In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. What happens to the width
1 answer:
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There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, 
, so we can write:
(1)
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find