At the point of fission, a nucleus of 235U that has 92 protons is divided into two smaller spheres, each of which has 46 protons
and a radius of 5.9 × 10−15 m. What is the magnitude of the repulsive force pushing these two spheres apart? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 .
1 answer:
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Answer:
her displacement <em>s=337.5m</em>
Explanation:
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Answer:
Explanation:is in the picture below
Answer:
108.37°C
Explanation:
P₁ = Initial pressure = 101 kPa
V₁ = Initial volume = 530 m³
T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K
P₂ = Final pressure = 101 kPa (because it is open to atmosphere)
V₂ = Final volume = 530 m³
P₁V₁ = n₁RT₁
⇒101×530 = n₁RT₁
⇒53530 J = n₁RT₁
P₂V₂ = n₂RT₂
⇒53530 J = n₂RT₂
Dividing the first two equations we get
∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C