Hmmm, no choices? No? Okay then... I believe it should be 9.0 J.
No. You have to lose a lot of momentum to slow down enough to survive the impact (obviously depending on the height of the building). In your scenario, you can only transfer that momentum to the chair, by pushing it downward with your legs.
Let's say you jump off a 10 metre tall building and have a mass of 75 kg. You will be travelling at about 14 m/s just before impact, with a momentum of 1050 kgm/s. You want to reduce that momentum to around 750 kgm/s (equivalent to falling from a height of 5 m, which is probably survivable and may leave you able to walk away), so you have to transfer 300 kgm/s of momentum to the chair just by pushing it with your legs. For a 10 kg chair that means accelerating it to 30 m/s (in addition to the 14 m/s velocity the chair and you are already falling at), which is rather difficult.
You'd probably be better off landing on the chair and hoping that the chair breaking absorbs enough of the impact.
Answer:
2.4 meters
Explanation:
Let
m1= 10 kg
m2= 6 kg
Distance of m2 from pivot = 4 m
Distance of m1 from pivot = x (suppose) =?
If system is not rotating then it should be in equilibrrium position. And in this case clock-wise and anti -clock wise torques should balance each other i-e
weight of m1 mass bolck × x = weight of m2 mass block × 4
m1×g× x = m2 × g × 4
==> 10×9.8×x= 6×9.8×4
==> x = 6×9.8×4 / 10×9.8
==> x = 24/10=2.4 m
Use v^2 = u^2 + 2 a s
Message me if you need help.
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km