<em>500 sec</em>
<em>8 min 20 sec</em>
<em>Hi there !</em>
<em />
<em>8 m ................ 1 s </em>
<em>4000 m ........ x s</em>
<em>x = 4000m×1s/8m = 500 sec = 8 min 20 sec</em>
<em />
<em>Good luck ! </em>
<u>Answer:</u>
The given statement is a True statement
<u>Explanation:</u>
All cars and trucks have load capacities marked on the jamb of the driver's door, as well as the owner's manual.
This is very significant for braking distance.
A heavier object will require much more braking force and distance to stop, due to New ton's law of motion. “ An object in motion will tend to stay in motion “ .A heavy object will remain in motion longer and require much more force to stop. Four wheel disc brakes are great compared to front disc and rear drum configuration, in dissipating heat, and stopping more efficiently.
By definition of average acceleration,
<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²
Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:
∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N
If you just want the magnitude, drop the negative sign.
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
