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:
distance cover is = 102.53 m
Explanation:
Given data:
speed of object is 17.1 m/s
from equation of motion we know that
where d_1 is distance covered in time t1
so
=
where d_2 is distance covered in time t2
distance cover is = 213.31 - 110.78 = 102.53 m
Answer: Velocity terminal = 0.093m/s
Explanation:
1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)
= (0.0604/2 - 0.06/2)m
= 2×10^-4
Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L
= (π×0.06×0.4)m²
= 0.075m²
Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.
Shearing stress = u×V.terminal/h = 0.86×V/0.0002
= 4300Vterminal
Therefore, Fw = shearing stress × A
30N = 4300Vterminal × 0.075
V. terminal = 30/4300 m.s
V. terminal = 0.093m/s
Answer:
The bottom one has the highest amplitude
Answer:
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