Answer:
Explanation:
Lead has an half life of 43.5 seconds
The radioactive decay law says
N(t) = No•2^(-t / t½)
Hence, the percent of a sample of the Lead-214 isotope that decay after a time equivalent to 7 half-lives
N(t) = No•2^(-t / t½)
N(t) / No = 2^(-t / t½)
N(t) / No = 2^-7
N(t) / No = 7.81 × 10^-3
To percentage
N(t) / No = 7.81 × 10^-3 × 100
N(t) / No = 0.781 %
Answer:
<h2><em>
0.165Tesla</em></h2>
Explanation:
The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;
B is the magnetic field (in Tesla)
I is the current (in amperes)
L is the length of the wire (in meters)
∝ is the angle that the conductor makes with the magnetic field.
Given parameters
L = 0.56 m
I = 2.6A
F = 0.24N
∝ = 90°
Required
magnitude of the magnetic field (B)
Substituting the given values into the formula given above we will have;
F = BILsin∝
0.24 = B * 2.6 * 0.56 sin90°
0.24 = B * 2.6 * 0.56 (1)
0.24 = 1.456B
1.456B = 0.24
Dividing both sides by 1.456 will give;
1.456B/1.456 = 0.24/1.456
B ≈ 0.165Tesla
<em>Hence the magnitude of the magnetic field is approximately 0.165Tesla</em>
The answer is 125 because if u add you'll get the answer
Just choose the direction and write down the force shown in the arrow. It's too blurry. Can't do it for u. Sorry.