<span>From the point of view of the astronaut, he travels between planets with a speed of 0.6c. His distance between the planets is less than the other bodies around him and so by applying Lorentz factor, we have 2*</span>√1-0.6² = 1.6 light hours. On the other hand, from the point of view of the other bodies, time for them is slower. For the bodies, they have to wait for about 1/0.6 = 1.67 light hours while for him it is 1/(0.8) = 1.25 light hours. The remaining distance for the astronaut would be 1.67 - 1.25 = 0.42 light hours. And then, light travels in all frames and so the astronaut will see that the flash from the second planet after 0.42 light hours and from the 1.25 light hours is, 1.25 - 0.42 = 0.83 light hours or 49.8 minutes.
Given: The mass of stone (m) = 0.5 kg
Raised from heights (h₁) = 1.0 m to (h₂) = 2.0 m
Acceleration due to gravity (g) = 9.8 m/s²
To find: The change in potential energy of the stone
Formula: The potential energy (P) = mgh
where, all alphabets are in their usual meanings.
Now, we shall calculate the change in potential energy of the stone
Δ P = P₂ - P₁ = mg (h₂ - h₁)
or, = 0.5 kg ×9.8 m/s² ×(2.0 m - 1.0 m)
or, = 4.9 J
Hence, the required change in the potential energy of the stone will be 4.9 J
b. A Boeing 747 airplane
Explanation:
The Boeing 747 airplane will have the most inertia of all. Inertia is the tendency of an object to remain at rest or uniform motion.
- Newton's first law of motion is also regarded as the law of inertia.
- It states that "an object will remain at rest or uniform motion motion unless acted upon by an external force".
- A body that has a large mass will be more resistant to any external force that acts on it.
- The Boeing 747 has the most mass of all and will have the most inertia.
Learn more:
Inertia brainly.com/question/691705
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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) =