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White raven [17]

3 years ago

10

Astronaut Rob leaves Earth in a spaceship at a speed of 0.960crelative to an observer onEarth. Rob's destination is a star syste

m 14.4 light-years away (one light-year is the distance lighttravels in one year). Relative to a frame of reference that is fixed with respect to Earth, how longdoes it take Rob to complete the trip?

1 answer:

Slav-nsk [51]3 years ago

5 0

To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to

$1Ly =9.4605284*10^{15}m \rightarrow 'Ly'$ means Light Year

Then

$14.4Ly = 1.36231609*10^{17} m$

If we have that

$v= \frac{x}{t} \rightarrow t = \frac{x}{t}$

Where,

v = Velocity

x = Displacement

t = Time

We have that

$t = \frac{1.36231609*10^{17}}{0.96c} \rightarrow c$ = Speed of light

$t = \frac{1.36231609*10^{17}}{0.96(3*10^8)}$

$t= 454105363 s (\frac{1hour}{3600s})$

$t= 126140 hours(\frac{1day}{24hours})$

$t= 5255.85 days(\frac{1 year}{365days})$

$t = 14.399 years$

Therefore will take 14.399 years

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How many stairs?
You can use this to find the work
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W=mgh
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If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is

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A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationary seal of m

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Answer:

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3 years ago

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

$P=\frac{E}{t}$

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

$E=1.15\cdot 10^7 J$

Here we want to find the time t' needed to convert an amount of chemical energy of

$E'=3.8\cdot 10^5 J$

So we can setup the following proportion:

$\frac{t}{E}=\frac{t'}{E'}$

And solving for t',

$t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min$

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2 years ago

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Answer: the answer is C

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3 years ago