Answer:
1) Q ’= 8 Q
, 2) q ’= 16 q
, 3) r ’= ¾ r
Explanation:
For this exercise we will use Coulomb's law
F = k q Q / r²
It asks us to calculate the change of any of the parameters so that the force is always F
Original values
q, Q, r
Scenario 1
q ’= 2q
r ’= 4r
F = k q ’Q’ / r’²
we substitute
F = k 2q Q ’/ (4r)²
F = k 2q Q '/ 16r²
we substitute the value of F
k q Q / r² = k q Q '/ 8r²
Q ’= 8 Q
Scenario 2
Q ’= Q
r ’= 4r
we substitute
F = k q ’Q / 16r²
k q Q / r² = k q’ Q / 16 r²
q ’= 16 q
Scenario 3
q ’= 3/2 q
Q ’= ⅜ Q
we substitute
k q Q r² = k (3/2 q) (⅜ Q) / r’²
r’² = 9/16 r²
r ’= ¾ r
Answer:
Part a)

Part b)

Explanation:
As we know that magnetic flux through the loop is given as

now we have

now rate of change in flux is given as

now we know that



Now plug in all data


Part b)
Now the radius of the loop after t = 1 s



Now plug in data in above equation


Answer:
because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits
Answer:
Explanation:
Given
Wheels are rotating with constant angular velocity let say 
Presence of constant angular velocity show that there is no angular acceleration thus there is no tangential acceleration.
But any particle on the rim will experience a constant acceleration towards center called centripetal acceleration.
(a) yes, there will be tangential velocity which is given by

where r=radial distance from center
(b)tangential acceleration
there would be no tangential acceleration as velocity is constant
(c)centripetal acceleration
Yes, there will be centripetal acceleration given by
