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3 years ago

Please help me, Engineering.

Engineering

1 answer:

antiseptic1488 [7]3 years ago

8 0

Answer:

15. R = R₁ + R₂ + R₃

16. $R = \dfrac{1}{R_1} +\dfrac{1}{R_2} = \dfrac{R_2 + R_1}{R_1 \cdot R_2}$

Explanation:

15. The equation for the sum of resistances arranged in series is given as follows;

$R_{Network}$ (Series) = R₁ + R₂ + R₃ + ... + $R_N$

Therefore, the equation for the total resistance, 'R' of 3 series resistors is given as follows;

R = R₁ + R₂ + R₃

Where;

R₁, R₂, and R₃ are the three resistances arranged in series

16. The equation for the sum of resistances arranged in parallel is given as follows;

$R_n = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + ... + \dfrac{1}{R_n}$

Therefore, the total resistance, R, of two resistance arranged in parallel is given as follows;

$R = \dfrac{1}{R_1} +\dfrac{1}{R_2} = \dfrac{R_2 + R_1}{R_1 \cdot R_2}$

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Consider the brass alloy for which the stress-strain behavior is shown in the Animated Figure 7.12. A cylindrical specimen of th

sleet_krkn [62]

Answer:

(a) 0.1509 mm

(b) 0.00525 mm

Explanation:

Stress, $\sigma$ is given by

$\sigma=\frac {F}{A}$ where F is force and A is area and area is given by $\frac {\pi d^{2}}{4}$ hence

$\sigma=\frac {4F}{\pi d^{2}}$ where d is the diameter. Substituting 9970 N for F and 10mm=0.01 m for d hence

$\sigma=\frac {4*9970 N}{\pi 0.01m^{2}}=126941982.6 N/m^{2}$

$\sigma \approx 127 Mpa$

From the attached stress-strain diagram, the stress of 127 Mpa corresponds to strain of 0.0015 and since strain is given by

$\epsilon=\frac {\triangle l}{l}$ where $\epsilon$ is the strain, $\triangle l$ is elongation and l is original length and making elongation the subject