Answer:
The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.
Solution:
As per the question:
Radius of an alveolus, R =
Gauge Pressure inside,
Blood Pressure outside,
Now,
Change in pressure,
Since the alveolus is considered to be a spherical shell
The surface tension can be calculated as:
And we know that the surface tension of water is 72.8 mN/m
Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.
Answer:
The minimum speed =
Explanation:
The minimum speed that the rocket must have for it to escape into space is called its escape velocity. If the speed is not attained, the gravitational pull of the planet would pull down the rocket back to its surface. Thus the launch would not be successful.
The minimum speed can be determined by;
Escape velocity =
where: G is the universal gravitational constant, M is the mass of the planet X, and R is its radius.
If the appropriate values of the variables are substituted into the expression, the value of the minimum speed required can be determined.
Answer:
v_{f} = 115.95 m / s
Explanation:
This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions
Thrust =
-v₀ = v_{e}
where v_{e} is the velocity of the gases relative to the rocket
let's apply these expressions to our case
the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units
M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg
The final mass is the mass of the engines + the mass of the rocket
M_{f} = 25.5 +54.5 = 80 g = 0.080 kg
thrust and duration of ignition are given
thrust = 5.26 N
t = 1.90 s
Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear
thrust = v_{e}
v_{e} = thrust
v_{e} = 5.26
v_{e} = - 786.93 m / s
the negative sign indicates that the direction of the gases is opposite to the direction of the rocket
now we look for the final speed of the rocket, which as part of rest its initial speed is zero
v_{f}-0 = v_{e}
we calculate
v_{f} = 786.93 ln (0.0927 / 0.080)
v_{f} = 115.95 m / s
Answer:
P = 40.7kPa
Explanation:
To find the pressure on a surface 6 meter below you use the following formula, which takes into account the heights in which pressures are measured and also the density of the fluid and the gravitational acceleration:
(1)
P2: pressure for a height of -6 m = ?
P1: pressure for a height of -2 m = 1.5kPa = 1500 Pa
ρ: density of water = 1000kg/m^3
g: gravitational acceleration = 9.8 ms^2
y2: -6m
y1: -2m
(the height is measure from the water level, because of that, the heights are negative)
You solve the equation (1) for P1:
(2)
Next, you replace the values of all variables in equation (2):
hence, the pressure on a surface 6 m below the water level is 40.7kPa