According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.
Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²
Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.
Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
Answer:
= 201.53 meters
Explanation:
A car started from rest and accelerated at 9.54 m/s^2 for 6.5 seconds. How much distance was covered by the car?
Use the formula d = 
where d is the distance, t is the time and "a" is the acceleration.
Answer:
temperature on left side is 1.48 times the temperature on right
Explanation:
GIVEN DATA:
T1 = 525 K
T2 = 275 K
We know that
n and v remain same at both side. so we have
..............1
let final pressure is P and temp 
..................2
similarly
.............3
divide 2 equation by 3rd equation
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)
thus, temperature on left side is 1.48 times the temperature on right
