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Zolol [24]
3 years ago
9

A planet is 15 light years from earth. At which of the following speeds will the crew of a spaceship complete the trip to the pl

anet in 8 years?
A) 0.78c

B) 0.85c

C) 0.88c

D) 0.92c
Physics
1 answer:
Svetradugi [14.3K]3 years ago
7 0

Answer:

option c) 0.88c is the correct answer

Explanation:

using the Lorrentz equation we have

t=\frac{d}{v}\sqrt{1-(\frac{v}{c})^2}

where,

t = time taken to cover the distance

d = Distance

v = velocity

c = speed of light

given

d = 15 light years

Now,

8years=\frac{15years\times c}{v}\sqrt{1-(\frac{v}{c})^2}

or

(\frac{v}{c})^2\times (\frac{8}{15})^2=1-(\frac{v}{c})^2

or

(\frac{v}{c})^2\times 0.284=1-(\frac{v}{c})^2

or

(\frac{v}{c})^2\times 0.284 + (\frac{v}{c})^2 =1

or

(1.284\times \frac{v}{c})^2 =1

or

(\frac{v}{c}) =\sqrt{\frac{1}{0.284}}

or

v=0.88c

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sdas [7]

Answer:

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Explanation:

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2 years ago
A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans
alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

                 y = (0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)

Now, we will compare above equation with the standard form of transeverse wave equation,

                 y = A sin(kx + \omega t)

where,    A is the amplitude = 0.09 m

              k is the wave vector = \frac{\pi}{11}

              \omega is the angular frequency = 4\pi

              x is displacement = 1.40 m

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Now, we will differentiate the equation with respect to t as follows.

The speed of the wave  will be:

                   v(t) = \frac{dy}{dt}

                v(t) = A \omega cos(kx + \omega t)

        v(t) = (0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)

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The acceleration of the particle in the location is

            a(t) = \frac{dv}{dt}

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           a(t) = -(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)

           a(t) = -9.49 m/s^{2}

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 m/s^{2} .

(b)  Wavelength of the wave is given as follows.

               \lambda = \frac{2\pi}{k}

              \lambda = (frac{2\pi}{\frac{\pi}{11})


              \lambda = 22 m

The period of the wave is

             T = \frac{2 \pi}{\omega}

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                = 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

                    v = \frac{\lambda}{T}

                       = \frac{22 m}{0.5 s}

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therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0
3 years ago
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Answer:

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Explanation:

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