Johannes Kepler was a main stargazer of the Scientific Revolution known for detailing the Laws of Planetary Motion. A stargazer, obviously, is a man who contemplates the sun, stars, planets and different parts of room. Kepler was German and lived in the vicinity of 1571 and 1630.
Despite the fact that Kepler is best known for characterizing laws in regards to planetary movement, he made a few other striking commitments to science. He was the first to discover that refraction drives vision in the eye and that utilizing two eyes empowers profundity recognition.
Answer:

Explanation:
As we know that amplitude of forced oscillation is given as

here we know that natural frequency of the oscillation is given as

here mass of the object is given as



angular frequency of applied force is given as


now we have


Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:

where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>
Answer:
More force
Explanation:
Object A has more mass than object B
For object A to accelerate at the same rate as object B, it will need more force.
According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".
Net force = mass x acceleration
Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.
Explanation:
The given data is as follows.
Velocity of bullet,
= 814.8 m/s
Observer distance from marksman, d = 24.7 m
Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.
t =
(velocity in air = 343 m/s)
= 0.072 sec
Now, before the observer hears the report the distance traveled by the bullet is as follows.

= 
= 58.66
= 59 (approx)
Thus, we can conclude that each bullet will travel a distance of 59 m.