Answer:
Option C. 16.6 m/s
Explanation:
To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.
So.
16.558 m/s = 16.6m/s to 3sf
Answer:
350 ft/s²
Explanation:
First, convert mph to ft/s.
58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s
Given:
v₀ = 85.1 ft/s
v = 0 ft/s
t = 0.24 s
Find: a
v = at + v₀
a = (v − v₀) / t
a = (0 ft/s − 85.1 ft/s) / 0.24 s
a = -354 ft/s²
Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².
Answer:
b) Betelgeuse would be 
times brighter than Sirius
c) Since Betelgeuse brightness from Earth compared to the Sun is 
the statement saying that it would be like a second Sun is incorrect
Explanation:
The start brightness is related to it luminosity thought the following equation:
(1)
where 
is the brightness, 
is the star luminosity and 
, the distance from the star to the point where the brightness is calculated (measured). Thus:
b) 
and 
where 
is the Sun luminosity (
) but we don't need to know this value for solving the problem. 
is light years.
Finding the ratio between the two brightness we get:
c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 
. Then
Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.
