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2 years ago

A 0.5 kg yellow bouncy ball rolls to the right at a speed of 8 m/s while a 0.25 kg blue bouncy ball rolls

Physics

1 answer:

inessss [21]2 years ago

3 0

Explanation:

Let yellow ball be m1 = 0.5kg with u1 = 8 m/s and blue ball be m2 = 0.25 kg with u2 = - 4 m /s respectively.

After collision, blue ball travels 12 m/s.

<u>Using conservation of Linear Momentum</u> :

m1u1 + m2u2 = m1v1 + m2v2

0.5* 8 + 0.25 * - 4 = 0.5 * v1 + 0.25 * 12

v1 = 0 m/sec i.e. <u>Yellow ball comes to rest</u>.

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2 years ago

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2 years ago

Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,

Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41 , RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω

Using law of Ohm

V = I * R

I = Vc / Rc = 364 V / 473 Ω

I = 0.76 A

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

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RL = 473 (+ / -) (129.58)

RL₁ = 34.41 , RL₂ = 602.58

The higher value have the less angular frequency

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

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3 years ago