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Amiraneli [1.4K]

3 years ago

7

Calculate the total resistance in a parallel circuit made up of resistances of 2Ω, 3Ω, and 4Ω. (Hint: When converting to a decim

al from a fraction, keep the same number of significant figures in the decimal as was present in the original measurement from which the fraction was derived.)

1 answer:

natta225 [31]3 years ago

6 0

Answer:

Total resistance = 0.92Ω

Explanation:

For parallel connected resistors we have effective resistance

$\frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+.....$

Here parallel circuit made up of resistances of 2Ω, 3Ω, and 4Ω.

That is

R₁ = 2Ω

R₂ = 3Ω

R₃ = 4Ω

Substituting

$\frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\\\frac{1}{R_{eff}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\\\\frac{1}{R_{eff}}=\frac{6+4+3}{12}\\\\R_{eff}=\frac{12}{13}=0.92ohm$

Total resistance = 0.92Ω

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$K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\$

$v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\$

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$\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i$

Thus, the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

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2 years ago