Answer:
7.89 7.91
Explanation:
The ranges of measurement lie between 7.92-0.05 and 7.92+0.05
7.87g and 7.97g
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
I’m not too sure but I think it’s 8,91 m/s2
C is correct answer.
Walking along a 6 meter beam without falling helps to develop balance.
Hope it helped.
-Charlie
Answer:
T=189.15 N
Explanation:
As we know that for downward motion
F acting = F (weight) - Tension T
m a = mg - T
⇒ T = m (g - a)
T = 29.1 kg ( 9.8 m/s² - 3.3 m/s²)
T=189.15 N