Noise can be measured in decibels.
Answer:
(D) the spot could have been caused by an object that emitted a flash that lasted for only a fraction of the time that the camera shutter was open
Explanation:
(A) the spot was not the brightest object in the photograph: The effect described and the brightness of the objects have no relation. Stars of very different brightness will be shown.
(B) the photograph contains many streaks that astronomers can identify as caused by noncelestial objects: Yes, but that doesn't explain the effect described. A plane could leave a streak.
(C) stars in the night sky do not appear to shift position relative to each other: True, at least for relative short times, but that has nothing to do with the effect described, which happens in a very short period of time.
(E) if the camera shutter had not been open for an extended period, it would have recorded substantially fewer celestial objects: True, but quantity of objects does not relate with the particular case described.
(D) the spot could have been caused by an object that emitted a flash that lasted for only a fraction of the time that the camera shutter was open: True, this can happen, for example, with Iridium satellites, they emit a flash (reflect solar light) that lasts a very short time as seen from one point on the surface (the place where the camera is), and something like this could have been captured by the camera shutter, appearing like a point compared to the streaks left by the stars.
Hovering their foot over the gas petal.
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
A) f = 1.8 rev/s = 2 Hz
<span>T = 1 / f = 0.55s
B) not really sure..srry
C) </span><span>T = 2 pi √ ( L / g ) </span>
<span>0.57 = 2 x 3.14 x √ ( 0.2 / g )
</span><span>
g = 25.5 m/s²
</span>
Hope this helps a little at least.. :)
