Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
Answer:
14778.29 N
Explanation:
Diameter, d=10.1mm= 
Since stress, 
where F is force, A is area and since specimen is cylindrical,
A= 
Therefore, 
Also strain 
where L is length
Poison’s ratio,v is the ratio of lateral strain to longitudinal strain hence
V= 
From Hooke’s law, 
Conclusively, 
F=
F= 
= 14778.29N
Therefore, required force F is 14778.29 N
Answer: The angle between the wire segment and the magnetic field 66.42°
Explanation:
Please see the attachment below
Answer:
0.60 m/s
Explanation:
The average velocity from t = a to t = b is:
v_avg = (x(b) − x(a)) / (b − a)
Given that x(t) = 0.36t² − 1.20t, and the time is from 1.0 to 4.0:
v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)
v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0
v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0
v_avg = [0.96 − (-0.84)] / 3.0
v_avg = 0.60
The average speed is 0.60 m/s.
