Answer:
2) c) give-way vessel
3) a) With one short blast
Explanation:
2) A vessel that is required to take early substantial action to ensure avoiding collision called Give way vessel
In overtaking, the vessel intending to overtake is the Give-Way Vessel the vessel that is going to be overtaken is the Stand-On Vessel
Therefore, the correct option is c) give-way vessel
3) When vessels use sound signals in a meeting head on situation both vessel are Give-Way vessels and both vessel pass the each other by turning to the starboard side therefore they intend to pass each other on their port side requiring one short blast
Therefore, the correct option is a) With one short blast.
solution:
radius of steel ball(r)=5cm=0.05m
density of ball =8000kgm
terminal velocity(v)=25m/s^2
density of air( d) =1.29 kgm
now
volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3
density of ball= mass of ball/Volume of ball
or, 8000=m/0.00052
or, m=4.16 kg
weight of the ball (W)= mg=4.16×10=41.6 N
viscous force(F)=6 × pi × eta × r × v
=6×3.14×eta×0.05×25
=23.55×eta
To attain the terminal velocity,
Fiscous force=Weight
or, 23.55× eta = 41.6
or, eta = 1.76
whete eta is the coefficient of viscosity.
BEING HONEST , OPTIMISTIC,DISCIPLINED, PUNCTUAL ,TRUTHFUL AND GOOD ORATOR(SPEAKER) , AND HOPEFUL ARE OTHER BASIC QUALITIES AN EFFECTIVE LEADER MUST POSSESS. ALSO,HE MUST NEVER DIFFERENCE BETWEEN ANY OF HIS FOLLOWERS OR DO PARTIALLY AT ANY TIME . HE SHOULD TREAT EVERY ONE EQUALLY.
Explanation:
Answer:
0.423m
Explanation:
Conversion to metric unit
d = 4.8 cm = 0.048m
Let water density be 
Let gravitational acceleration g = 9.8 m/s2
Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.
Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:
Where 
N is the spring force, 
is the buoyancy force, which equals to the weight 
of the water displaced by the submerged portion of the cylinder, which is the product of water density 
, submerged volume 
and gravitational constant g. W = mg is the weight of the metal cylinder.
The submerged volume would be the product of cross-section area and the submerged length x
Plug that into our force equation and we have
Answer:
P₁ = 2.3506 10⁵ Pa
Explanation:
For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
A₁ v₁ = A₂ v₂
Let's look for the areas
r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm
r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm
A₁ = π r₁²
A₁ = π 1.125²
A₁ = 3,976 cm²
A₂ = π r₂²
A₂ = π 0.1²
A₂ = 0.0452 cm²
Now with the continuity equation we can look for the speed of water inside the hose
v₁ = v₂ A₂ / A₁
v₁ = 11.2 0.0452 / 3.976
v₁ = 0.1273 m / s
Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)
P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂
P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂
Let's calculate
P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25
P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴
P₁ = 2.3506 10⁵ Pa
