The stretching force acting on the second wire, given the data is 588 N
<h3>Data obtained from the question</h3>
- Radius of fist wire (r₁) = 3.9×10⁻³ m
- Force of first wire (F₁) = 450 N
- Radius of second wire (r₂) = 5.1×10⁻³ m
- Force of second wire (F₂) =?
<h3>How to determine the force of the second wire</h3>
F₁ / r₁ = F₂ / r₂
450 / 3.9×10⁻³ = F₂ / 5.1×10⁻³
cross multiply
3.9×10⁻³ × F₂ = 450 × 5.1×10⁻³
Divide both side by 3.9×10⁻³
F₂ = (450 × 5.1×10⁻³) / 3.9×10⁻³
F₂ = 588 N
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<span>Nuclear fission produces elements that are heavier than helium.
The elements that are used in nuclear fission and their products are much heavier than helium.</span>
Answer: 200 joules I think.
Explanation:
Positive
In an LED, the longer lead is always connected to the positive terminal