Answer:
the maximum allowable current is 7302.967 amperl
Explanation:
The computation of the maximum allowable current is shown below;
Force F = mean ÷ 4π 2 I_1 I_2 ÷d × ΔL
200 N = (10)^-7 (2I × I) ÷ 0.08 × 1.5
200 = 3.75 × 10^-6 I^2
I = √200 ÷ √ 3.75 × 10^-6
= 7302.967 amperl
Hence, the maximum allowable current is 7302.967 amperl
Basically we applied the above formula
If the mass of the original sample of Galium-68 is 10 mg, then . . .
== After one half-life, (1/2)(10 mg) = 5 mg remain
== After the second half-life, (1/2)(5 mg) = 2.5 mg remain
== After the third half-life, (1/2)(2.5 mg) = 1.25 mg remain
It doesn't matter how long the half-life is.
Answer:
B.lenses bounce light from their surface, mirrors do not
Answer:
Long question good luck:)..............
Explanation:
