What type of OS is very fast and small, and is often embedded, or built into the circuitry of a device?
1 answer:
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1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):
Rearranging the equation, we can write
In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is
2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:
Answer:
the radii of curvature is 30 cm.
Explanation:
given,
object is place at = 45 cm
image appears at = 90 cm
focal length = ?
refractive index = 1.5
radii of curvature = ?
f = 30 cm
using lens formula
R = 30 cm
hence, the radii of curvature is 30 cm.
Answer:
w = 3.2 rev / min
Explanation:
For this exercise we will use the centrine acceleration equal to the acceleration of gravity
a = v² / r
Angular and linear variables are related.
v = w r
Let's replace
a = w² r = g
w = √ g / r
r = d / 2
r = 175/2 = 87.5 m
w = √( 9.8 / 87.5)
w = 0.3347 rad / s
Let's reduce to rotations per min
w = 0.3347 rad / s (1 rov / 2pi rad) (60 s / 1 min)
w = 3.2 rev / min
Suppose the space station rotates counterclockwise, we have two possibilities for the car
The first car turns counterclockwise (same direction of the station
=
r
[texwv_{c}[/tex] = / r
[texwv_{c}[/tex] = 25.0 / 87.5
[texwv_{c}[/tex] = 0.286 rad / s
When the two rotate in the same direction their angular speeds are subtracted
w total = w -[texwv_{c}[/tex]
w total = 0.3347 - 0.286
w total= 0.487 rad / s
The car goes in the opposite direction of the station the speeds add up
w = 0.3347 + 0.286
w = 0.62 rad / s
From this values we can see that the person feels a variation of the acceleration of gravity, feels that he has less weight when he goes in the same direction of the season and that his weight increases when he goes in the opposite direction to the season.