Answer:
Explanation:
separation between slit d = 1.2 x 10⁻⁴ m
distance of screed D = 3.5 m
distance of second order right fringe
= 2λD / d where λ is wavelength of light .
2λD / d = .0375
2 x λ x 3.5 / 1.2 x 10⁻⁴ = .0375
λ = 6428 x 10⁻¹⁰ m
= 643 nm .
You do 0.5 x 2kg x 4 squared so
0.5x2x16
So the answer is 16
Answer:
volume : {l}^3
speed: (l)^1*(t)^-1
Explanation:
Volume is a measure of 3 dimensional space. It is expressed with 3 orthogonal lengths. The volume of a box would be the product of it's height, width and length. These 3 are longitudes that can be expressed in meters, feet, inches, etc. Because these are 3 longitudes multiplied the result will be a cubic longitude (l)^3.
A more general method for finding a volume is to use integral calculus:
This is for Cartesian coordinates. Cylindrical and spherical coordinates can also be used.
Speed is defined as the rate of change in position respect of time:
For movement in one dimension.
For movement in 3 dimensions you calculate the speed component of each space direction and express them as components of a speed vector:
This is a vector of velocity components, each one is expressed as a division of a longitude over a time, so speed components have dimensions of (l)^1*(t)^-1
The speed vector has a magnitude that is obtained with the Pitagoras theorem:
Since each component is squared, added together and then the square root is taken this magnitude is also in (l)^1*(t)^-1
Answer:
Explanation:
Given that,
Current in wire are 1.3A and 3.15A
Distance between wire is d= 2.25cm
d = 2.25/100 = 0.025m
Force per unit length F/l?
Let us consider the field produced by wire 1 and the force it exerts on wire 2 (call the force F2).
The field due to I1 at a distance r is given to be
B1 = μo• I1 / 2πr
This field is uniform along wire 2 and perpendicular to it, and so the force F2 it exerts on wire 2 is given by
F=ILBsinθ
with sinθ=1:
F2=I2 • L •B1
By Newton’s third law, the forces on the wires are equal in magnitude, and so we just write F for the magnitude of F2. (Note that F1=−F2.) Since the wires are very long, it is convenient to think in terms of F/l, the force per unit length. Substituting the expression for B1 into the last equation and rearranging terms gives
F/l = μo• I1• I2 / 2πr
Where μo is constant
μo = 4π×10^7 Tm/A
Then,
F/l = μo• I1• I2 / 2πr
F/l = 4π ×10^-7 × 1.3×3.15/(2π×0.025)
F/l = 3.276×10^-5 N/m
the magnitude of the force per unit length that one wire exerts on the other is 3.276×10^-5 N/m
Answer:
0.5 hours or 30 minutes
Explanation:
each hour travels 80km
so, time taken= 40/80 hours
= 0.5 hours
= 30 minutes