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Ratling [72]
3 years ago
12

Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth. Sarah's destination is a star-system 12.5 li

ght-years away (one light-year is the distance light travels in one year). According to Sarah, how far did she travel during the trip?
Physics
1 answer:
Usimov [2.4K]3 years ago
8 0

Answer:

L= 12 light years

Explanation:

for length dilation we use the formula

L=L_0\sqrt{1-\frac{v^2}{c^2} }

now calculating Lo

Lo = 12.5×365×24×3600×3×10^8

= 1.183×10^17 m

now putting the values of v and Lo in the above equation we get

L=1.183\times10^{17}\sqrt{1-\frac{0.28c^2}{c^2} }

= 1.136×10^17 m

L=  = \frac{1.136\times10^{17}}{365\times24\times3600\times3\times10^8}m

so L= 12 light years

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The distance a toy car travels over time is shown in the graph.
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Answer:

a toy car speed is about 2.5 to 3.5 mph

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A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v
gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, d_{\frac{1}{2}}.

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

                                          d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, d_{\text{remain}} , in which the driver reduces the speed to 40km/hr is

                                             d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}.

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by  t_{\text{remian}}.

                                              t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}.

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

                                                     \text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}

Therefore, the average speed of the car is 50 km/hr.

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A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick
DerKrebs [107]
Dispersion angle = 0.3875 degrees. 
Width at bottom of block = 0.09297 cm 
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 n1*sin(θ1) = n2*sin(θ2)
 where
 n1, n2 = indexes of refraction for the different mediums
 Î¸1, θ2 = angle of incident rays as measured from the normal to the surface. 
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 n1*sin(θ1) = n2*sin(θ2)
 n1*sin(θ1)/n2 = sin(θ2)
 asin(n1*sin(θ1)/n2) = θ2 
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 Red:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.641) = θ2
 asin(1.00029*0.653420604/1.641) = θ2
 asin(0.398299876) = θ2
 23.47193844 = θ2 
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 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.667) = θ2
 asin(1.00029*0.653420604/1.667) = θ2
 asin(0.39208764) = θ2
 23.08446098 = θ2 
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 tan(θ) = X/11.6
 11.6*tan(θ) = X
 So for Red:
 11.6*tan(θ) = X
 11.6*tan(23.47193844) = X
 11.6*0.434230136 = X
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 And violet:
 11.6*tan(θ) = X
 11.6*tan(23.08446098) = X
 11.6*0.426215635 = X
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 0.092968218*cos(θ) = X 
 0.092968218*cos(40.80) = X
 0.092968218*0.756995056 = X
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