Answer:
The decelerating force is ![3\times 10^{- 11}\ N](https://tex.z-dn.net/?f=3%5Ctimes%2010%5E%7B-%2011%7D%5C%20N)
Solution:
As per the question:
Frontal Area, A = ![10\ m^{2}](https://tex.z-dn.net/?f=10%5C%20m%5E%7B2%7D)
Speed of the spaceship, v = ![1\times 10^{6}\ m/s](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B6%7D%5C%20m%2Fs)
Mass density of dust, ![\rho_{d} = 3\times 10^{- 18}\ kg/m^{3}](https://tex.z-dn.net/?f=%5Crho_%7Bd%7D%20%3D%203%5Ctimes%2010%5E%7B-%2018%7D%5C%20kg%2Fm%5E%7B3%7D)
Now, to calculate the average decelerating force exerted by the particle:
(1)
Volume, ![V = A\times v\times t](https://tex.z-dn.net/?f=V%20%3D%20A%5Ctimes%20v%5Ctimes%20t)
Thus substituting the value of volume, V in eqn (1):
![m = \rho_{d}(Avt)](https://tex.z-dn.net/?f=m%20%3D%20%5Crho_%7Bd%7D%28Avt%29)
where
A = Area
v = velocity
t = time
(2)
![Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t](https://tex.z-dn.net/?f=Momentum%2C%5C%20p%20%3D%20%5Crho_%7Bd%7D%28Avt%29v%20%3D%20%5Crho_%7Bd%7DAv%5E%7B2%7Dt)
From Newton's second law of motion:
![F = \frac{dp}{dt}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bdp%7D%7Bdt%7D)
Thus differentiating w.r.t time 't':
![F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}](https://tex.z-dn.net/?f=F_%7Bavg%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%28%5Crho_%7Bd%7DAv%5E%7B2%7Dt%29%20%3D%20%5Crho_%7Bd%7DAv%5E%7B2%7D)
where
= average decelerating force of the particle
Now, substituting suitable values in the above eqn:
![F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N](https://tex.z-dn.net/?f=F_%7Bavg%7D%20%3D%203%5Ctimes%2010%5E%7B-%2018%7D%5Ctimes%2010%5Ctimes%201%5Ctimes%2010%5E%7B6%7D%20%3D%203%5Ctimes%2010%5E%7B-%2011%7D%5C%20N)
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
Measurements of the Earths magnetic field has been recorded since 1829