Answer:
Linear mass density,
Explanation:
Given that,
Mass of the string, m = 0.3 g = 0.0003 kg
Length of the string, l = 1.5 m
The linear mass density of a string is defined as the mass of the string per unit length. Mathematically, it is given by :
or
So, the linear mass density of a string is . Hence, this is the required solution.
Answer: (a) α =
(b) For r≤R: B(r) = μ_0.
For r≥R: B(r) = μ_0.
Explanation:
(a) The current I enclosed in a straight wire with current density not constant is calculated by:
where:
dA is the cross section.
In this case, a circular cross section of radius R, so it translates as:
For these circunstances, α =
(b) <u>Ampere's</u> <u>Law</u> to calculate magnetic field B is given by:
μ_0.
(i) First, first find for r ≤ R:
Calculating B(r), using Ampere's Law:
μ_0.
.μ_0
B(r) = .μ_0
B(r) = .μ_0
For r ≤ R, magnetic field is B(r) = .μ_0
(ii) For r ≥ R:
So, as calculated before:
I
Using Ampere:
B.2.π.r = μ_0.I
B(r) = .μ_0
For r ≥ R, magnetic field is; B(r) = .μ_0.
For us to understand the missing item that would complete beta decay reaction, we need to achieve in depth understanding of chemical formulas and nuclear symbols. Next is to have great comprehension of the following points:
<span>1.) Neutron in nucleus breaks down and changes into a proton.
2) Then it emits an electron, as well as an anti-neutrino which go into space.
3) Lastly, atomic number continuously goes UP while mass number remains unchanged.</span>
Answer:
1.424 μC
Explanation:
I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),
tension T (acting along the string - to the pivot point), and
F (electric force – acting along the line connecting the charges).
We then have something like this
x: T•sin α = F,
y: T•cosα = mg.
Dividing the first one by the second one we have
T•sin α/ T•cosα = F/mg, ultimately,
tan α = F/mg.
Since we already know that
q1=q2=q, and
r=2•L•sinα,
k=9•10^9 N•m²/C²
Remember,
F =k•q1•q2/r², if we substitute for r, we have
F = k•q²/(2•L•sinα)².
tan α = F/mg =
= k•q²/(2•L•sinα)² •mg.
q = (2•L•sinα) • √(m•g•tanα/k)=
=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =
q = 0.486 • √(8.61•10^-12)
q = 0.486 • 2.93•10^-6
q = 1.424•10^-6 C
q = 1.424 μC.