The electric force between the two particles are calculated through the equation,
F = kQ₁Q₂ / d²
where F is the force, k is a constant called Coulomb's law constant, Q₁ and Q₂ are the charges, and d is the distance. This equation is called the Coulomb's law.
It can be seen from the equation above that the electric forces between the objects are majorly affected by the substance's charges and distance.
The answer to this item is therefore letter A.
Shadows are formed when an opaque object or an object that doesn't allow light to pass through is in the way or infront of etc. a source of light.
64 miles/hour
Therefore 1/64 hours/mile
68 miles * 1/64 hours/mile (notice how miles cancels out)
Therefore the answer is 68/64 hours = 1.0625 hours = 1 hour 3min and 45sec.
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
Answer:
alpha=53.56rad/s
a=5784rad/s^2
Explanation:
First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)
Now, we can calculate the angular acceleration (w0=0rad/s)
with this value we can compute the angular velocity
and the tangential velocity of point B, and then the acceleration of point B:
hope this helps!!
