In a parallel circuit, the total resistance calculated from the individual resistances is computed from the formula: 1/Rt = 1/R1 + 1/R2. substituting R1 and R2, then
1/Rt = 1/7 + 1/49
1/Rt = 1/6.125 = 1/ 49/8
Rt = 49/8 <span>Ω
The total resistance hence is </span>49/8 Ω
Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period
Thank you
Answer:
Explanation:
Work done on the lever ( input energy ) = force applied x input distance
= 24 N x 2m = 48 J
Work done by the lever ( output energy ) = load x output distance
= 72 N x 0.5m = 36 J
efficiency = output energy / input energy
= 36 J / 48 J
= 3 / 4 = .75
In percentage terms efficiency = 75 % .
The formula is=1/2(m x v^2)
so = 1/2*(0.05)*(310)^2
ans is =2402.5 joules
