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Zinaida [17]
1 year ago
10

Two gliders on an air track collide in a perfectly elastic collision. Glider A has a mass of 1.1 kg and is initially travelling

at a velocity of 2.7 m/s [E]. It collides head-on with glider B with a mass of 2.4 kg, travelling at a velocity of 1.9 m/s [W]. Determine the final velocity of glider A.

Physics
1 answer:
Eva8 [605]1 year ago
6 0

m1= mass 1 = 1.1 kg

Vi1 = initial velocity 1 = 2.7 m/s

m2= 2.4 kg

V2i = -1.9 m/s

We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2

Replacing:

Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9Vf1=-1-2.6=-3.6\text{ m/s}

Answer: 3.6 m/s west

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Part b)

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   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

  The current density in both wires is  J  =  1750 \  A/m^2

Considering the first wire

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The  cross-sectional area of the second wire is

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Generally the resistance of the first wire is mathematically represented as

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=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

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