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LekaFEV [45]
3 years ago
9

Which of the following systems reflects a decrease in entropy? A. burning match B. freezing water C. evaporating alcohol D. brea

king glass **Thank you!!
Physics
2 answers:
Bad White [126]3 years ago
7 0
It would be B)  freezing water because this change gets less chaotic
torisob [31]3 years ago
4 0
The answer is B. because introducing heat to something (A and C) increases entropy, and so does breaking something (D)
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A wooden rod of negligible mass and length 84.0 cm is pivoted about a horizontal axis through its center. A white rat with mass
Mandarinka [93]

Answer:

v = 2.029 m/s

Explanation:

Given

L = 84.0 cm   ⇒  R = 0.5*L = 0.5*84 cm = 42 cm = 0.42 m

m₁ = 0.600 kg

m₂ = 0.200 kg

g = 9.8 m/s²

u₁ = u₂ = 0 m/s

v₁ = ?

v₂ = ?

Due to gravity, the bar oscillates and becomes vertical. The mass that occupies the lower position is the one with the highest torque. The one that reduces the potential energy (the system tends to the position of minimum energy). This is achieved if the mass that goes down is 0.6kg (that goes down 42cm) and the one that goes up is 0.2kg (goes up 42cm).

In this system mechanical energy is conserved, so we can match its value in the horizontal position with the one in the vertical.

then

Ei = Ki + Ui = 0.5*(m₁+m₂)*(0)² + (m₁+m₂)*9.8*(0) = 0 J

Ef = Kf + Uf

⇒ Kf = 0.5*(m₁+m₂)*v² = 0.5*(0.6+0.2)*v² = 0.4*v²

⇒ Uf = m₁*g*h₁ + m₂*g*h₂ = 0.6*9.8*(-0.42) + 0.2*9.8*0.42 = - 1.6464

⇒ Ef = Kf + Uf = 0.4*v² - 1.6464

Since

0 = 0.4*v² - 1.6464  ⇒  v = 2.029 m/s

v is the same value due to the wooden rod is pivoted about a horizontal axis through its center and the masses are on opposite ends.

v₁ = v₂ = v    ⇒  ω₁*R₁ = ω₂*R₂  ⇒ ω₁*R = ω₂*R  ⇒ ω₁ = ω₂ = ω

⇒ v = ω*R

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3 years ago
How do the forces in nuclear and chemical reactions differ?
kvasek [131]

B. Chemical reactions must overcome the strong nuclear force

8 0
3 years ago
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The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

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