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>
83 km/h * 2.5 hours (3:30 - 1:00) = 207.5 km
207.5 km - 15 km = 192.5 km
Definitely the last one, brain stem
Answer:
Δx = 39.1 m
Explanation:
- Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:
where vf is the final velocity (0 in our case), v₀ is the initial velocity
(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance
traveled since the brakes are applied.
- Solving (1) for Δx, we have:
