Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity 
= 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity is proportional to 1/(distance)²
i.e
∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e ₂ = ₁/2
Hence,
₂/₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
Nuclear fission formula by the looks of it. Possibly how Professor Lisa Meitner realised that she had split the atomic nucleus. The Xenon and the Strontium (Xe and Sr) would presumably show up in a radio chemical assaying test at her university.
A few years later, Professor J Robert Oppenheimer watched a nuclear test somewhere near Los Alamos, US and lamented "I am become death, the destroyer of worlds". Shortly thereafter, Hiroshima and Nagasaki were razed to the ground and annihilated by nuclear bombs. Professor Meitner, probably inadvertently, had got the keys to the doors to "nuclear hell", and JRO ended up turning them. Something like that maybe, and a very harrowing and tumultuous period in human history.
Note in the fission equation, that out come two neutrons. They go off and produce a similar fission in another U235 nucleus into a chain reaction which, i not moderated by, say, Boron, can end up as a "mushroom cloud".
Answer:
Magnets come in a variety of shapes and one of the more common is the horseshoe (U) magnet. The horseshoe magnet has north and south poles just like a bar magnet but the magnet is curved so the poles lie in the same plane. The magnetic lines of force flow from pole to pole just like in the bar magnet.
Answer:
Explanation:
As we know that initial speed of the fall of the stone is ZERO
also the acceleration due to gravity on Mars is g
so we have
now we have
now if the same is dropped for 4t seconds of time
then again we will use above equation
We will make the comparison between each of the sizes against the known wavelengths.
In the case of the <em>hydrogen atom</em>, we know that this is equivalent to 
m on average, which corresponds to the wavelength corresponding to X-rays.
In the case of the <em>Virus</em> we know that it is oscillating in a size of 30nm to 200 nm, so the size of the virus is equivalent to the range of the wavelength of an ultraviolet ray.
In the case of <em>height</em>, it fluctuates in a person around 
to 
m, which falls to the wavelength of a radio wave.
