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3 years ago

Why sodium &pottasium are kept under kerosine

1 answer:

5 0

As they are Alkali metals (group 1), they are extremely reactive in both air and water so must be stored in kerosine to stop them from reacting with the air or water.

You might be interested in

Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:

Deffense [45]

Answer:

a) $T=0.01s$

b) $T=0.001s$

c) $T=0.00001s$

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

$T=\frac{1}{f}$

Therefore

For

$T=100 Hz$

$T=\frac{1}{100}$

$T=0.01s$

For

$F=1kHz$

$T=\frac{1}{1000}$

$T=0.001s$

For

$F=100kHz$

$T=\frac{1}{100*100}$

$T=0.00001s$

6 0

3 years ago

Which object has the larger magnitude of its momentum?

Charra [1.4K]

The object that had the most 1000 ton weight has the most momentum

5 0

3 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

3 years ago

Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani

Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

$n = \frac{t}{t_{1/2}}$