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:
Your answer should be 2. Short
Explanation:
firstly you get your acceleration with the formula, a=v-u/t. Then you use the formula for kinetic energy 1/2mv^2
then you can finally get the answer for power by dividing your previous answer by the time
Answer:
F= 5.71 N
Explanation:
width of door= 0.91 m
door closer torque on door= 5.2 Nm
In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.
so wee need to exert 5.2 Nm torque on the door.
If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.
T= r x F
T= r F sin∅
F= T/ (r * sin∅)
F= 5.2/ (0.91 * 1)
F= 5.71 N
