Answer:
(A). The order of the bright fringe is 6.
(B). The width of the bright fringe is 3.33 μm.
Explanation:
Given that,
Fringe width d = 0.5 mm
Wavelength = 589 nm
Distance of screen and slit D = 1.5 m
Distance of bright fringe y = 1 cm
(A) We need to calculate the order of the bright fringe
Using formula of wavelength


Put the value into the formula


(B). We need to calculate the width of the bright fringe
Using formula of width of fringe

Put the value in to the formula



Hence, (A). The order of the bright fringe is 6.
(B). The width of the bright fringe is 3.33 μm.
Answer is b that is Heat energy from below the ground converts water to steam to drive a steam turbine attached to an electrical generator.. .
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Answer:
c is the one that makes the most sense
Explanation:
B: Extension Lines! You could have just searched this up on google