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Natalija [7]
3 years ago
12

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

rs, as measured by you? You can ignore the time it takes to get up to speed or stop. b. how long would your trip take as measured by someone who stayed on Earth?
Physics
1 answer:
viva [34]3 years ago
4 0

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}                           (1)

time taken = 5 years

We know that :

time = \frac{distance}{speed}

5 = \frac{L''}{v}                                                      (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

Solving the above eqn:

t'' = 20.56 years

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I'm very confused. Thanks for whoever helps me :)
sergij07 [2.7K]
(C). Remember gravity provides an acceleration of 9.81m/s^2, so the y component of velocity initial is zero because it isn’t already falling, and we have the height, so basically we use the kinematic equation vf^2=vi^2+2ad, substitute given values and you get vf^2=2(9.81)(65) which is 1275, when you take the square root you get 35.7m/s for final velocity
(B). Then you use vf=vi+at to get the equation 35.7=(9.81)t, when you divide out you get 3.64s for time t
(A). Finally, since we assume that there is no acceleration or deceleration horizonatally, we just multiply the time taken for it to hit the ground and the initial speed ((3.64)(35.7)) to get 129.96, with significant figures I would round that to 130 metres.
**this is in the order that I felt was easiest to answer**
6 0
3 years ago
A​ hot-air balloon is 150 ft above the ground when a motorcycle​ (traveling in a straight line on a horizontal​ road) passes dir
Katyanochek1 [597]

Answer:

 dR/dt = 10.2 ft / s

Explanation:

Let's work this problem by finding the distance between the balloon and the motorcycle and then drift for the speed change of the distance

Balloon

      y = y₀ +v_{oy} t

Motorcycle

      x = v₀ₓ t

Distance, let's use Pythagoras' theorem

      R² = x² + y²

      R² = (v₀ₓ t)² + (y₀ + v_{oy} t)²

     v₀ₓ = 88 ft / s

     v_{oy} = 8 ft / s

     y₀ = 150 ft

     R² = (8 t)² + (150 + 8 t )²

     R² = 64 t² + (150 + 8t )²

This is the expression for the distance between the two bodies, the rate of change is the derivative with respect to time (d / dt)

         2RdR / dt = 64 2 t + 2 (150 + 8t) 8

        dR / dt = [64 t + (1200 + 64t )] / R

dR/dt = (1200 +128 t)/R

Let's calculate for the time of 10 s

        dR / dt = (1200 + 128 10) / R = 2480 /R

       R = √ [64 10² + (150 + 8 10)²

       R = √ [6400 + 52900]

       R = 243.5 ft

       dR / dt = (2480) / 243.5

       dR / dt = 10.2 ft / s

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2 years ago
A wave oscillates 50 times per second. What is its frequency?
Advocard [28]

"per second" is exactly the definition of "Hertz".

50 per second is 50 Hz. (B)

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