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valentinak56 [21]
4 years ago
9

Express the following quantities to the nearest standard prefix using no more than three digits.(a) 20,000,000 Hz(b) 1025 W(c) 0

:333 × 10−8 s(d) 33 × 10−12 F
Engineering
1 answer:
bija089 [108]4 years ago
8 0

Answer:

(a) 20 MHz

(b) 1.025 KW

(c) 3.33 ns

(d) 33 pF

Explanation:

(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>

(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>

(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>

(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>

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The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. Ifh=3ft, determin
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Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

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FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

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A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
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Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

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Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

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                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

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- Design Torque T for Steel rod:

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                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

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