Actually, they're not. There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around. And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night.
Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.
Why does the night sky change at various times of the year ? Here's how to
think about it:
The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.
Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.
In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ? Now they're straight in the
direction of the sun. So they're directly overhead at Noon, not at Midnight.
THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
Answer:
There will be 1800 W power consumption in heater
Explanation:
We have given current flowing in the heater I = 15 A
Voltage on which heater is operating V = 120 volt
We have to find the power consumption in the heater
We know that power consumption is given by P = VI
So power consumption in heater = 120 × 15 = 1800 W
So there will be 1800 W power consumption in heater
Answer:
See answers below
Explanation:
a.
F = mg,
15.5 N = m(9.8 m/s²)
m = 1.58 kg
b.
Fnet = Applied force - resistance,
Fnet = 18 N - 4.30 N,
Fnet = 13.70 N
Fnet = ma
13.70 N = (1.58 kg)a
a = 8.67 m/s²
For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.
A) 1 rev = 2π rad. Using this ratio, you can find the rad/s: 1160 rev/min x 2π rad/rev x 1 min/60 s = 121.5 rad/s
b) You can find linear speed from angular speed using this equation (note the radius is half the diameter given in the question): v = ωr = 121.5 rad/s x 1.175 m = 142.8 m/s
c) You can find centripetal acceleration using this equation: a = v^2/r = (142.8 m/s)^2 / 1.175 m = 17 355 m/s^2
