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3 years ago

*Please Help!

1 answer:

8 0

We calculate by using the equation as follows:

Free Energy = DeltaG

The formula is DeltaG = Enthalpy - (Temperature)*(Entropy) 

DeltaG = 1.25 * 10^5 J - (290)(300) 

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

This means that this system is not spontaneous at 290 K

Hope this answers the question. Have a nice day.

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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 t

frutty [35]

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 !!!

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8 0

3 years ago

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Oksana_A [137]

Answer:

av=0.333m/s, U=3.3466J

$v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_{B2}-v_{A2}$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}$

$2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s$

7 0

4 years ago

A 10 Ω resistor is connected to a 120-V ac power supply. What is the peak current through the resistor?

11111nata11111 [884]

Answer:

Peak current = 16.9 A

Explanation:

Given that

RMS voltage = 120 Volts

$V_{rms} = 120 V$

AC is connected across resistance

$R = 10 ohm$

now by ohm's law

$V = i R$

$120 = i (10)$

$i_{rms} = \frac{120}{10} = 12 A$

now peak value of current will be given as

$i_{peak} = \sqrt{2} i_{rms}$

$i_{peak} = \sqrt2 (12) = 16.9 A$

8 0

3 years ago

Which object(s) formed last in our solar system?

alexira [117]

Answer:

first option bro

Explanation:

by the way if you play free fire give me your uid on the comment section

5 0

4 years ago

On being introduced to the laws of thermodynamics, a student retorted, “When the brakes of a moving car are applied, the kinetic

inn [45]

Answer:

No, not at all. This kinetic energy of car is converted/lost in the form of sound and heat under the tyre when driver apply brakes.

Explanation:

This law states that energy can neither be created nor destroyed but it changes from one form to the other.

In this case reduction of kinetic energy of car does not contradict the law but it changes into heat and sound energy. When driver apply brakes we hear the sound of Tyre due to friction with the road. The tyre become hot too.

So it is not in contradiction to the law of conservation of energy (First Law of Thermodynamics).

Please mark branliest if you are satisfied with the answer. Thanking you in anticipation.

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3 years ago