What is the speed of a truck traveling 10km in 10 minutes
1 answer:
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Answer:
The correct option is C
Explanation:
From the question we are told that
The initial speed of the rocket is
The speed of the rocket engine sound is
The final speed of the rocket is
The speed of the sound at would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).
A clear example when lightning strikes you will first see (that is because it travels at the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the lightning
Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates
<h2>Answer:</h2>
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