Answer:
the speed of the center of mass stays the same
Explanation:
In a system with no energy loss, momentum is conserved if the mass remains constant. The system described has no change in mass, and energy loss is considered negligible. Hence the product of the total mass and the velocity of its center will be a constant. The center of mass stays the same speed.
Answer: Heat will transfer from the water to the air. When a mass of air moves on a warmer surface it is heated by its base. Then thermal instability develops in the lower layers and then extends upwards. If the air initially contained inversions, these are destroyed and a strong gradient is established uniformly in the lower troposphere temperature.
The nuclear fusion of hydrogen atoms releases a huge amount of energy. So the correct choice is C. Conversion of mass to energy.
What is nuclear fusion?
When two small nuclei join to form a new nucleus, then this process is termed nuclear fusion. A huge amount of energy is released when there occurs nuclear fusion between the two nuclei. And a new element is formed.
It has been observed that the amount of energy released in nuclear fusion is equal to the mass difference between the mass of the formed nucleus and the total mass of old nuclei. Hence in the nuclear fusion of hydrogen nuclei to form a helium nucleus, the energy is released due to the conversion of mass into energy.
The pressure is increased to make the hydrogen atoms fuse but this change in pressure does not contribute to the energy released in the fusion of hydrogen.
The magnitude of the gravitational field is too low and it does not contribute to the energy released in the fusion of hydrogen.
The gravitational collapse does not occur between the two hydrogen atoms. This phenomenon occurs in celestial bodies so this also does not contribute to the energy released in the fusion of hydrogen.
Learn more about nuclear fusion here:
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The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.
2. This is immediate from (2.2.7).
3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =
