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vaieri [72.5K]
2 years ago
12

Help me with the following problem

Physics
1 answer:
timofeeve [1]2 years ago
5 0

The rope will remain taut until the particle makes 79⁰ angle.

<h3>Change in kinetic energy of the particle</h3>

The change in kinetic energy of the particle is calculated as follows;

ΔK.E = K.Ei - K.Ef

Before the particle will achieve the given angular displacement, it will touch two new corners. Total kinetic energy lost = 30%

ΔK.E = 100%K.E - 30%K.E = 70%K.E = 0.7K.E

  • let the vertical displacement of the particle = h
  • horizontal length = side of the prism = a
  • hypotenuse side  = length of the pendulum = L
<h3>Apply principle of conservation of energy</h3>

K.E = P.E

0.7K.E = mgh

0.7(¹/₂mv²) = mg(Lsinθ)

0.7(v²) = 2g(Lsinθ)

from third kinematic equation;

v² = u² + 2gh

v² = 0 + 2gh

v² = 2g(a tanθ)

0.7(2g(a tanθ)) = 2g(Lsinθ)

0.7(a tanθ) = Lsinθ

0.7a/L = sinθ/tanθ

0.7a/L = cosθ

(0.7 x 0.8)/(3) = cosθ

0.1867 = cosθ

θ = cos⁻¹(0.1867)

θ = 79⁰

Thus, the rope will remain taut until the particle makes 79⁰ angle.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

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A)

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v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

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velocity of oxygen = √(3KT)/(16 m)

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B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

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v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

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