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2 years ago

This group of elements is in group 2?

Physics

1 answer:

asambeis [7]2 years ago

5 0

Answer:

If you meant the Periodic Table it's the alkaline-earth metal

beryllium (Be)

magnesium (Mg)

calcium (Ca)

strontium (Sr)

barium (Ba)

radium (Ra)

Explanation:

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What are some charactersitics of pressure waves

True [87]

Soft target by impact and its contribution to indirect bone fractures.

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3 years ago

Asteroids are between 1000 km and less than 10 m in diameter. What is the diameter of most asteroids?

e-lub [12.9K]

This question needs research to be answered. From the given information alone it can't be answered without making wild assumptions.

Ideally, you need to take a look at a distribution (or a histogram) of asteroid diameters, identify the "mode" of such a distribution, and find the corresponding diameter. That value will be the answer.

I am attaching one such histogram on asteroid diameters from the IRAS asteroid catalog I could find online. (In order to get a single histogram, you need to add the individual curves in the figure first). Eyeballing this sample, I'd say the mode is somewhere around 10km, so the answer would be: the diameter of most asteroid from the IRAS asteroid catalog is about 10km.

7 0

3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

3 years ago

⦁ Match the following terms:

ki77a [65]

Answer:

Mass number - ⦁ The number of protons and neutrons in the nucleus of an atom.

Isotopes - ⦁ Atoms with the same number of protons, but different number of neutrons.

Nitrogen - ⦁ The name of the element with atomic number 7.

Atomic number - ⦁ The number of protons in the nucleus of an atom.

4 0

3 years ago

Please Help!!

kotegsom [21]

Answer:

Potential energy of spring = 24 Joules.

Explanation:

Given the following data;

Spring constant = 85N/m

Extension, e = 0.75m

Mass = 25kg

To find the potential energy of a spring

Potential energy of a spring is given by the formula;

P.E = ½ke²

Substituting into the equation, we have

P.E = ½*85*0.75²

P.E = 42.5 * 0.5625

P.E = 23.91 ≈ 24 Joules

P.E = 24 Joules

8 0

3 years ago