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
To solve this problem, use the ratio given by the total number of electrons or protons that exist as a function of the total charge, and inversely proportional to the value of the fundamental charge. The number of fundamental unit
that constitutes a charge of 40.0C can be calculated as

Here,
= Value of charge and it is the fundamental charge
Q = Total Charge
N = Total number of electron or protons
The number of fundamental units is calculated as follows


Therefore the number of fundamental charge units moved by lightning bolt is 
The complementary base pair is:
TTC, CTG, AGT, CTA.
Answer:
a car
A sled sliding across snow or ice.
a ball down a hill
mercury
Explanation: