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levacccp [35]
3 years ago
6

The diagram is being used to illustrate the second law of thermodynamics, where Qh represents a hot object and Qc represents a c

old object.
A black box on the left labeled Q Subscript h Baseline and a touching white box on the right labeled Q Subscript c Baseline.

Which will best complete the diagram to illustrate the law?

adding in the boxes an arrow that points from Qh to Qc
adding in the boxes an arrow that points from Qc to Qh
adding in the boxes an arrow that points in both directions between Qh and Qc
adding in each box an arrow that points upward from Qh and from Qc

The correct answer is A. adding in the boxes an arrow that points from Qh to Qc
Physics
2 answers:
sdas [7]3 years ago
9 0

Answer:

Adding in the boxes an arrow that points from Q_{h} to Q_{c}.

Explanation:

According to the Second Law of Thermodynamics,<em> the heat transfer occur from a higher temperature body to a lower temperature body</em>.

So, applying the Second Law of Thermodynamics, the correct answer is the first one: Adding in the boxes an arrow that points from Q_{h} to Q_{c}; because this direction is pointing the right one, it doesn't violate the second law of the thermodynamics, the heat is gonna go from Q_{h} to Q_{c}, and the arrow indicates that.

Remember that the problem states that Q_{c} is a cold object and Q_{h} is a hot object.

katrin [286]3 years ago
6 0

Answer:

The answer is A. on edgen.

Explanation:

A. adding in the boxes an arrow that points from Qh to Qc

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A ball is thrown off the top of a building and lands on the ground below.
natita [175]

Answer:

Mass and velocity.

Explanation:

Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:

K=\frac{1}{2} mv^2

where m is the mass of the object, and v is its velocity at a given point in time.

So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:

  • mass of the ball
  • velocity of the ball before it hits the ground

With the knowledge of these two quantities the kinetic energy of the ball  before touching the gound can be determined.

4 0
3 years ago
Show solution for # 4
velikii [3]
M1 = 750Kg, v1 = 10m/s
m2 = 2500Kg , v2= 0 (because in problem say cuz that object don t move).

The momentum before colision is equal with the momentum after colision:

m1v1 + m2v2 = (m1+m2)v3 => v3 is the velocity after colison and that s u want to caluclate for your problem

=> m1v1 = (m1+m2)v3 => v3 = m1v1/(m1+m2) now u should do the math i think v3 prox 2,4 but not sure u should caculate
7 0
3 years ago
A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t
andreyandreev [35.5K]

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be (  R+ h )

orbital velocity  V₀ =  \sqrt{\frac{GM}{( R+h)} }

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x \sqrt{\frac{( R+h)}{GM } }

\frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } = R +h

h = \frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } - R.

7 0
3 years ago
Read 2 more answers
The cunninghams are moving across the country. Mr.cunningham leaves 2.5 hours before mrs.cunningham. If he averages 40 mph and s
Eva8 [605]

When she starts out, he is (40x2.5)= 100 miles ahead of her.

She gains (65-40)= 25 miles on him every hour.

It takes her (100/25)= 4 hours to catch up to him.

5 0
3 years ago
A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a
Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0
3 years ago
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