Answer:
False
Explanation:
The formula of force that exists between two charges is expressed as;
F = kq1q2/r²
If two charges separated by one meter exert a 9 N force on each other, the;
9 = kq1q2/1²
9 = kq1q2 ..... 1
If the charges are pushed to a 3 meter separation, then;
F = kq1q2/3²
F = kq1q2/9 .... 2
Divide both equations;
9/F = (kq1q2)/ kq1q2/9
9/F = kq1q2 * 9/ kq1q2
9/F = 9
F = 9/9
F = 1N
Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False
The correct answer is A hemoglobin
Answer:
Twice
Explanation:
From the formula for velocity in a circle
V= 2πr/T
Where V is velocity
r is raduis
T is period
We see that as r increases V increases so if r is doubled V becomes doubled
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v = 
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀