What kind of electromagnetic radiation could be used to see molecules

Physics

1 answer:

Vera_Pavlovna [14]3 years ago

4 0

Answer:

X ray

Explanation:

To see an object the light used must have the same or smaller wavelength than the object. The size of an atom is about $10^{10}$ m, smaller than the wavelength, therefore we'll need radiation with a shorter wavelength <em>since X-ray wavelengths are about the same size as atoms this characteristic makes it ideal to use.</em>

I hope you find this information useful nd interesting! Good luck!

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Artyom0805 [142]

Answer:

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3 years ago

How does inertia affect someone who isn't wearing a seatbelt

Paul [167]

Basically, when someone is resting in an accelerated vehicle without restraint from a seatbelt, the force of stopping the vehicle will be when inertia occurs, and that force of the vehicle coming to a stop will affect the passenger (without a seatbelt/restraint from another force or object) greatly by throwing them.
For example;
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I hope this helps!

7 0

4 years ago

Two simple pendulums are in two different places. The length of the second pendulum is 0.4 times the length of the first pendulu

faltersainse [42]

Answer:

$\sqrt{\frac{4}{9}}$

Explanation:

The frequency of a simple pendulum is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}$

where

g is the acceleration of gravity

L is the length of the pendulum

Calling $L_1$ the length of the first pendulum and $g_1$ the acceleration of gravity at the location of the first pendulum, the frequency of the first pendulum is

$f_1=\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}$

The length of the second pendulum is 0.4 times the length of the first pendulum, so

$L_2 = 0.4 L_1$

while the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum, so

$g_2 = 0.9 g_1$

So the frequency of the second pendulum is

$f_2=\frac{1}{2\pi}\sqrt{\frac{g_2}{L_2}}=\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}$

Therefore the ratio between the two frequencies is

$\frac{f_1}{f_2}=\frac{\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}}{\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}}=\sqrt{\frac{0.4}{0.9}}=\sqrt{\frac{4}{9}}$