A bullet of 10g strikes a sand bag at a speed of 100 m/s and gets embedded after travelling
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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:
Explanation:
using the law of the conservation of energy:
where K is the spring constant, x is the spring compression, N is the normal force of the block, is the coefficiet of kinetic friction and d is the distance.
Also, by laws of newton, N is calculated by:
N = mg
N = 3.35 kg * 9.81 m/s
N = 32.8635
So, Replacing values on the first equation, we get:
solving for :