masses of the objects, r = distance between the objects

F is proportional to both M and m:

F ∝ M, F ∝ m

F is proportional to the inverse square of r:

F ∝ 1/r²

Calculate the scaling factor of F due to the change in M:

k₁ = 2M/M = 2

Calculate the scaling factor of F due to the change in m:

k₂ = 2m/m = 2

Calculate the scaling factor of F due to the change in r:

k₃ = 1/(4r/r)² = 1/16

Multiply the original force F by the scaling factors to obtain the new force:

Fk₁k₂k₃

= F(2)(2)(1/16)

= F/4

4 0
3 years ago
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