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
13 year old and the light year mean cell of moucles that can use simple year light diffusion
Answer:
33.33 m/sec
Explanation:
A baseball travels 200 metes in 6 seconds,
what is the baseball’s velocity?
use the formula: velocity = distance over time
where (d) distance = 200 m
and (t) time = 6 sec.
plugin values into the formula:
v = d / t
= 200 m / 6 sec
= 33.33 m/sec.
therefore, the baseball's velocity is 33.33 m/sec
Answer:
400 g
Explanation:
The computation of the number of grams in the original sample is shown below:
Given that
half-life = 5.26 years
total time of decay = 15.8 years
final amount = 50.0 g
Now based on the above information
number of half-lives past is
= 15.8 ÷ 5.26
= 3 half-lives
Now
3 half-lives = 1 ÷ 8 remains = 50.0 g
So, the number of grams would be
= 50.0 g × 8
= 400 g
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
