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forsale [732]
3 years ago
13

Is gravity keeps the planets in orbit around the sun a. True b. False

Physics
1 answer:
salantis [7]3 years ago
3 0
True. It's a bit more complex than simply that, but for the question you asked the answer is A.
Inertia also works in tandem with gravity, and plants orbit they don't revolve (kudos on that! Small pet peeve of mine). Newtons laws of physics will help you understand this more; remember to keep in mind we are talking about space so the suns mass has gravitational pull, but space has no gravity- So inertia plays an important part as well.
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During which lunar phase is the moon s far side entirely dark?
bearhunter [10]
The lunar phase in which the moon's far side (the side that does not get any sunlight) is entirely dark is during a full moon. Since the side that faces the Earth is fully illuminated, the other side would be the exact opposite.
8 0
4 years ago
1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it
spin [16.1K]

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. KE=\frac{1}{2}mv^2 so

   KE=\frac{1}{2}(2.5)(14)^2 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:

   380=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(380)}{2.5} } 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

   310=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(310)}{2.5} } 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:

   v=v_0+at 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 = v_0t+\frac{1}{2}at^2 and filling in the time the object was at 26 m/s:

   Δx = 0t + \frac{1}{2}(-9.8)2.7)^2 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.

4 0
3 years ago
Which elements are not noble gases?
Ivan
Helium lithium and calcium
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During the water cycle, solar energy causes water to change from a _____ to a ______.
tankabanditka [31]
Liquid to a gas hence the name water vapor 
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3 years ago
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A ball is dropped from rest from a tower and strikes the ground 125meters below. Approximately how many seconds does it take for
kogti [31]
Oh ok thanks i i miss me calf and she is going on the ground
4 0
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