Answer:
The angle for the forward Mach line is 19.47°
The angle for the rearward Mach line is 5.21°
Explanation:
From table A-1 (Modern Compressible Flow: with historical perspective):
(M₁ = 3)
If Po₁ = Po₂
Table A-1:
Table A-5:
v₁ = 49.76°
μ₁ = 19.47°
v₂ = 60.55°
μ₂ = 16°
θ = 60.55 - 49.76 = 10.79°
The angle for the forward Mach line is:
μ₁ = 19.47°
The angle for the rearward Mach line is:
θr = μ₂ - θ = 16 - 10.79 = 5.21°
Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>
Answer:
The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.
Explanation:
Let us call the mass of the balloon and the mass of the basket
, then according to newton's second law:
,
where is the upward acceleration, and
is the net propelling force (counts the gravitational force).
Also, the tension in the rope is 79.8 N more than the basket's weight; therefore,
and this tension must equal
Combining equations (2) and (3) we get:
since , we have
Putting this into equation (1) and substituting the numerical values of and
, we get:
Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
Distance traveled by the ball is given by
here we know that
speed = 20 m/s
times = 0.25 s
now we have
so ball will travel 5 m distance in the given interval of time