Which classification scheme is based on the brightness of a star as it appears to you in the night sky?

1 answer:

3 0

The apparent magnitude scale is a classification scheme which is based on the brightness of stars. The range of brightness values is from 1 to 6.
The stars which are the most brightest are ranked as number 1 and also called first magnitude stars, stars which are little dimmer than number 1 are ranked as number 2 and also called second magnitude stars. Similarly the most faintest stars are ranked number 6 and also called as the sixth magnitude stars.

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A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106hz and 1hz=1

lapo4ka [179]

In order to determine the wavelength, we use the wave equation:

speed = frequency * wavelength

speed of light c = 3 x 10⁸ m/s

Frequency f = 104.1 MHz = 1.041 x 10⁸ Hz

c = fλ
λ = c / f

λ = 3 x 10⁸ / 1.041 x 10⁸
λ = 2.88 meters

The wavelength of the waves is 2.88 meters.

5 0

3 years ago

LiRa [457]

The horizontal speed of the object 1.0 seconds later is 1) 5.0 m/s.

Explanation:

The motion of an object thrown horizontally off a cliff is a projectile motion, which follows a parabolic path that consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

This means that the horizontal speed of an object in projectile motion does not change, and remains constant during the whole motion.

Since in this case the object has been launched with a horizontal speed of

v = 5.0 m/s

this means that this speed will remain constant during the motion, so its horizontal speed 1.0 s later is also 5.0 m/s.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

A particle with positive charge q = 9.61 10-19 C moves with a velocity v = (5î + 4ĵ − k) m/s through a region where both a unifo

user100 [1]

Answer:

$\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N$

Explanation:

The total force on the particle is given by

$\vec{F}=q\vec{v}\ X\ \vec{B}+q\vec{E}$

Then, by replacing we have:

$q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N$