True. These are the characteristics of a social drinker.
Answer:
the <em>ratio F1/F2 = 1/2</em>
the <em>ratio a1/a2 = 1</em>
Explanation:
The force that both satellites experience is:
F1 = G M_e m1 / r² and
F2 = G M_e m2 / r²
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- r is the orbital radius
- M_e is the mass of Earth
Therefore,
F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]
F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]
F1/F2 = m1/m2
F1/F2 = 1000/2000
<em>F1/F2 = 1/2</em>
The other force that the two satellites experience is the centripetal force. Therefore,
F1c = m1 v² / r and
F2c = m2 v² / r
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- v is the orbital velocity
- r is the orbital velocity
Thus,
a1 = v² / r ⇒ v² = r a1 and
a2 = v² / r ⇒ v² = r a2
Therefore,
F1c = m1 a1 r / r = m1 a1
F2c = m2 a2 r / r = m2 a2
In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,
F1 = F1c
G M_e m1 / r² = m1 a1
a1 = G M_e / r²
also
a2 = G M_e / r²
Thus,
a1/a2 = [G M_e / r²] / [G M_e / r²]
<em>a1/a2 = 1</em>
Answer:
No
Explanation:
Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):
A = 12 - 12t
B = 9t
Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

For the ships to sight each other, distance must be 5 or smaller







Since
then

So our equation has no solution, the answer is no, the 2 ships never sight each other.
Answer:
Explanation:
An object is thrown up from a height of 520m
h_o = 520m
Initial velocity of thrown is 18m/s
u = 18m/s
What is the position of the object after 3seconds
t = 3s
Acceleration due to gravity
g = 9.81 m/s²
Let calculated the height the object will reached when thrown from the top.
Using equation of motion
h = ut + ½gt²
Since the body is thrown upward, it is acting against gravity then, gravity will be negative
Then,
h = ut — ½gt²
h = 18 × 3 — ½ × 9.81 × 3²
h = 54 — 44.145
h = 9.855 m
So, the body is above the top of the building at a distance of 9.855m
So, the total distance from the bottom is
Position = h + h_o
x = 9.855 + 520
x = 529.855m
x ≈ 530m,
The position of objects after 3seconds is 520m