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
1)
HCl: hydrogen, chloride
3CO2: carbon, oxygen
2Na2SO4:sodium, sulphur, oxygen.
2)
-HCl: 1 hydrogen atom, 1 chlorine atom
-CO2: 1 carbon atom, 2 oxygen atoms
-Na2SO4: 2 sodium atoms, 1 sulphur atom, 4 oxygen atoms.
3)
-HCl: 2 atoms
-3CO2: 9 atoms
-2Na2SO4: 14 atoms.
Answer:
B. 80 m/s²
Explanation:
F = ma
a = F/m = (40 N)/(0.5 kg) = 80 m/s²
Answer:
1.8 m/s
Explanation:
momentum = mass × velocity
initial momentum = m1v1+m2v2
= 3×3 +2×0 = 9+0= 9 kg m/s
let combined velocity be V
HENCE
final momentum = total mass × velocity
= (3+2) × V = 5V
According to law of conservation of momentum
final momentum = initial momentum
5V = 9
V =9/5
V = 1.8 m/s
<span>Gravitational force is affected by: a. mass c. distance b. weight d. both a and c
Mass.
</span>When an object is above the Earth's surface it hasgravitational potential<span> energy (GPE). The amount of GPE an object has depends on its mass and its height above the Earth's surface</span><span>
</span>
