Answer:
it is atoms and molecules
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:
a. slope=rise/run
rise=0.02
run=-2
determined using the point (3,0.08) and (1,0.1) on the graph
slope=0.02/-2
= -0.01 or -1/100
b.area= area of trapizoid+ rectangle
((0.07+0.11)÷2)×4+1×0.07
0.36+0.07
=0.43$
c. the area represent the total cost after 5 hours
PLEASE MARK BRAINLIEST
<h2>momentum(p) = mass(m)×acceleration (a)</h2>
<h3>p = 5.6 × 22</h3><h3>p = 123.2 kg m/s</h3>
Answer:
1.)1.265+or minus 0.0006m
2).0.71%
Explanation:
See attached file