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
Answer:
x=?
dt=?
vi=23m/s
vf=0m/s (it stops)
d=0.25m/s^2
time =
vf=vi+d: 0=23m/s+(0.25m/s^2)t
t=92s
displacement=
vf^2=vi^2+2a(dx)
23^2=0^2+2(0.25m/s^2)x =-1058m
Explanation:
you can find time from vf = vi + a(Dt): 0 = 23 m/s + (0.25 m/s/s)t so t = 92 s and you can find the displacement from vf2 = vi2 + 2a(Dx) and find the answer in one step: 232 = 02 + 2(0.25 m/s/s)x so x = -1058 m
ANSWER: THE ANSWER IS SUMMER