Training behaviors..........................................
Answer:
Explanation:
y1 = 16.2/2 = 8.1 cm
d = 1.22 lambda D/y1
d = 1.22 x 657 x 10^-9 x 23.7/0.081
d = 2.35 x 10^-4 m = 234.52 um
Refer to the diagram shown below.
The force, F, is applied at 5 cm from the elbow.
For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N
Answer: 1500 N
Answer:
Maximum weight that can be lifted = 18,000 N
Explanation:
Given:
Cross-sectional area of input (A1) = 0.004 m²
Cross-sectional area of the output (A2) = 1.2 m
²
Force (F) = 60 N
Computation:
Pressure on input piston (P1) = F / A1
Assume,
Maximum weight lifted by piston = W
Pressure on output piston (P2) = W / A2
We, know that
P1 = P2
[F / A1] = [W / A2]
[60 / 0.004] = [W / 1.2]
150,00 = W / 1.2
Weight = 18,000 N
Maximum weight that can be lifted = 18,000 N
Answer:
for an object to be in equilibrium, it must be experiencing no acceleration. Both the net force and the net torque must be zero.
Hope I helped