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:
4.4 seconds
Explanation:
Given:
a = -5.5 m/s²
v₀ = 0 m/s
y₀ = 53 m
y = 0 m
Find: t
y = y₀ + v₀ t + ½ at²
0 = 53 + 0 + ½ (-5.5) t²
0 = 53 − 2.75 t²
t = 4.39
Rounded to two significant figures, it takes 4.4 seconds for the object to land.
We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
Answer:
while using brainly app. there is an option of uploading image of diagram, graph or plot. plz use that feature to upload image of the standing wave so that your question can be properly answeres
