Answer:
Examples of reciprocating motion in daily life are;
1) The needles of a sewing machine
2) Electric powered reciprocating saw blade
3) The motion of a manual tire pump
Explanation:
A reciprocating motion is a motion that consists of motion of a part in an upward and downwards 
or in a backward and forward (↔) direction repetitively
Examples of reciprocating motion in daily life includes the reciprocating motion of the needles of a sewing machine and the reciprocating motion of the reciprocating saw and the motion of a manual tire pump
In a sewing machine, a crank shaft in between a wheel and the needle transforms the rotary motion of the wheel into reciprocating motion of the needle.
Answer:
Only Technician B is right.
Explanation:
The cylindrical braking system for a car works through the mode of pressure transmission, that is, the pressure applied to the brake pedals, is transmitted to the brake pad through the cylindrical piston.
Pressure applied on the pedal, P(pedal) = P(pad)
And the Pressure is the applied force/area for either pad or pedal. That is, P(pad) = Force(pad)/A(pad) & P(pedal) = F(pedal)/A(pedal)
If the area of piston increases, A(pad) increases and the P(pad) drops, Meaning, the pressure transmitted to the pad reduces. And for most cars, there's a pressure limit for the braking system to work.
If the A(pad) increases, P(pad) decreases and the braking force applied has to increase, to counter balance the dropping pressure and raise it.
This whole setup does not depend on the length of the braking lines; it only depends on the applied force and cross sectional Area (size) of the piston.
The equations are based on the following assumptions
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)
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Answer:
work done by electric field is 0.06 J
Explanation:
Given data:
Two point charge is 
0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)
-1 charge positioned is (0 cm , -1 cm, 0.00 cm)
E = 3.0\times 10^6 N/C
From above information, the distance between given two charges d = 2 cm
then d = 0.02m
work needed is W = q E d
W = 0.06 J
Therefore work done by electric field is 0.06 J
