Calculate the pressure due to sea water as density*depth.
That is,
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa
Atmospheric pressure is 101.3 kPa
Total pressure is 94423 + 101.3 = 94524 kPa (approx)
The area of the window is π(0.44 m)^2 = 0.6082 m^2
The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx
Answer:
Mass = 386 kg
Explanation:
<u><em>Density = Mass / Volume</em></u>
Mass = Density × Volume
Where D = 19300 kg/m³ , V = 0.02 m³
<em>Putting the given in the above formula</em>
Mass = 19300 × 0.02
Mass = 386 kg
Answer: The height of the cloud = 394.55 m
Explanation:
The observer is 500m away from the spotlight.
Let x be the distance from the observer to the interception of the segment of the height, h with the floor. The equations are thus:
Tan 45° = h/x ... eq1
Tan 75° = h/(500- x ) ... eq2
From eq 1, Tan 45° = 1, therefore eq1 becomes:
h = x ... eq3
Put eq3 into eq2
Tan 75° = h/(500- h)
h = ( 500 - h ) Tan 75°
h = 500Tan 75° - hTan75°
h + h Tan 75° = 500 Tan 75°
h ( 1 + Tan 75° ) = 500 Tan75°
h = 500Tan75°/ (1 + Tan 75°)
h= 1866.02 / 4.73
h = 394.55m
