Answer:
The earth's pull on the moon
Explanation:
Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth.
Answer:
The scientific questions here are:
<em>a) How will climate change affect forests</em>
<em>b) How did life on Earth begin</em>
<em>c) Why did dinosaurs go extinct</em>
Explanation:
Scientific question are logical quantifiable questions, whose answers can be measured. A good scientific question must have answers that can be tested by a carefully designed experiment or measurement. Some qualities like "prettiest" and "amazing" cannot be tested for nor are they measurable, and hence, they do not make a testable component of good scientific question.
<span> d = r*t is the basic distance equation
d = 6000 km
t with the tail wind = 6 hr
r with the tail wind = speed of the plane + wind speed = s + w
t with the head wind = 7.5 hr
r with the head wind = speed of the plane - wind speed = s-w
(s+w)*6 = 6000
(s-w)*7.5 = 6000
s + w = 1000
s - w = 800
</span><span> 2s = 1800
s = 900 km/h
s + w = 1000
w = 100
Check the anwer by calculating the return trip.
(900-100) * 7.5 = 800 * 7.5
800 * 7.5 = 6000 km
Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.</span>
Energy of the waves are redistributed to form a resultant wave with amplitude given by the summation of individual wave's amplitude.
<span>If the two waves are of same frequency, speed and amplitude and travelling in opposite direction den stationary waves are form.</span>
Answer:
Explanation:
I got everything but i. Don't know why but it's eluding me. So let's do everything but that.
a. PE = mgh so
PE = (2.5)(98)(14) and
PE = 340 J
b.
so
and
KE = 250 J
c. TE = KE + PE so
TE = 340 + 250 and
TE = 590 J
d. PE at 8.7 m:
PE = (2.5)(9.8)(8.7) and
PE = 210 J
e. The KE at the same height:
TE = KE + PE and
590 = KE + 210 so
KE = 380 J
f. The velocity at that height:
and
so
v = 17 m/s
g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:
590 = KE + PE and
PE = (2.5)(9.8)(11.6) so
PE = 280 then
590 = KE + 280 so
KE = 310 then
and
so
v = 16 m/s
h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:
and
26 = 0 + 9.8t and
26 = 9.8t so the time at 26 m/s is
t = 2.7 seconds. Now we use that in the equation for displacement:
Δx =
and filling in the time the object was at 26 m/s:
Δx = 0t +
so
Δx = 36 m
i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.