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Schach [20]
3 years ago
7

*Please Help!

Physics
1 answer:
Ierofanga [76]3 years ago
8 0
We calculate by using the equation as follows:

Free Energy = DeltaG 

<span>The formula is DeltaG = Enthalpy - (Temperature)*(Entropy) </span>

<span>DeltaG = 1.25 * 10^5 J - (290)(300) </span>

<span>DeltaG or Free Energy = 38000 J/mol or 38kJ/mol. </span>

<span>This means that this system is not spontaneous at 290 K
</span>
Hope this answers the question. Have a nice day.
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Small cubes that are 10 cm on a side and larger ones that are 12 cm on a side are submerged in water. Cubes A and B are made of
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Answer

given,

given,

small cube side = 10 cm

larger cube side = 12 cm

density of steel = 7 g/cm³

density of aluminium = 2.7 g/cm³

density of the water (ρ₁)= 1 g/cm³

Cube A and B made of steel

buoyant force of Cube A

 B₁ = ρ₁ V g = 1 x 10 x 10 x 10 x g=  1000 g

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 B₂ = ρ₁ V g = 1 x 12 x 12 x 12 x g=  1728 g

buoyant force of Cube C

 B₃ = ρ₁ V g = 1 x 10 x 10 x 10 x g=  1000 g

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 B₄ = ρ₁ V g = 1 x 12 x 12 x 12 x g=  1728 g

buoyant force acting on the cube depends on the density of the fluid

hence,

B₂ = B₄  > B₁ = B₃

 

8 0
3 years ago
Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A
boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴  22.7 × v₁'  = -3.63 × 3.05

v₁' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law,  m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴  22.7 × v₂'  = -3.63 × 3.05

v₂' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = F_{average} × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}

∴ F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s}  = 922.625 \ kg\cdot m/s^2 = 922.625 \ N  

The average force on the right sled applied by the cat while landing, \mathbf{F_{average}} = 922.625 N

Learn more about conservation of momentum here:

brainly.com/question/7538238

brainly.com/question/20568685

brainly.com/question/22257327

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