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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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Paul [167]
Average speed = (distance covered) / (time to cover the distance)

Tissa covered 60 meters in 10 seconds. Her average speed was

(60 m) / (10 sec) = 6 m/s.

That's the slope of the dotted line.

Lilly covered 60 meters in 8 seconds. Her average speed was

(60 m) / (8 sec) = 7.5 m/s .

That's the slope of the solid line.

Lilly covered the same distance in less time, and both girls
arrived at the finish line together. Technically, in science talk,
we would say that Lilly ran "faster", and her average speed
was "greater".

We can detect that by looking at the graph, because Lilly's line
has the characteristic of being "steeper", and we know that the
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3 0
3 years ago
A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
3 years ago
A small person was running way faster than a bigger person that weighed more collided in a football game. Who would be pushed ba
Semenov [28]

Answer:

smaller one

Explanation:

even though he is moving quicker doesn't mean he will be packing more force in the collision

5 0
3 years ago
A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?​
Ilya [14]

The frequency of the wave is 1\cdot 10^9 Hz

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

c=f \lambda

where

c is the speed of the wave

f is the frequency

\lambda is the wavelength

For the radio wave in this problem,

\lambda = 0.3 m

c=300,000,000 m/s = 3\cdot 10^8 m/s

Therefore, the frequency is:

f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz

Learn more about waves here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0
3 years ago
Please help on this one?
qwelly [4]

Answer:

D. An image that is smaller than the object and is behind the mirror

5 0
3 years ago
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