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Vlad [161]
2 years ago
8

Two point charges are separated by 6 cm. The attractive force between them is 20 N. Find the force between them when they are se

parated by 12 cm. (Why can you solve this problem without knowing the magnitudes of the charges?)
Physics
1 answer:
Scorpion4ik [409]2 years ago
7 0

Answer:

5 N

Explanation:

F_1 = 20 N

r_1 = 6 cm

r_2 = 12 cm

k = Coulomb constat

q = Charge

\dfrac{F_1}{F_2}=\dfrac{\dfrac{kq_1q_2}{r_1^2}}{\dfrac{kq_2q_2}{r_2^2}}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{12^2}{6^2}\\\Rightarrow \dfrac{F_1}{F_2}=4\\\Rightarrow F_2=\dfrac{F_1}{4}\\\Rightarrow F_2=\dfrac{20}{4}\\\Rightarrow F_2=5\ N

The force is 5 N

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A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

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c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

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