Answer:
a) 0.0130 m
b') w' = =6.46*10^{-3] m
Explanation:
given data:
\lambda of light = 633 nm
width of siit a =0.360 mm
distance from screen = 3.75 m
a) the first minima is located at

=


with of central fringe = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m
b)
width of the first bright fringe on either side of the central one = 
calculation for y_2

= 



w' = =6.46*10^{-3] m
In a non accelerated frame, also called inertial frame, when no additional forces are acting upon the body.
For instance, a scale in free fall will not record the weight, but zero, like the astronauts. But also, if someone is pulling the object or pushing it, the scale will record some other value.
Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density 
Drag force FD is given as


Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D


T = 16653.32 N
I think it is equal for your question