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umka2103 [35]
4 years ago
5

A large number of very industrious people make a very long pole. It is 10.0 light years long! ( As they measure it. ) Soon a spa

ceship flies along the length of the pole at 90% the speed of light. How much time passes on the spaceship from the moment the ship passes the first end of the pole to the moment the ship passes the second end of the pole
Physics
1 answer:
Pie4 years ago
5 0

Answer:

L = L0 ( 1 - v^2/c^2))1/2     where L0 is the proper length

L = 10 L-y (1 - .9^2)^1/2 = 4.36 L-y   length of pole measured by ship

t = 4.36 L-y / .9 c = 4.84 y  since the ship travels at .9 c

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The answer is a 11.25m/s

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Dynamo converts the
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The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of
kolbaska11 [484]

Answer:

a) A = 4.0 m , b)   w = 3.0 rad / s , c)  f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

          x = A cos (wt + fi)

In the exercise we are told that the expression is

          x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

         A = 4.0 m

b) the frequency or angular velocity

         w = 3.0 rad / s

c) angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 3 / 2π

           f = 0.477 Hz

d) the period

frequency and period are related

           T = 1 / f

           T = 1 / 0.477

           T = 20.94 s

e) the phase constant

          Ф = 0.10 rad

f) velocity is defined by

          v = dx / dt

         

         v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

         v = A w

          v = 4 3

          v = 12 m / s

g) the angular velocity is

          w² = k / m

          k = m w²

          k = 1.2 3²

          k = 10.8 N / m

h) the total energy of the oscillator is

          Em = ½ k A²

           Em = ½ 10.8 4²

          Em = 43.2 J

i) the potential energy is

           Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

               x = 3.98 m

j) kinetic energy

           K = ½ m v²

for t = 00.1 ²

    v = A w sin 0.10

    v = 4 3 sin 0.10

    v = 1.98 m / s

3 0
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V125BC [204]

Answer:

A longitudinal wave

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