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Pepsi [2]
3 years ago
6

Mainly need help with part A. No links, thanks!​

Physics
1 answer:
Vlad1618 [11]3 years ago
4 0

Answer:

R = 4Ω

Explanation:

If we have two resistors with resistances R1 and R2 in series the total resistance is R = R1 + R2

If the resistances are in parallel, the total resistance is given by:

1/R = 1/R1 + 1/R2.

First, we have a resistor with R1 = 1.5Ω

This resistor is connected in series with a parallel part (let's find the resistance of this parallel part), in one branch we have two resistors in series with resistances:

R2 = 8Ω and R3 = 4Ω

Because these are in series, the resistance of that branch is:

R = 8Ω + 4Ω = 12Ω

In the other branch, we have a single resistor of R4 = 4Ω

The resistance of the parallel part is:

1/R = 1/12Ω + 1/4Ω = 1/12Ω + 3/12Ω = 4/12Ω = 1/3Ω

1/R = 1/3Ω

R = 3Ω

Then we have a resistor (the first one, R1 = 1.5Ω) in series with a resistor of 3Ω.

Then the total resistance is:

R = 1Ω + 3Ω = 4Ω

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A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
andriy [413]

Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

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