PLEASE HELP
NASA scientists have determined that the chance a single person would get hit by a piece of falling satellite is one in 3,200. What can you infer based on this probability?
A. Satellites that enter Earth’s atmosphere land mostly in the ocean.
B. NASA does a good job informing people of when satellites will enter the atmosphere.
C. Satellites that enter Earth’s atmosphere land mostly in unpopulated areas.
D. Much of a satellite is destroyed during the process of entering Earth’s atmosphere.
Answer:
Sorry it doesnt tall me anythikng
Explanation:
Answer:
increases by a factor of 6.
Explanation:
Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:
Initial flow rate = area * velocity = A * V = AV m³/s
The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:
Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate
Hence, the volume flow rate of the water passing through it increases by a factor of 6.
Answer:
K
Explanation:
For installations where the nonlinear load is huge, most consulting engineers will specify K-rated transformers.
The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
<h3>
Thickness of the aluminum</h3>
The thickness of the aluminum can be determined using from distance of closest approach of the particle.
![K.E = \frac{2KZe^2}{r}](https://tex.z-dn.net/?f=K.E%20%3D%20%5Cfrac%7B2KZe%5E2%7D%7Br%7D)
where;
- Z is the atomic number of aluminium = 13
- e is charge
- r is distance of closest approach = thickness of aluminium
- k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>
![r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B2KZe%5E2%7D%7BK.E%7D%20%5C%5C%5C%5Cr%20%3D%20%5Cfrac%7B2%20%5Ctimes%209%5Ctimes%2010%5E9%20%5Ctimes%2013%5Ctimes%20%281.6%5Ctimes%2010%5E%7B-19%7D%29%5E2%7D%7B2.5%20%5Ctimes%2010%5E6%20%5Ctimes%201.6%20%5Ctimes%2010%5E%7B-19%7D%7D%20%5C%5C%5C%5Cr%20%3D%201.5%20%5Ctimes%2010%5E%7B-14%7D%20%5C%20m)
<h3>For 2.5 MeV protons</h3>
Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.
<h3>For 10 MeV alpha-particles</h3>
Charge of alpah particle = 2e
![r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B2KZe%5E2%7D%7BK.E%7D%20%5C%5C%5C%5Cr%20%3D%20%5Cfrac%7B2%20%5Ctimes%209%5Ctimes%2010%5E9%20%5Ctimes%2013%5Ctimes%20%282%20%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%29%5E2%7D%7B10%20%5Ctimes%2010%5E6%20%5Ctimes%201.6%20%5Ctimes%2010%5E%7B-19%7D%7D%20%5C%5C%5C%5Cr%20%3D%201.5%20%5Ctimes%2010%5E%7B-14%7D%20%5C%20m)
Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
Learn more about closest distance of approach here: brainly.com/question/6426420