Answer:
15mm
Explanation:
We know that for circular holes first dark spot is given by
sin စ = 1.22 λ/D
Also we know that at the same time
tan စ = r/L
So
r = L tanစ = 6 x tan( arcsin(1.22x 500 x10^9/0.50 x 10^ 3))
= 0.0073 m = 7.3 mm
However since the size is twice that so 14.6 mm which is approx 15mm
Writing the acceleration as a function of time:
a(t) = 1 + 3√t
Integrating acceleration, we obtain velocity:
v(t) = t + 2(t)^(3/2) + c;
object at rest so velocity at t = 0 is 0 so c = 0.
v(t) = t + 2(t)^(3/2)
Integrating velocity to obtain an equation for displacement:
d(t) = t²/2 + 4/5 t^(5/2) + c
Applying limits from t = 0 to t = 9
d = 9²/2 + 4/5 9^(5/2)
d = 234.9 m
Question is not complete and the missing part is;
A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.
Answer:
0.828 m/s
Explanation:
Resolving vertically, we have;
Fn and Fg act vertically. Thus,
Fn - Fg = 0 - - - - eq(1)
Resolving horizontally, we have;
Ff = ma - - - - eq(2)
Now, Fn and Fg are both mg and both will cancel out in eq 1.
Leaving us with eq 2.
So, Ff = ma
Now, Frictional force: Ff = μmg where μ is coefficient of friction.
Also, a = v²/r
Where v is linear speed or velocity
Thus,
μmg = mv²/r
m will cancel out,
Thus, μg = v²/r
Making v the subject;
rμg = v²
v = √rμg
Plugging in the relevant values,
v = √0.14 x 0.5 x 9.8
v = √0.686
v = 0.828 m/s
