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Feliz [49]
2 years ago
6

A spaceship with a constant speed of 0. 800c relative to earth travels to a star that is 4. 30 light-years from earth. how much

time for this trip would elapse on a clock on board the spaceship?
Physics
1 answer:
Strike441 [17]2 years ago
3 0

Main Answer:

Given  speed of spaceship v = 0.8c = 0.8 * 3 x 10^8 m/s

Speed of spaceship v = 2.4 x 10^8 m/s

Distance need to be travelled d = 4.3 light years

we know that 1 light year  = 9.461 x 10^15 m

Distance need to be travelled d = 4.3 x 9.461 x 10^15

d = 40.6823 x 10^15 m

Time taken for the trip would elapse on a clock on board the spaceship

t = distance/ velocity

t = 40.6823 x 10^15 / 2.4 x 10^8

t = 16.95 x 10^7 sec

t = 4.71 x 10^4 hours

Explanation:

What is light year?

Light year is defined as the distance travelled by the light in one year. In a year, light travels through 300000 km per sec in the interstellar space.

To know more about spaceship, please visit: brainly.com/question/14423386

#SPJ4

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At 4.00 l, an expandable vessel contains 0.864 mol of oxygen gas. how many liters of oxygen gas must be added at constant temper
svet-max [94.6K]

Givens
=====
V = 4.00 L
T = 273oK We're assuming the temperature does not change, just the pressure.
n = 0.864 moles
R = 8.314 joules / mole  * oK
P = ?????

Formula
======
PV = n*R*T
P = n*R*T/V
P = 0.864 * 8.314 * 273 / 4
P = 490 kpa


You have to add 1.6 – 0.864 = 0.736 moles of gas.


We have to assume that the temperature and pressure remain the same when we add the 0.736 moles of gas. We are now looking for the volume.


PV = n*R*T

<span> V = 0.736 * 8.314 * 273 / 490</span>

V = 3.41 L Remember this is at about 4 atmospheres so we have to convert to Standard Pressure.

Total Volume = 3.41 + 4.00 = 4.41


V1 * P1 = V2 * P2

P1 = 490 kPa

P2 = 101 kPa

V1 = 7.41 L

V2 = ????


<span> <span> 7.41* 490 = V2 * 101 V2 = 7.41 * 490 / 101 V2 = 35.94 L </span> </span>


<span>You had 4 L now you need 31.94 more.</span> 
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