Answer:
Explanation:
wave length of light λ = 623 x 10⁻⁹ m .
Distance of screen D = 76.5 x 10⁻² m
width of slit = d
Distance on the screen between the second order minimum and the central maximum = 2 λ D / d
1.11 x 10⁻² = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d
d = ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²
= 85872.97 x 10⁻⁹
= 85.87297 x 10⁻⁶
= 85.87 μm
width a of the slit is = 85.87 μm
Answer:
Explanation:
For answer this we will use the law of the conservation of the angular momentum.
so:
where 
is the moment of inertia of the merry-go-round, 
is the initial angular velocity of the merry-go-round, 
is the moment of inertia of the merry-go-round and the child together and 
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where 
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:
Finally we replace all the data:
Solving for :
The Kelvin scale has no negatives on it.
Zero Kelvin is 'Absolute Zero', and nothing can get colder than that.
I can’t answer without any graph options
