Body mass index (bm) is a measure of body weight relative

Physics

1 answer:

kow [346]2 years ago

8 0

BMI is a measure of body fat determined by ones Height, Weight, and Gender.

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What characteristics distinguish the inner planets in the solar system from the outer planets?

Julli [10]

The four inner plants have shorter orbits slower spin ,no rings ,and they are made up of rock and metal
The outer plants are made up of hydrogen and helium, so they are called gas giants

Explanation:

hope it is the right answer

5 0

2 years ago

A laser used to dazzle the audience at a rock concert emits blue light with a wavelength of 463 nm . calculate the frequency of

ZanzabumX [31]

The wavelength of light is given as 463 nm or can also be written as 463 x 10^-9 m. [wavelength = ʎ]

We know that the speed of light is 299 792 458 m / s or approximately 3 x 10^8 m / s. [speed of light = c]

Given the two values, we can calculate for the frequence (f) using the formula:

f = c / ʎ

Substituting the given values:

f = (3 x 10^8 m / s) / 463 x 10^-9 m

f = 6.48 x 10^14 / s = 6.48 x 10^14 s^-1

5 0

3 years ago

A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge

Kisachek [45]

Answer:

$\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r$

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

$\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$