3 years ago

a coiled spring is stretched 31.50 cm by a 2.00N weight. How far is it stretched by a 10.00 N weight?

Engineering

1 answer:

Maksim231197 [3]3 years ago

5 0

K = 2/0.315 =6.35 N/m

x = 10/6.35 = 1.57m

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3 years ago

PLEASE HELP QUICK!!

ivolga24 [154]

R01= 14.1 Ω

R02= 0.03525Ω

<h3>Calculations and Parameters</h3>

Given:

K= E2/E1 = 120/2400

= 0.5

R1= 0.1 Ω, X1= 0.22Ω

R2= 0.035Ω, X2= 0.012Ω

The equivalence resistance as referred to both primary and secondary,

R01= R1 + R2

= R1 + R2/K2

= 0.1 + (0.035/9(0.05)^2)

= 14.1 Ω

R02= R2 + R1

=R2 + K^2.R1

= 0.035 + (0.05)^2 * 0.1

= 0.03525Ω