Two objects with a mass of 10,000-kg each and a distance of 1 meter between them.
1 answer:
You might be interested in
Complete question:
A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.
(a) Find the available kinetic energy of the bullet. J
(b) Find the heat required to melt the bullet. J
Answer:
Part (a) the available kinetic energy of the bullet is 323.32 J
Part (b) the heat required to melt the bullet is 216.17 J
Explanation:
Given;
mass of the bullet = 3.53 g = 0.00353 kg
velocity of the bullet = 428 m/s
initial temperature of the bullet = 40.0°C
final temperature of the bullet = 327°C
specific heat capacity, c= 128 J/(kg · °C)
latent heat of fusion, Hf = 24.5 kJ/kg
Part (a) the available kinetic energy of the bullet. J
KE = ¹/₂ × mv²
KE = ¹/₂ × 0.00353 × 428²
= 323.32 J
Part (b) the heat required to melt the bullet. J
This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.
Quantity of heat required to raise the temperature of the bullet:
Q = mcΔT
= 0.00353 × 128 × (327-40)
= 0.00353 × 128 × 287
= 129.68 J
Quantity of heat required to melt the bullet:
Q = 0.00353 × 24500
= 86.49 J
TOTAL energy required to melt the bullet = 129.68 J + 86.49 J
= 216.17 J
<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