Respuesta :
Answer:
The decelerating force is [tex]3\times 10^{- 11}\ N[/tex]
Solution:
As per the question:
Frontal Area, A = [tex]10\ m^{2}[/tex]
Speed of the spaceship, v = [tex]1\times 10^{6}\ m/s[/tex]
Mass density of dust, [tex]\rho_{d} = 3\times 10^{- 18}\ kg/m^{3}[/tex]
Now, to calculate the average decelerating force exerted by the particle:
[tex]Mass,\ m = \rho_{d}V[/tex] (1)
Volume, [tex]V = A\times v\times t[/tex]
Thus substituting the value of volume, V in eqn (1):
[tex]m = \rho_{d}(Avt)[/tex]
where
A = Area
v = velocity
t = time
[tex]m = \rho_{d}(A\times v\times t)[/tex] (2)
[tex]Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t[/tex]
From Newton's second law of motion:
[tex]F = \frac{dp}{dt}[/tex]
Thus differentiating w.r.t time 't':
[tex]F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}[/tex]
where
[tex]F_{avg}[/tex] = average decelerating force of the particle
Now, substituting suitable values in the above eqn:
[tex]F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N[/tex]
