Answer:
spacing between the slits is 405.32043 ×
m
Explanation:
Given data
wavelength = 610 nm
angle = 2.95°
central bright fringe = 85%
to find out
spacing between the slits
solution
we know that spacing between slit is
I = 4
× cos²∅/2
so
I/4 = cos²∅/2
here I/4 is 85 % = 0.85
so
0.85 = cos²∅/2
cos∅/2 = √0.85
∅ = 2 ×
0.921954
∅ = 45.56°
∅ = 45.56° ×π/180 = 0.7949 rad
and we know that here
∅ = 2π d sinθ / wavelength
so
d = ∅× wavelength / ( 2π sinθ )
put all value
d = 0.795 × 610× / ( 2π sin2.95 )
d = 405.32043 × m
spacing between the slits is 405.32043 × m
Answer:
A) The net force
Explanation:
If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement.
Answer:
Explanation:
Stored energy in spring = 1/2 k x² , k is spring constant , x is compression.
= 1/2 x 8 x (5.7 x 10⁻²)²
= 129.96 x 10⁻⁴ J
Energy lost due to friction = force x distance
= .035 x .17
= .00595 J
Energy used in providing kinetic energy to projectile.
129.96 x 10⁻⁴ - .00595
.012996 - .00595
= .007046 J
So
1/2 m v² = .007046
v² = .007046 x 2 / .0059
= 2.3885
v = 1.545 m /s
Answer:
(C) length / height of the plane
Explanation:
The mechanical advantage of an inclined plane can be determined using different variables. In this case, the geometry of the setup is relevant. The advantage is proportional to the length of the plane, and inversely proportional to the height: it is the ratio (length) / (height) of the plane. For example, given a desired, fixed height, a long inclined plane gives you a bigger mechanical advantage than a short inclined plane. In this example, pushing an object up the long plane will require a smaller force, than it would on the short plane.
Strictly speaking, (D) would also "allow you to determine the mechanical advantage" because you could simply invert the ratio listed under (D). However, (C) is the best, direct, answer.
