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Neko [114]
2 years ago
12

Suppose that a car weighing 4000 pounds is supported by four shock absorbers Each shock absorber has a spring constant of 6500 l

bs/foot, so the effective spring constant for the system of 4 shock absorbers is 26000 lbs/foot. Assume no damping and determine the period of oscillation of the vertical motion of the car.
Required:
a. Assume no damping and determine the period of oscillation of the vertical motion of the car.
b. After 10 seconds the car body is 1/3 foot above its equilibrium position and at the high point in its cycle. What were the initial conditions?
Physics
1 answer:
scoundrel [369]2 years ago
7 0

Answer:

0.43622 seconds

0.9158 foot

-4.43 ffot/sec

Explanation:

we first find the period of oscillation

= 2π√w/gk

= 2π√4000/32x2600

= 2π√0.00481

= 2π0.0694

= 0.43622

b. we find the angular velocity

2π/T

= 2π/0.43622

= 14.41 rad/sec

we find displacement

rom the calculation in the attachment

Ф = -144.1

initial condition

1*cos(-144.1 rad)

= 0.9158 foot

initial velocity of the car

= (-1)(14.41)sin(ω(0)-144.1)

= -4.44 foot/sec

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Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

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Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

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Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

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         N  = 8.46 * 10^{28}  \  electrons

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