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3 years ago

Why won't a chain reaction occur in natural uranium?

1 answer:

5 0

<span>The density of fissionable uranium is not high enough. Basically more neutrons are absorbed than are produced so any chain reaction dies. hope this helps</span>

You might be interested in

When you must add three vectors together, what is not true this process? You must only give a magnitude of the resultant vector

vovangra [49]

The addition of any numbers of vector provide the magnitude as well as the direction of the resultant vector, hence the mentioned first option is not true.

The addition of vector required to connect the head of the one vector with the tail of the other vector and any vector can be moved in the plane parallet to the previous location, so, the mentioned second and third options are true.

4 0

3 years ago

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit

Ainat [17]

Answer:

The number of fringe is z = 3 fringes

The ratio is $I = 0.2545I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 600 nm$

The distance between the slit is $d = 0.117mm = 0.117 *10^{-3} m$

The width of the slit is $a = 35.7 \mu m = 35.7 *10^{-6}m$

let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

$z = \frac{d}{a}$

Substituting values

$z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}$

z = 3 fringes

From the question we are told that the order of the bright fringe is n = 3

Generally the intensity of a pattern is mathematically represented as

$I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]$

Where $I_o$ is the intensity of the central fringe

And Generally $sin \theta = \frac{n \lambda }{d}$

$I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]$

$I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]$

$I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]$

$I = I_o (1)(0.2545)$

$I = 0.2545I_o$

6 0

3 years ago

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

ValentinkaMS [17]

Answer:

$2.02672\times 10^{-11}\ V$

Explanation:

f = Frequency = 75 Hz

B = Magnetic field = $1\times 10^{-3}\ T$

d = Diameter of cell = $7.4\ \mu m$

t = Time taken

$\omega$ = Angular frequency

The expression of emf is

$E=NAB\omega sin\omega t$

Emf is maximum when $sin\omega t=1$ and N = number of turns = 1

$E_m=AB\omega\\\Rightarrow E_m=\pi \left(\frac{d}{2}\right)^2 \times B \times 2\pi\times f\\\Rightarrow E_m=\pi \left(\frac{7.4\times 10^{-6}}{2}\right)^2\times 1\times 10^{-3}\times 2\pi\times 75\\\Rightarrow E_m=2.02672\times 10^{-11}\ V$