Answer:
(a)
(b) 15 hours
Explanation:
half life, T = 12 hours
No = 19 g
(a) Let N be the amount remaining after time t.
Let λ be the decay constant.
The equation of radioactivity used here is given by
(b) N = 8 gram
Substitute the values in above equation
λ = 0.0577 per hour
So,
Take natural log on both the sides
- 0.0577 t = - 0.865
t = 15 hours
Answer:
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
Equality Properties
<u>Geometry</u>
<u>Algebra I</u>
<u>Calculus</u>
Derivatives
Differentiating with respect to time
Basic Power Rule:
Explanation:
<u>Step 1: Define</u>
Area is A = lw
2w = l
w = 300 m
<u>Step 2: Rewrite Equation</u>
<u>Step 3: Differentiate</u>
<em>Differentiate the new area formula with respect to time.</em>
<u>Step 4: Find Rate</u>
<em>Use defined variables</em>
Answer:
The electron tends to go to the region of 4. higher electric potential.
Explanation:
When a charged particle is immersed in an electric field, it experiences a force given by
where
q is the charge of the particle
E is the electric field
The direction of the force depends on the sign of the charge. In particular:
- The force and the electric field have the same direction if the charge is positive
- The force and the electric field have opposite directions if the charge is negative
Therefore, an electron (negative charge) moves in the direction opposite to the electric field lines.
However, electric field lines go from points at higher potential to points at lower potential: so, electrons move from regions at lower potential to regions of higher potential.
Therefore, the correct answer is
The electron tends to go to the region of 4. higher electric potential.
Answer:
(b) the point charge is moved outside the sphere
Explanation:
Gauss' Law states that the electric flux of a closed surface is equal to the enclosed charge divided by permittivity of the medium.
According to this law, any charge outside the surface has no effect at all. Therefore (a) is not correct.
If the point charge is moved off the center, the points on the surface close to the charge will have higher flux and the points further away from the charge will have lesser flux. But as a result, the total flux will not change, because the enclosed charge is the same.
Therefore, (c) and (d) is not correct, because the enclosed charge is unchanged.