Answer:
695800 N/m^2 or Pa
Explanation:
Height of the water from the ground H = 71 m
Acceleration due to gravity g =9.8 m/s^2
density of water ρ= 1000 kg/m^3
The minimum output gauge pressure to make water reach height H
P= ρgH
= 1000×9.8×71= 695800 N/m^2 or Pa
Answer:
If it is not an object in motion, all forces are balanced.
Answer:
8.874
Explanation:
You need to times 5.22 kg and 1.7 m/s to get 8.874.
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03
Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:

where

is the coefficient of friction, while N is the normal force. So we have:

since we know that F=300 N and

, we can find N, the magnitude of the normal force:

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is

due to the presence of the oil. Therefore, we have:
(A)energy lost in the lever due to friction
(C)
visual estimation of height of the beanbag
(E)position of the fulcrum for the lever affecting transfer of energy