it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram
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:
C. 30.6m
Explanation:
To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:
Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.
Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)
Answer: You do not specify what is being asked for. ∆E? ∆H?
∆E = (430 - 238) J = 192 J
∆H = 430 J
Explanation:
If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.
Therefore ∆H = 430 J
If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w
The question states that 238 J of work are done AND the system expanded
(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)
Therefore, ∆E = (430 - 238) J = 192 J
