Answer:
0.3967mg of sodium-24 are left
Explanation:
If the half life of radioactive element (which is sodium-24 here) is , then at the end of this time the number of atoms in the sample will become half i.e. 1/2 .
if the original amount of atoms is N°. The formula for calculating the remaining atoms (N) left in the sample after t half lives would be:
Here, in this question,
Given amount of sodium-24 = N° = 13 grams
Half life of sodium-24 = t = 15
and Remaining amount of sodium after 45 hours = N = ?
Therefore, using the given Forumla:
ANSWER
EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:
where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:
That is the answer.
0.46Ω
<h2>Explanation:</h2>The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;
E = V + Ir --------------------(a)
Where;
I = current flowing through the circuit
But;
V = I x Rₓ ---------------------(b)
Where;
Rₓ = effective or total resistance in the circuit.
<em>First, let's calculate the effective resistance in the circuit:</em>
The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.
Let;
R₁ = resistance in the first bulb
R₂ = resistance in the second bulb
Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;
P =
=> R = -------------------(ii)
Where;
P = Power of the bulb
V = voltage across the bulb
R = resistance of the bulb
To get R₁, equation (ii) can be written as;
R₁ = --------------------------------(iii)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iii) as follows;
R₁ =
R₁ =
R₁ = 36Ω
Following the same approach, to get R₂, equation (ii) can be written as;
R₂ = --------------------------------(iv)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iv) as follows;
R₂ =
R₂ =
R₂ = 36Ω
Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;
=
+
-----------------(v)
Substitute the values of R₁ and R₂ into equation (v) as follows;
=
+
=
Rₓ =
Rₓ = 18Ω
The effective resistance (Rₓ) is therefore, 18Ω
<em>Now calculate the current I, flowing in the circuit:</em>
Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;
11.7 = I x 18
I =
I = 0.65A
<em>Now calculate the battery's internal resistance:</em>
Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;
12.0 = 11.7 + 0.65r
0.65r = 12.0 - 11.7
0.65r = 0.3
r =
r = 0.46Ω
Therefore, the internal resistance of the battery is 0.46Ω