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: Point A is the answer for the potential energy. Point D is the answer for the kinetic energy.
Explanation:
Answer: - 452.088joule
Explanation:
Given the following :
Mass of water = 12g
Change in temperature(Dt) = (11 - 20)°C = - 9°C
Specific heats capacity of water(c) = 4.186j/g°C
Q = mcDt
Where Q = quantity of heat
Q = 12g × 4.186j/g°C × - 9°C
Q = - 452.088joule
Answer:
When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.
Explanation:
Newton’s third law motion states that for every action there will be an equal and opposite reaction.
Thrust reversal is also known as reverse thrust. It acts opposite to the motion of the aircraft by providing the deceleration.
Commercial aircraft moves the ejected air in the forward direction means that the thrust will acts opposite to the motion of the aircraft that is backward direction due to thrust reversal. This thrust force might be used to decelerate the craft.
Uses of thrust reversal in practice:
When the ejected air is moving forward direction then the thrust force moving backward direction due to reversal thrust the speed of the craft slows down.
When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.
Answer:
P₂ = 375 kPa.
Explanation:
Given that,
Initial volume, V₁ = 3 m³
Initial pressure, P₁ = 150 kPa
Final volume, V₂ = 1.2 m³
We need to find the final pressure.
At constant temperature,
So, the new pressure is 375 kPa.
