A. Ask other people for their opinion. :)
Answer:
Explanation:
T = 2π√(L/g)
If you increase L to 2L, the period is increased by a factor of √2
T = 3.5√2 ≈ 4.9 s
Answer:
r = 0.05 m = 5 cm
Explanation:
Applying ampere's law to the wire, we get:
![B = \frac{\mu_oI}{2\pi r}\\\\r = \frac{\mu_oI}{2\pi B}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_oI%7D%7B2%5Cpi%20r%7D%5C%5C%5C%5Cr%20%3D%20%20%5Cfrac%7B%5Cmu_oI%7D%7B2%5Cpi%20B%7D)
where,
r = distance of point P from wire = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = 2 A
B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T
Therefore,
![r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%284%5Cpi%5C%20x%5C%2010%5E%7B-7%7D%5C%20N%2FA%5E2%29%282%5C%20A%29%7D%7B2%5Cpi%288%5C%20x%5C%2010%5E%7B-6%7D%5C%20T%29%7D%5C%5C%5C%5C)
<u>r = 0.05 m = 5 cm</u>
Answer:
Lifetime = 4.928 x 10^-32 s
Explanation:
(1 / v2 – 1 / c2) x2 = T2
T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225
T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6
T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s