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
Resistance is current x potential difference. So therefor run wafff
Answer:
Whats the question/word problem or where is the graph (if included) representing this problem?
It's a subsection of Geology and Biology
Answer:
The answer is 49m/s.
Explanation:
Given,
acceleration (a) =9.8m/s^2
initial velocity (u)=0 (as it was in rest condition in helicopter)
time (t) =5 s
now, final velocity (v) after 5s =?
we have,


or, v = 49m/s
Therefore, the velocity after 5s is 49m/s.
<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>