Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
Answer:
<h2>
128.61 Watts</h2>
Explanation:
Average power done by the torque is expressed as the ratio of the workdone by the toque to time.
Power = Workdone by torque/time
Workdone by the torque =
= 
I is the rotational inertia = 16kgm²



To get the angular acceleration, we will use the formula;


Workdone by the torque = 16 * 1.28 * 12.56
Workdone by the torque = 257.23 Joules
Average power done by the torque = Workdone by torque/time
= 257.23/2.0
= 128.61 Watts
C because it’s not a or B so 50/50 c or d and d is def not the answer so c
Answer: vf = 51 m/s
d = 112 m
Explanation: Solution attached:
To find vf we use acceleration equation:
a = vf - vi / t
Derive to find vf
vf = at + vi
Substitute the values
vf = 3.5 m/s² ( 8.0 s) + 23 m/s
= 51 m/s
To solve for distance we use
d = (∆v)² / 2a
= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)
= (28 m/s)² / 7 m/s²
= 784 m/s / 7 m/s²
= 112 m