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maria [59]
2 years ago
10

Two carts are free to slide along the frictionless track shown in figure below. Cart A of mass m1 = 8 kg is released from 12m. A

cart B of mass m2 = 4 kg, initially at rest. The two carts combine together and move as one object. Calculate the height reached
by the two carts after collision.
Physics
1 answer:
Harrizon [31]2 years ago
3 0

The height reached by the two carts after collision is determined as 5.34 m.

<h3>Initial velocity of Cart A</h3>

Apply the principle of conservation of mechanical energy.

K.E = P.E

v = √2gh

v = √(2 x 9.8 x 12)

v = 15.34 m/s

<h3>Final velocity of the two carts after the collision</h3>

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

8(15.34) + 4(0) = v(8 + 4)

122.72 = 12v

v = 10.23 m/s

<h3>Height reached by both carts</h3>

Apply the principle of conservation of mechanical energy.

P.E = K.E

mgh = ¹/₂mv²

h = v²/(2g)

h = (10.23²) / (2 x 9.8)

h = 5.34 m

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

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Given:-

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We know,

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Putting the values,

s = (40 m/s)(6 s)

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6 0
3 years ago
The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
elixir [45]

Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Hope this Helps!!!

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KE=TE-PE=0.918 J - 0.054 J =0.864 J
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Answer:

C. f = 440 Hz

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        where v = speed of the wave

        λ = wavelength

        f = frequency

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