Finding out the acceleration 12/3 = 4m/s^2
thus it is descending so the actual acceleration would be 9.8-4 = 5.8 m/s^2
the weight will be 90*5.8 = 522 N
522/9.8 = 53.2 kg
O2 refers to two oxygen atoms bonded together, while 2 O refers to two oxygen atoms that are not bonded to each other. They both have two oxygen atoms, but in O2, the oxygen atoms are bonded, while in 2O, the atoms are not.
Answer:
The ratio is
Explanation:
From the question we are told that
The radius of Phobos orbit is R_2 = 9380 km
The radius of Deimos orbit is 
Generally from Kepler's third law

Here M is the mass of Mars which is constant
G is the gravitational constant
So we see that 
=> ![[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%5E2%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E3)
Here
is the period of Deimos
and
is the period of Phobos
So
![[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D)
=> ![\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7BT_1%7D%7BT_2%7D%20%20%3D%20%20%5B%5Cfrac%7B23500%20%7D%7B9380%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D)
=>
I have no idea what that is, but all of your answers right
Answer:
v = 16.11 m / s
Explanation:
For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation
starting point. When the spring is compressed
Em₀ = K_e + U = ½ k x² + m g x ’
final point. The point where it leaves the platform
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
½ k x² + m g x ’= ½ m v²
v² =
x² + g x
let's calculate
v² =
1.25² + 9.8 1.25
v² = 247.159 + 12.25 = 259.409
v = 16.11 m / s