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3 years ago

15

What is the ecliptic, and why is it tilted with respect to the celestial equator?

1 answer:

Dahasolnce [82]3 years ago

4 0

The ecliptic is the line in the celestial sphere traced out by the sun as the rotates (the path in the sky that the sun takes over head during one 24hr day). The celestial equator is just the extension of the regular equator on the surface of the earth.

This means that there will be a "tilt" between the ecliptic and the celestial equator that depends on your latitude of the observers location on the earths surface.

source: astrophysicist

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A block with mass M = 3 kg is moving on a flat surface with constant speed v1 =

Alchen [17]

Answer:

this makes no since so i cant help you here sorry

5 0

2 years ago

A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?

rosijanka [135]

Answer:

a

$l_o =52 \ m$

b

$l = 37.13 \ LY$

Explanation:

From the question we are told that

The speed of the spaceship is $v = 0.800c$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

The length is $l = 31.2 \ m$

The distance of the star for earth is $d = 145 \ light \ years$

The speed is $v_s = 2.90 *10^{8}$

Generally the from the length contraction equation we have that

$l = l_o \sqrt{1 -[\frac{v}{c } ]}$

Now the when at rest the length is $l_o$

So

$l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }$

$l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }$

$l_o=52 \ m$

Considering b

Applying above equation

$l =l_o \sqrt{1 - [\frac{v}{c } ]}$

Here $l_o =145 \ LY(light \ years )$

So

$l=145 * \sqrt{1 - \frac{v_s^2}{c^2 } }$

$l =145 * \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }$

$l = 37.13 \ LY$

4 0

3 years ago

Могут ли быть сообщающиеся сосуды неодинаковы пр своей форме

Rudik [331]

Answer:

no

Explanation:

they cannot because they contain the same amount of liquid

8 0

3 years ago

Read 2 more answers

What is the formula for work?

babunello [35]

W = F • d • cos(theta)

6 0

3 years ago

A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc

max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

$f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }$

$f = \frac{1}{2\pi } ( 13.522)$

$f = 2.1535 Hz$

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

#SPJ4

6 0

2 years ago