A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the
1 answer:
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Answer:
0.82 mm
Explanation:
The formula for calculation an bright fringe from the central maxima is given as:
so for the distance of the second-order fringe when wavelength = 745-nm can be calculated as:
where;
n = 2
= 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
= 0.00276 m
= 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength = 660-nm is as follows:
= 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
= 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
= 10 ⁻³ (2.76 - 1.94)
= 10 ⁻³ (0.82)
= 0.82 × 10 ⁻³ m
= 0.82 × 10 ⁻³ m
= 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
Answer:
0.47 N
Explanation:
Here we have a ball in motion along a circular track.
For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.
This force is called centripetal force, and its magnitude is given by:
where
m is the mass of the object
is the angular velocity
r is the radius of the circle
For the ball in this problem we have:
m = 40 g = 0.04 kg is the mass of the ball
is the angular velocity
r = 30 cm = 0.30 m is the radius of the circle
Substituting, we find the force: