A space transportation vehicle releases a 470-kg communications satellite while in a circular orbit 350 km above the surface of
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Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 × m
dynamic viscosity = 1.75 × Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ
so
= µ ............1
put here value
= 1.75× ×
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 × m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 ×
force = 1.374 × v
and now apply newton second law
force = mass × acceleration
- force =
- 1.374 × v =
t =
time = 2.18
so time required after impact for a puck is 2.18 seconds
Answer:
1.38 x 10^-18 J
Explanation:
q = - 1.6 x 10^-19 C
d = 5 x 10^-10 m
the potential energy of the system gives the value of work done
The formula for the potential energy is given by
So, the total potential energy of teh system is
As all the charges are same and the distance between the two charges is same so the total potential energy becomes
K = 9 x 10^9 Nm^2/C^2
By substituting the values
U = 1.38 x 10^-18 J